Calculus

Limits and Continuity

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Exercise 1.

Find the limit: \(\lim_{x\to 0} \frac{\sin(2x)}{tan(x)}\)

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Exercise 2.

Determine the interval of continuity of the function: \(f(x) = \sqrt{\frac{x^2 - 4}{x-1}}\)

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Exercise 3.

Find the limit: \(\lim_{x\to 1} \frac{x^3 - 1}{x^2 - 1}\):

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Exercise 4.

Prove that the function \(f(x) = \sqrt{ x^3 - 3x - 2}\) is continuous at \(x = 3\).

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Exercise 5.

Find the limit: \(\lim_{x\to\infty} \frac{3x^3 - 2x^2 + x - 1}{x^3 + 4x^2 - 3x}\)

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Exercise 6.

Find the limit: \(\lim_{x\to 0} \frac{\ln(1+x)}{x}\)

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Exercise 7.

Find the limit: \(\lim_{x\to 0} \frac{1 - \cos x}{x^2}\)

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Exercise 8.

Determine the point of discontinuity for the function \(f(x) = \frac{x^2 - 4}{x - 2}\).

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Exercise 9.

Find the limit: \(\lim_{x\to\infty} (1 + \frac{1}{x})^x\)

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Exercise 10.

Determine the value of \(a\) such that the function \(f(x) = \begin{cases} ax^2 - 2x + 1, & x < 2 \\ a + x, & x \geq 2 \end{cases}\) is continuous everywhere.

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Exercise 11.

Determine the value of \(b\) such that the function \(g(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & x \neq 3 \\ b, & x = 3 \end{cases}\) is continuous everywhere.

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Exercise 12.

Determine if the function \(h(x) = \begin{cases} \sin x, & x < \frac{\pi}{2} \\ \cos x, & x \geq \frac{\pi}{2} \end{cases}\) is continuous everywhere .

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Limits and Continuity

Introduction

In calculus, understanding the concept of limits is essential, as it provides the foundation for derivatives and integrals. Limits help us analyze the behavior of a function as its input approaches a specific value. Continuity, on the other hand, deals with the smoothness of a function and its unbroken nature over its domain.

Definition of a Limit

The limit of a function \(f(x)\) as \(x\) approaches a certain value \(c\) is denoted as \(\lim_{x \to c} f(x) = L\), if for every number \(\varepsilon > 0\) there exists a number \(\delta > 0\) such that if \(0 < |x - c| < \delta\), then \(|f(x) - L| < \varepsilon\).

Properties of Limits

  1. Sum/Difference of Limits: \(\lim_{x \to c} (f(x) \pm g(x)) = \lim_{x \to c} f(x) \pm \lim_{x \to c} g(x)\)
  2. Product of Limits: \(\lim_{x \to c} (f(x) \cdot g(x)) = \lim_{x \to c} f(x) \cdot \lim_{x \to c} g(x)\)
  3. Quotient of Limits: \(\lim_{x \to c} \frac{f(x)}{g(x)} = \frac{\lim_{x \to c} f(x)}{\lim_{x \to c} g(x)}\) provided that \(\lim_{x \to c} g(x) \neq 0\)
  4. Constant times a Limit: \(\lim_{x \to c} (k \cdot f(x)) = k \cdot \lim_{x \to c} f(x)\) for any constant \(k\)

Definition of Continuity

A function \(f(x)\) is continuous at a point \(c\) if the following three conditions are met:

  1. \(f(c)\) is defined.
  2. \(\lim_{x \to c} f(x)\) exists.
  3. \(\lim_{x \to c} f(x) = f(c)\)

If a function is continuous at every point in its domain, it is considered to be a continuous function.

Exercises

Find the following limits:

  1. \(\lim_{x \to 2} (3x - 1)\)
  2. \(\lim_{x \to 0} \frac{x^2 - 1}{x - 1}\)
  3. \(\lim_{x \to -3} \frac{x^3 + 27}{x + 3}\)
  4. \(\lim_{x \to 0} \frac{\sin{x}}{x}\)
  5. \(\lim_{x \to \infty} \frac{3x^2 - 2x + 1}{x^2 + 5x - 8}\)
  6. \(\lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1}\)
  7. \(\lim_{x \to 0} \frac{1 - \cos{x}}{x^2}\)
  8. \(\lim_{x \to 0^+} x \ln{x}\)
  9. \(\lim_{x \to 0} (1 + x)^{\frac{1}{x}}\)
  10. \(\lim_{x \to 0} \frac{\tan{x}}{x}\)

Differentiation

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  1. Exercise 1

    Differentiate the function \(y = \sqrt[3]{x^4}\).
    Click to show the solution Rewrite the function as \(y = x^{\frac{4}{3}}\), then use the power rule: \(y' = \frac{4}{3}x^{\frac{1}{3}}\).
  2. Exercise 2

    Differentiate the function \(y = e^{2x} \sin(3x)\).
    Click to show the solution Use the product rule: \(y' = (e^{2x})'sin(3x) + e^{2x}sin(3x)'\). So, \(y' = 2e^{2x}\sin(3x) + 3e^{2x}\cos(3x)\).
  3. Exercise 3

    Differentiate the function \(y = \ln(x^2 + 1)\).
    Click to show the solution Use the chain rule: \(y' = \frac{1}{x^2 + 1}(2x)\). So, \(y' = \frac{2x}{x^2 + 1}\).
  4. Exercise 4

    Differentiate the function \(y = x^3 \sin x\).
    Click to show the solution Use the product rule: \(y' = (x^3)'(\sin x) + (x^3)(\sin x)'\), which gives: \(y' = 3x^2 \sin x + x^3 \cos x\).
  5. Exercise 5

    Differentiate the function \(y = \ln(x^3 + 1)\).
    Click to show the solution Use the chain rule: \(y' = (\ln u)'u'(x)\), where \(u = x^3 + 1\). This gives: \(y' = \frac{1}{x^3 + 1} (3x^2)\), and \(y' = \frac{3x^2}{x^3 + 1}\).
  6. Exercise 6

    Differentiate the function \(y = e^{2x}\cos x\).
    Click to show the solution Use the product rule: \(y' = (e^{2x})'(\cos x) + (e^{2x})(\cos x)'\), which gives: \(y' = 2e^{2x}\cos x - e^{2x}\sin x\).
  7. Exercise 7

    Differentiate the function \(y = \frac{1}{x^2 + 1}\).
    Click to show the solution Use the quotient rule: \(y' = \frac{-(x^2 + 1)'}{(x^2 + 1)^2}\), which gives: \(y' = \frac{-2x}{(x^2 + 1)^2}\).
  8. Exercise 8

    Differentiate the function \(y = \frac{x^2 - 4}{x^2 + 4}\).
    Click to show the solution Use the quotient rule: \(y' = \frac{(x^2 - 4)'(x^2 + 4) - (x^2 - 4)(x^2 + 4)'}{(x^2 + 4)^2}\), which gives: \(y' = \frac{16x}{(x^2 + 4)^2}\).
  9. Exercise 9

    Differentiate the function \(y = \tan^{-1}(x)\).
    Click to show the solution Use the chain rule: \(y' = (\tan^{-1} u)'u'(x)\), where \(u = x\). This gives: \(y' = \frac{1}{1 + x^2}\).
  10. Exercise 10

    Differentiate the function \(y = x^2 e^{-x}\).
    Click to show the solution Use the product rule: \(y' = (x^2)'(e^{-x}) + (x^2)(e^{-x})'\), which gives: \(y' = 2x e^{-x} - x^2 e^{-x}\).
  11. Exercise 11

    Differentiate the function \(y = \sqrt[3]{x^4}\).
    Click to show the solution Rewrite the function as \(y = x^{\frac{4}{3}}\), then use the power rule: \(y' = \frac{4}{3}x^{\frac{1}{3}}\).
  12. Exercise 12

    Differentiate the function \(y = \sin x\ln(x)\).
    Click to show the solution Use the product rule: \(y' = (\sin x)'(\ln x) + (\sin x)(\ln x)'\), which gives: \(y' = \cos x\ln(x) + \frac{\sin x}{x}\).
  13. Exercise 13

    Differentiate the function \(y = \frac{\ln x}{x}\).
    Click to show the solution Use the quotient rule: \(y' = \frac{(\ln x)'(x) - (\ln x)(x)'}{x^2}\), which gives: \(y' = \frac{1 - \ln x}{x^2}\).

Differentiation

Introduction

Differentiation is the process of finding the derivative of a function. Derivatives are a crucial concept in calculus and have applications in various fields, such as physics, engineering, and economics. The derivative of a function represents the rate of change of the function with respect to its independent variable.

Definition of the Derivative

The derivative of a function \(f(x)\) at a point \(x = a\) is defined as:

\(f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h}\)

Basic Rules of Differentiation

There are several basic rules of differentiation that can be applied to find the derivative of a function. Some of the most important ones are listed below:

1. Constant Rule: If c is a constant, then \(\frac{d}{dx}(c) = 0\) 2. Power Rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\) for any real number n 3. Sum/Difference Rule: \(\frac{d}{dx}(f(x) ± g(x)) = f'(x) ± g'(x)\) 4. Product Rule: \(\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)\) 5. Quotient Rule: \(\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\) if \(g(x) ≠ 0\) 6. Chain Rule: If \(y = f(u)\) and \(u = g(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\)

Examples

Let's work through some examples of finding derivatives using the rules of differentiation:

1. Find the derivative of \(f(x) = 5x^3 - 2x^2 + 7x - 3\) Solution: Using the sum/difference and power rules, we get: \( f'(x) = 3(5x^2) - 2(2x) + 7 = 15x^2 - 4x + 7\) 2. Find the derivative of \(f(x) = \frac{x^2 - 4}{x + 2}\) Solution: Using the quotient rule, we get: \( f'(x) = \frac{(2x)(x + 2) - (x^2 - 4)(1)}{(x + 2)^2} = \frac{2x^2 + 4x - x^2 + 4}{(x + 2)^2} = \frac{x^2 + 4x + 4}{(x + 2)^2}\) 3. Find the derivative of \(f(x) = \sqrt{4x^2 + 3x}\) Solution: First, rewrite the function using exponents: \(f(x) = (4x^2 + 3x)^{1/2}\) Using the chain rule, we get: \(f'(x) = \frac{1}{2}(4x^2 + 3x)^{-1/2} \cdot (8x + 3) = \frac{8x + 3}{2\sqrt{4x^2 + 3x}}\)

Higher Order Derivatives

Higher order derivatives are derivatives of derivatives. The second derivative, denoted by \(f''(x)\), is the derivative of the first derivative \(f'(x)\). Similarly, the third derivative, denoted by \(f'''(x)\), is the derivative of the second derivative \(f''(x)\), and so on.

Example: Find the second derivative of \(f(x) = x^3 - 6x^2 + 9x + 1\) Solution: First, find the first derivative: \(f'(x) = 3x^2 - 12x + 9\) Now, find the second derivative: \(f''(x) = 6x - 12\)

Applications of Derivatives

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  1. Exercise 1

    Find the critical points and determine their nature (local maximum, local minimum, or neither) for the function \(f(x) = x^3 - 6x^2 + 9x\).
    Click to show the solution First, find the derivative: \(f'(x) = 3x^2 - 12x + 9\). Set \(f'(x) = 0\) to find the critical points: \(3x^2 - 12x + 9 = 0\). Factoring, we get \((3x - 9)(x - 1) = 0\), so the critical points are \(x = 1\) and \(x = 3\). Next, use the second derivative test: \(f''(x) = 6x - 12\). At \(x = 1\), \(f''(1) = -6\), so there is a local maximum at \(x = 1\). At \(x = 3\), \(f''(3) = 6\), so there is a local minimum at \(x = 3\).
  2. Exercise 2

    Find the intervals where the function \(g(x) = x^4 - 4x^3\) is increasing and decreasing.
    Click to show the solution First, find the derivative: \(g'(x) = 4x^3 - 12x^2\). Set \(g'(x) = 0\) to find the critical points: \(4x^3 - 12x^2 = 0\). Factoring, we get \(4x^2(x - 3) = 0\), so the critical points are \(x = 0\) and \(x = 3\). To determine the intervals of increase and decrease, analyze the sign of \(g'(x)\) in each interval: \((-\infty, 0), (0, 3), (3, \infty)\). Since \(g'(x) = 4x^2(x - 3)\), we have \(g'(x) < 0\) on \((0, 3)\) and \(g'(x) > 0\) on \((-\infty, 0)\) and \((3, \infty)\). So, \(g(x)\) is decreasing on \((0, 3)\) and increasing on \((-\infty, 0)\) and \((3, \infty)\).
  3. Exercise 3

    Find the equation of the tangent line to the curve \(y = x^2 + 2x - 1\) at the point \((1, 2)\).
    Click to show the solution First, find the derivative: \(y'(x) = 2x + 2\). Then, find the slope of the tangent line at the point \((1, 2)\): \(m = y'(1) = 2(1) + 2 = 4\). Finally, use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), so \(y - 2 = 4(x - 1)\). Simplifying, we get the equation of the tangent line: \(y = 4x - 2\).
  4. Exercise 4

    Given the function \(y = \sqrt[3]{x^2}\), find \(y'\) and the equation of the tangent line at the point \((8, 4)\).
    Click to show the solution Rewrite the function as \(y = x^{\frac{2}{3}}\). Then, find the derivative using the power rule: \(y'(x) = \frac{2}{3}x^{-\frac{1}{3}}\). Next, find the slope of the tangent line at the point \((8, 4)\): \(m = y'(8) = \frac{2}{3}(8)^{-\frac{1}{3}} = \frac{2}{3}(1/2) = \frac{1}{3}\). Finally, use the point-slope form of a line: \(y - y_1 = m(x - x_1)\), so \(y - 4 = \frac{1}{3}(x - 8)\). Simplifying, we get the equation of the tangent line: \(y = \frac{1}{3}x + \frac{4}{3}\).
  5. Exercise 5

    Find the global maximum and minimum values of the function \(f(x) = x^3 - 9x^2 + 24x\) on the interval \([0, 6]\).
    Click to show the solution First, find the critical points of the function by setting the derivative equal to zero: \(f'(x) = 3x^2 - 18x + 24 = 3(x - 2)(x - 4)\). The critical points are \(x = 2\) and \(x = 4\). \(f"(x)=6x-18\) and \(f"(2)=-6\) and \(f"(4)=6\) Evaluate the function at the critical points and the endpoints of the interval: \(f(0) = 0\), \(f(2) = 20\), \(f(4) = 16\), and \(f(6) = 36\). The global maximum value is 36 at \(x = 6\), and the global minimum value is 0 at \(x = 0\).
  6. Exercise 6

    Given a right triangle with legs of length \(x\) and \(y\), and hypotenuse of length \(z\), find the dimensions of the triangle that maximize the area, subject to the constraint \(z = 10\).
    Click to show the solution First, express the area A of the triangle in terms of x and y: \(A = \frac{1}{2}xy\). Use the Pythagorean theorem to express y in terms of x and the constraint: \(x^2 + y^2 = z^2\) or \(y^2 = 100 - x^2\). Now, express A in terms of x only: \(A(x) = \frac{1}{2}x\sqrt{100 - x^2}\). To maximize A, find the critical points by setting the derivative equal to zero: \(A'(x) = \frac{1}{2}(\sqrt{100 - x^2} - \frac{x^2}{\sqrt{100 - x^2}})\). Set \(A'(x) = 0\): \(\sqrt{100 - x^2} = \frac{x^2}{\sqrt{100 - x^2}}\). Squaring both sides and simplifying, we get \(100 - x^2 = x^2\), or \(x^2 = 50\). Thus, \(x = \sqrt{50}\) and \(y = \sqrt{50}\). The dimensions of the triangle that maximize the area are legs of length \(\sqrt{50}\) and a hypotenuse of length 10.
  7. Exercise 7

    A particle moves along the x-axis with its position \(x(t)\) given by \(x(t) = t^3 - 6t^2 + 9t + 3\). Find the time(s) when the particle is at rest and the time(s) when the particle changes direction.
    Click to show the solution To find when the particle is at rest, differentiate the position function to find the velocity function: \(v(t) = 3t^2 - 12t + 9\). Set the velocity equal to zero: \(3t^2 - 12t + 9 = 0\). Factoring, we get \((3t - 3)(t - 3) = 0\), so the particle is at rest at \(t = 1\) and \(t = 3\). To find when the particle changes direction, differentiate the velocity function to find the acceleration function: \(a(t) = 6t - 12\). Set the acceleration equal to zero: \(6t - 12 = 0\), so \(t = 2\). Now, check the sign of the velocity on either side of \(t = 2\). For \(t < 2\), \(v(t)*a(t) > 0\); for \(t > 2\), \(v(t)*a(t) < 0\). Therefore, the particle changes direction at \(t = 2\).
  8. Exercise 8

    Find the equation of the tangent line to the curve \(y = e^{x^2}\) at the point \((1, e)\).
    Click to show the solution First, find the derivative of the function: \(y'(x) = \frac{d}{dx}(e^{x^2}) = 2xe^{x^2}\). Now, find the slope of the tangent line at the point \((1, e)\): \(m = y'(1) = 2(1)e^{(1)^2} = 2e\). The equation of the tangent line is given by \(y - y_1 = m(x - x_1)\), where \((x_1, y_1) = (1, e)\) and \(m = 2e\). Therefore, the equation of the tangent line is \(y - e = 2e(x - 1)\), or \(y = 2ex - e\).
  9. Exercise 9

    Determine the intervals on which the function \(f(x) = x^4 - 4x^3 + 4x^2\) is increasing and decreasing.
    Click to show the solution First, find the derivative of the function: \(f'(x) = 4x^3 - 12x^2 + 8x\). Factor out a common factor of \(4x\): \(f'(x) = 4x(x^2 - 3x + 2)\). Further factoring, we get \(f'(x) = 4x(x - 1)(x - 2)\). Set the derivative equal to zero: \(4x(x - 1)(x - 2) = 0\), so \(x = 0, 1, 2\). Now, use a number line to test the intervals between the critical points. Choose a test point in each interval and evaluate the derivative at that point:
    1. Interval \((-\infty, 0)\): Choose \(x = -1\), \(f'(-1) = 4(-1)(-2)(-3) < 0\), so the function is deccreasing.
    2. Interval \((0, 1)\): Choose \(x = 0.5\), \(f'(0.5) = 4(0.5)(-0.5)(-1.5) > 0\), so the function is increasing.
    3. Interval \((1, 2)\): Choose \(x = 1.5\), \(f'(1.5) = 4(1.5)(0.5)(-0.5) < 0\), so the function is decreasing.
    4. Interval \((2, \infty)\): Choose \(x = 3\), \(f'(3) = 4(3)(2)(1) > 0\), so the function is increasing.
    Therefore, the function is increasing on \((0, 1) \cup (2, \infty)\) and decreasing on \( (-\infty, 0) \cup (1, 2)\).
  10. Exercise 10

    A particle moves along the x-axis with position function \(s(t) = t^3 - 6t^2 + 9t + 1\). Determine the intervals of time during which the particle is moving to the right and the intervals during which it is moving to the left.
    Click to show the solution To determine the direction of the particle's movement, we need to find the velocity function, which is the derivative of the position function: \(v(t) = s'(t) = 3t^2 - 12t + 9\). Factor the quadratic: \(v(t) = 3(t^2 - 4t + 3)\). Further factoring, we get \(v(t) = 3(t - 1)(t - 3)\). Set the derivative equal to zero: \(3(t - 1)(t - 3) = 0\), so \(t = 1, 3\). Now, use a number line to test the intervals between the critical points. Choose a test point in each interval and evaluate the velocity at that point:
    1. Interval \((-\infty, 1)\): Choose \(t = 0\), \(v(0) = 3(-1)(-3) > 0\), so the particle is moving to the right.
    2. Interval \((1, 3)\): Choose \(t = 2\), \(v(2) = 3(1)(-1) < 0\), so the particle is moving to the left.
    3. Interval \((3, \infty)\): Choose \(t = 4\), \(v(4) = 3(3)(1) > 0\), so the particle is moving to the right.
    Therefore, the particle is moving to the right on \((-\infty, 1) \cup (3, \infty)\) and moving to the left on \((1, 3)\).

Lesson: Applications of Derivatives

In this lesson, we will explore some common applications of derivatives in calculus. These applications include determining rates of change, finding critical points, analyzing increasing and decreasing functions, solving optimization problems, and analyzing the concavity and inflection points of a function.

1. Rates of Change

Derivatives can be used to determine the rate of change of a function with respect to its independent variable. In real-world problems, this concept is often used to calculate velocity, acceleration, and other rates.

For example, if \(y = f(x)\) represents the position of an object as a function of time \(x\), then the first derivative \(f'(x)\) gives the velocity of the object and the second derivative \(f''(x)\) gives its acceleration.

2. Critical Points and Extrema

A critical point of a function occurs when its derivative is either zero or undefined. These points can correspond to local maximums, local minimums, or points of inflection. To classify a critical point, we can use the first or second derivative test.

First Derivative Test: If \(f'(x)\) changes sign from positive to negative at a critical point, the point is a local maximum. If it changes from negative to positive, it's a local minimum. If the sign does not change, the test is inconclusive.

Second Derivative Test: If \(f''(x) > 0\) at a critical point, the point is a local minimum. If \(f''(x) < 0\), it's a local maximum. If \(f''(x) = 0\), the test is inconclusive.

3. Increasing and Decreasing Functions

A function is increasing on an interval if its derivative is positive on that interval, and it is decreasing if the derivative is negative. To determine where a function is increasing or decreasing, find its critical points and analyze the sign of the derivative between those points.

4. Optimization Problems

Optimization problems involve finding the maximum or minimum value of a function subject to certain constraints. To solve such problems, first express the quantity to be optimized as a function of a single variable. Then, find the critical points of the function and use the first or second derivative test to determine if they correspond to maximum or minimum values.

5. Concavity and Inflection Points

The concavity of a function is determined by its second derivative. If \(f''(x) > 0\), the function is concave up (shaped like a U). If \(f''(x) < 0\), it's concave down (shaped like an inverted U). A point of inflection occurs when the concavity of a function changes. To find inflection points, set the second derivative equal to zero or determine where it is undefined, and analyze the sign change of the second derivative around those points.

6. Related Rates

Related rates problems involve finding the rate at which one quantity changes with respect to another, given their relationship as a function. To solve a related rates problem, follow these steps:

  1. Identify the given rates and the rate you need to find.
  2. Write an equation that relates the quantities involved.
  3. Differentiate both sides of the equation with respect to time, using the chain rule as needed.
  4. Substitute the given rates and solve for the unknown rate.

For example, if \(x\) and \(y\) are related by \(x^2 + y^2 = 25\) and \(x\) is increasing at a rate of \(3\) units per second, we can find the rate at which \(y\) is changing when \(x = 4\) and \(y = 3\).

7. Mean Value Theorem

The Mean Value Theorem states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a number \(c\) in \((a, b)\) such that:

\(f'(c) = \frac{f(b) - f(a)}{b - a}\)

In other words, at some point within the interval, the instantaneous rate of change (the derivative) equals the average rate of change of the function.

8. Rolle's Theorem

Rolle's Theorem is a special case of the Mean Value Theorem. It states that if a function \(f(x)\) is continuous on the closed interval \([a, b]\), differentiable on the open interval \((a, b)\), and \(f(a) = f(b)\), then there exists at least one number \(c\) in \((a, b)\) such that:

\(f'(c) = 0\)

This theorem can be used to show the existence of critical points and analyze the behavior of a function within a given interval.

Integration

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  1. Exercise 1

    Evaluate the integral: \(\int (3x^2 - 4x + 2) \, dx\)
    Click to show the solution To evaluate the integral, integrate each term of the polynomial: \(\int (3x^2 - 4x + 2) \, dx = 3 \int x^2 \, dx - 4 \int x \, dx + 2 \int 1 \, dx\) Now, apply the power rule for integration to each term: \(= 3 \frac{x^3}{3} - 4 \frac{x^2}{2} + 2x + C = x^3 - 2x^2 + 2x + C\)
  2. Exercise 2

    Evaluate the integral: \(\int_0^2 (2x^3 - x^2 + 3x) \, dx\)
    Click to show the solution To evaluate the definite integral, first find the antiderivative: \(\int (2x^3 - x^2 + 3x) \, dx = \frac{1}{2}x^4 - \frac{1}{3}x^3 + \frac{3}{2}x^2 + C\) Now, use the Fundamental Theorem of Calculus to evaluate the integral: \(\int_0^2 (2x^3 - x^2 + 3x) \, dx = \left[\frac{1}{2}x^4 - \frac{1}{3}x^3 + \frac{3}{2}x^2\right]_0^2\) \(= \left(\frac{1}{2}(2^4) - \frac{1}{3}(2^3) + \frac{3}{2}(2^2)\right) - \left(\frac{1}{2}(0^4) - \frac{1}{3}(0^3) + \frac{3}{2}(0^2)\right) = 34/3\)
  3. Exercise 3

    Evaluate the integral: \(\int_1^e \frac{1}{x} \, dx\)
    Click to show the solution The antiderivative of \(\frac{1}{x}\) is \(\ln|x|\). So: \(\int_1^e \frac{1}{x} \, dx = \left[\ln|x|\right]_1^e = \ln{e} - \ln{1} = 1 - 0 = 1\)
  4. Exercise 4

    Evaluate the integral: \(\int \sin^2{x} \, dx\)
    Click to show the solution First, use the double-angle identity for cosine to rewrite the integrand: \(\sin^2{x} = \frac{1 - \cos{2x}}{2}\) Now, integrate: \(\int \sin^2(x)dx=x/2-(1/4)sin(2x)+C\)
  5. Exercise 5

    Evaluate the integral: \(\int e^{2x} \cos{3x} \, dx\)
    Click to show the solution To solve this integral, we use integration by parts twice. Let \(u = e^{2x}\) and \(dv = \cos{3x} \, dx\). Then, we have: \(du = 2e^{2x} \, dx\), and \(v = \frac{1}{3}\sin{3x}\) Apply integration by parts: \(\int e^{2x} \cos{3x} \, dx = e^{2x} \frac{1}{3}\sin{3x} - \frac{2}{3} \int e^{2x} \sin{3x} \, dx\) Now, let \(u = e^{2x}\) and \(dv = \sin{3x} \, dx\) for the remaining integral. Then, we have: \(du = 2e^{2x} \, dx\), and \(v = -\frac{1}{3}\cos{3x}\) Apply integration by parts again: \(-\frac{2}{3} \int e^{2x} \sin{3x} \, dx = -\frac{2}{3} \left( -e^{2x} \frac{1}{3}\cos{3x} + \frac{2}{3} \int e^{2x} \cos{3x} \, dx \right)\) Now, notice that we have the original integral in our expression. Solve for it: \(\int e^{2x} \cos{3x} \, dx = e^{2x} \frac{1}{3}\sin{3x} + \frac{2}{9} e^{2x} \cos{3x} - \frac{4}{9} \int e^{2x} \cos{3x} \, dx\) \(\left(1 + \frac{4}{9}\right) \int e^{2x} \cos{3x} \, dx = e^{2x} \frac{1}{3}\sin{3x} + \frac{2}{9} e^{2x} \cos{3x}\) \(\int e^{2x} \cos{3x} \, dx = \frac{9}{13} \left( e^{2x} \frac{1}{3}\sin{3x} + \frac{2}{9} e^{2x} \cos{3x} \right) + C\)
  6. Exercise 6

    Evaluate the integral: \(\int \frac{x^2}{\sqrt{9 - x^2}} \, dx\)
    Click to show the solution To solve this integral, we use a trigonometric substitution. Let \(x = 3\sin{u}\), so \(dx = 3\cos{u} \, du\). Then, we have: \(\int \frac{x^2}{\sqrt{9 - x^2}} \, dx = \int \frac{(3\sin{u})^2}{\sqrt{9 - (3\sin{u})^2}} \cdot 3\cos{u} \, du = \int \frac{9\sin^2{u}}{\sqrt{9 - 9\sin^2{u}}} \cdot 3\cos{u} \, du\) The denominator simplifies to \(\sqrt{9\cos^2{u}} = 3\cos{u}\). So, the integral becomes: \(\int \frac{9\sin^2{u}}{3\cos{u}} \cdot 3\cos{u} \, du = \int 9\sin^2{u} \, du\) Now, use the double angle identity for sine: \(\sin^2{u} = \frac{1}{2} - \frac{1}{2}\cos{2u}\). Then, we have: \(\int 9\sin^2{u} \, du = 9 \int \left(\frac{1}{2} - \frac{1}{2}\cos{2u}\right) \, du\) Integrate: \(9 \int \left(\frac{1}{2} - \frac{1}{2}\cos{2u}\right) \, du = 9 \left( \frac{1}{2}u - \frac{1}{4}\sin{2u} \right) + C\) Finally, substitute back for \(x\): \(u = \arcsin{\frac{x}{3}}\), so the solution is: \(9 \left( \frac{1}{2}\arcsin{\frac{x}{3}} - \frac{1}{4}\sin{2\arcsin{\frac{x}{3}}} \right) + C\)
  7. Exercise 7

    Evaluate the integral: \(\int \ln{x} \, dx\)
    Click to show the solution To solve this integral, we use integration by parts. Let \(u = \ln{x}\) and \(dv = dx\). Then, we have: \(du = \frac{1}{x} \, dx\), and \(v = x\) Apply integration by parts: \(\int \ln{x} \, dx = x\ln{x} - \int x \cdot \frac{1}{x} \, dx = x\ln{x} - \int dx\) Integrate: \(x\ln{x} - \int dx = x\ln{x} - x + C\)
  8. Exercise 8

    Evaluate the integral: \(\int \frac{1}{x^2 + 4x + 13} \, dx\)
    Click to show the solution First, complete the square for the denominator: \(x^2 + 4x + 13 = (x^2 + 4x + 4) + 9 = (x + 2)^2 + 3^2\) Now, the integral becomes: \(\int \frac{1}{(x + 2)^2 + 3^2} \, dx\) This is an arctangent form. Use the substitution: \(x + 2 = 3\tan{\theta}\), and \(dx = 3\sec^2{\theta} \, d\theta\). Then, \(\int \frac{1}{(x + 2)^2 + 3^2} \, dx = \int \frac{1}{(3\tan{\theta})^2 + 3^2} \cdot 3\sec^2{\theta} \, d\theta = \int \frac{3\sec^2{\theta}}{9\tan^2{\theta} + 9} \, d\theta\) Simplify: \(\int \frac{3\sec^2{\theta}}{9(\tan^2{\theta} + 1)} \, d\theta = \int \frac{3\sec^2{\theta}}{9\sec^2{\theta}} \, d\theta = \int \frac{1}{3} \, d\theta\) Integrate: \(\frac{1}{3}\theta + C\) Finally, substitute back for \(x\): \(\theta = \arctan{\frac{x + 2}{3}}\), so the solution is: \(\frac{1}{3}\arctan{\frac{x + 2}{3}} + C\)
  9. Exercise 9

    Evaluate the integral: \(\int_0^{\pi/2} x\sin{x} \, dx\)
    Click to show the solution To solve this integral, we use integration by parts. Let \(u = x\) and \(dv = \sin{x} \, dx\). Then, we have: \(du = dx\), and \(v = -\cos{x}\) Apply integration by parts: \(\int_0^{\pi/2} x\sin{x} \, dx = \left[-x\cos{x}\right]_0^{\pi/2} - \int_0^{\pi/2} -\cos{x} \, dx\) Integrate: \(\left[-x\cos{x}\right]_0^{\pi/2} + \int_0^{\pi/2} \cos{x} \, dx = \left[-\frac{\pi}{2}\cos{\frac{\pi}{2}} - 0\right] + [\sin{x}]_0^{\pi/2}\) Evaluate: \(-\frac{\pi}{2}\cos{\frac{\pi}{2}} - 0 + \sin{\frac{\pi}{2}} - \sin0 = -\frac{\pi}{2}(0) - 0 + 1 - 0\) Simplify: \(1\)
  10. Exercise 10

    Evaluate the integral: \(\int \frac{1}{\sqrt{4 - x^2}} \, dx\)
    Click to show the solution To evaluate the integral, use the substitution \(x = 2\sin{u}\) and \(dx = 2\cos{u} \, du\): \(\int \frac{1}{\sqrt{4 - x^2}} \, dx = \int \frac{1}{\sqrt{4 - (2\sin{u})^2}} \cdot 2\cos{u} \, du = \int \frac{2\cos{u}}{\sqrt{4 - 4\sin^2{u}}} \, du\) Simplify: \(\int \frac{2\cos{u}}{2\cos{u}} \, du = \int 1 \, du\) Integrate: \(u + C\) Finally, substitute back for \(x\): \(u = \arcsin{\frac{x}{2}}\), so the solution is: \(\arcsin{\frac{x}{2}} + C\)

Integration

1. Indefinite Integrals

An indefinite integral, also known as an antiderivative, represents a family of functions that all have the same derivative. The process of finding the antiderivative is called integration. The notation for an indefinite integral is:

\(\int f(x) \, dx = F(x) + C\)

where \(f(x)\) is the function to be integrated, \(F(x)\) is the antiderivative, and \(C\) is the constant of integration.

2. Basic Integration Rules

Some basic integration rules include:

  • \(\int k \, dx = kx + C\), where \(k\) is a constant.
  • \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\), for \(n \neq -1\).
  • \(\int \frac{1}{x} \, dx = \ln |x| + C\), for \(x \neq 0\).
  • \(\int e^x \, dx = e^x + C\).

3. Integration Techniques

There are various techniques for integration, such as:

  • Integration by Substitution
  • Integration by Parts
  • Trigonometric Substitution
  • Partial Fractions

4. Definite Integrals

Definite integrals are used to calculate the area under a curve between two points. The notation for a definite integral is:

\(\int_a^b f(x) \, dx\)

where \(a\) and \(b\) are the limits of integration.

5. Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects the process of differentiation and integration, and it consists of two parts:

Part 1: If \(F(x)\) is an antiderivative of \(f(x)\) on the interval \([a, b]\), then:

\(\int_a^b f(x) \, dx = F(b) - F(a)\)

Part 2: If \(f(x)\) is continuous on the interval \([a, b]\) and \(F'(x) = f(x)\), then:

\(\frac{d}{dx} \int_a^x f(t)dt = f(x)\) for all \(x\) in \([a, b]\).

6. Applications of Integration

Integration has various applications in mathematics and physics, such as:

  • Finding the area under a curve
  • Calculating volumes of solids
  • Computing work done by a force
  • Solving problems in physics, like kinematics and dynamics

Applications of Integrals

Back to curriculum

  • Exercise 1

    Find the area between the curves \(y = x^2\) and \(y = x^3\) from \(x = 0\) to \(x = 1\).
    Click to show the solution To find the area between the curves, we need to find the integral of the difference of the functions from \(x = 0\) to \(x = 1\): \(\int_{0}^{1} (x^2 - x^3) \, dx\) Integrate: \(\left[\frac{1}{3}x^3 - \frac{1}{4}x^4\right]_0^1\) Evaluate: \(\left(\frac{1}{3} - \frac{1}{4}\right) - \left(0 - 0\right) = \frac{1}{12}\) The area between the curves is \(\frac{1}{12}\).
  • Exercise 2

    Calculate the volume of the solid formed by rotating the region bounded by \(y = \sqrt{x}\), \(y = 0\), and \(x = 4\) about the \(y\)-axis.
    Click to show the solution We will use the disk method to find the volume. We need to find the integral of the area of each disk along the \(y\)-axis from \(y = 0\) to \(y = 2\). The radius of each disk is \(x\), and we need to express it in terms of \(y\): \(x = y^2\). The area of each disk is \(\pi x^2 = \pi y^4\). Thus, the volume is: \(\int_{0}^{2} \pi y^4 \, dy\) Integrate: \(\left[\frac{\pi}{5}y^5\right]_0^2\) Evaluate: \(\frac{32\pi}{5} - 0 = \frac{32\pi}{5}\) The volume of the solid is \(\frac{32\pi}{5}\).
  • Exercise 3

    Find the length of the curve \(y = (x^2 + 1)\) from \(x = 0\) to \(x = 2\).
    Click to show the solution To find the length of the curve, we need to find the integral of the square root of the sum of the squares of the derivatives of the function from \(x = 0\) to \(x = 2\): \(\int_{0}^{2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\) Differentiate: \(\frac{dy}{dx} = 2x\) Square and add 1: \(1 + 4x^2 \) Take the square root: \((\sqrt{1 + 4x^2 })\) Integrate: \(\int_{0}^{2} \sqrt{1 + 4x^2 } \, dx\) Let \( 2x=sinh(\theta)\), then \(1+4x^2=1+sinh^2(\theta)=cosh^2(\theta)\) and \(dx=(1/2)cosh(\theta)d\theta\) Then \(\int_{0}^{2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx\) \(=\frac{1}{2}\int_{0}^{sinh^(-1)(4)}cosh^2(\theta)d\theta=\int_{0}^{sinh^(-1)(4)}\frac{1+cosh(2\theta)}{4}d\theta \) \(=(1/4)[\theta+(1/2)sinh(2\theta)]_{0}^{sinh^{-1}(4)}\)
  • Exercise 4

    Find the centroid of the region bounded by the curve \(y = x^2\) and the \(x\)-axis from \(x = 0\) to \(x = 1\).
    Click to show the solution To find the centroid, we need to find the coordinates \((\bar{x}, \bar{y})\), where: \(\bar{x} = \frac{1}{A} \int_{0}^{1} x \cdot y \, dx\) and \(\bar{y} = \frac{1}{2A} \int_{0}^{1} y^2 \, dx\) Here, \(y = x^2\), and the area of the region is: \(A = \int_{0}^{1} x^2 \, dx = \left[\frac{1}{3}x^3\right]_0^1 = \frac{1}{3}\) Calculate \(\bar{x}\): \(\bar{x} = \frac{1}{\frac{1}{3}} \int_{0}^{1} x \cdot x^2 \, dx = 3 \int_{0}^{1} x^3 \, dx = 3 \left[\frac{1}{4}x^4\right]_0^1 = \frac{3}{4}\) Calculate \(\bar{y}\): \(\bar{y} = \frac{1}{2\cdot\frac{1}{3}} \int_{0}^{1} (x^2)^2 \, dx = \frac{3}{2} \int_{0}^{1} x^4 \, dx = \frac{3}{2} \left[\frac{1}{5}x^5\right]_0^1 = \frac{3}{10}\) The centroid of the region is \((\bar{x}, \bar{y}) = \left(\frac{3}{4}, \frac{3}{10}\right)\).
  • Exercise 5

    Find the work done by a force \(F(x) = 5x^2\) that moves an object along the \(x\)-axis from \(x = 0\) to \(x = 3\).
    Click to show the solution The work done by the force is given by the integral: \(W = \int_{0}^{3} F(x) \, dx = \int_{0}^{3} 5x^2 \, dx\) Integrate: \((W = \left[\frac{5}{3}x^3\right]_0^3)\) Evaluate: \(W = \frac{5}{3}(3^3) - \frac{5}{3}(0) = 45\) The work done by the force is \(45\) units of work.
  • Exercise 6

    Find the mass of a rod of length \(3\) meters with a density function \(\rho(x) = 2 + 3x\), where the density is given in kilograms per meter.
    Click to show the solution The mass of the rod is given by the integral: \(M = \int_{0}^{3} \rho(x) \, dx = \int_{0}^{3} (2 + 3x) \, dx\) Integrate: \(M = \left[2x + \frac{3}{2}x^2\right]_0^3\) Evaluate: \(M = (6 + \frac{27}{2}) - (0) = 6 + 13.5 = 19.5\) The mass of the rod is \(19.5\) kilograms.
  • Exercise 7

    Calculate the electric charge \(Q\) on a wire of length \(5\) meters, with a charge density function \(\lambda(x) = 4x^3\) coulombs per meter.
    Click to show the solution The electric charge on the wire is given by the integral: \(Q = \int_{0}^{5} \lambda(x) \, dx = \int_{0}^{5} 4x^3 \, dx\) Integrate: \(Q = \left[x^4\right]_0^5\) Evaluate: \(Q = 5^4 - 0 = 625\) The electric charge on the wire is \(625\) coulombs.
  • Exercise 8

    Find the total distance traveled by a particle moving along the \(x\)-axis with a velocity function \(v(t) = 3t^2 - 6t\) from \(t = 0\) to \(t = 3\).
    Click to show the solution To find the total distance traveled, we need to find the integral of the absolute value of the velocity function: \(D = \int_{0}^{3} |3t^2 - 6t| \, dt\) We first need to find the intervals where the velocity is positive or negative. To do this, we can find the zeros of the velocity function: \(3t^2 - 6t = 0\) Factor: \(3t(t - 2) = 0\) The zeros are \(t = 0\) and \(t = 2\). The velocity is negative on the interval \([0, 2)\) and positive on the interval \((2, 3]\). We can split the integral into two parts: \((D = \int_{2}^{3} (3t^2 - 6t) \, dt + \int_{0}^{2} (6t - 3t^2) \, dt)\) Integrate: \(D = \left[\frac{1}{3}t^3 - 3t^2\right]_2^3 + \left[3t^2 - \frac{1}{3}t^3\right]_0^2\) Evaluate: The total distance traveled by the particle is \(2/3\) units.
  • Exercise 9

    A tank has the shape of an inverted cone with a height of \(10\) meters and a radius of \(4\) meters at the top. If the tank is filled with water, find the work required to pump the water out of the tank through a pipe at the top. The density of water is \(1000\) kilograms per cubic meter, and the acceleration due to gravity is \(9.8\) meters per second squared.
    Click to show the solution Let \(x\) be the height from the top of the cone, and \(r(x)\) be the radius of the cone at height \(x\). Since the cone is similar to another cone with height \(10\) and radius \(4\), we have: \(\frac{r(x)}{x} = \frac{4}{10}\) So, \(r(x) = \frac{2}{5}x\) The volume of a thin disk at height \(x\) with thickness \(dx\) is: \(dV = \pi (r(x))^2 dx = \pi \left(\frac{2}{5}x\right)^2 dx\) The mass of the water in the disk is: \(dm = \rho dV = 1000 \pi \left(\frac{2}{5}x\right)^2 dx\) The work required to lift the water in the disk to the top of the tank is: \(dW = dm \cdot g \cdot x = 1000 \pi \left(\frac{2}{5}x\right)^2 \cdot 9.8 \cdot x \, dx\) The total work required is the integral of \(dW\) from \(x = 0\) to \(x = 10\): \(W = \int_{0}^{10} 1000 \pi \left(\frac{2}{5}x\right)^2 \cdot 9.8 \cdot x \, dx\) Integrate: \(W = 1000 \pi \cdot \frac{4}{25} \cdot 9.8 \cdot \left[\frac{1}{4}x^4\right]_0^{10}\) Evaluate: \(W = 1000 \pi \cdot \frac{4}{25} \cdot 9.8 \cdot \frac{1}{4}(10^4 - 0)\) The work required to pump the water out of the tank is \( 3920000\pi\) joules.
  • Exercise 10

    Find the probability that a point chosen randomly inside a circle of radius \(1\) lies in the area between the two circles with equations \(x^2 + y^2 = 1\) and \(x^2 + y^2 = \frac{1}{2}\).
    Click to show the solution We can find the areas of both circles using the formula for the area of a circle, \(A = \pi r^2\). The area of the larger circle is: \(A_1=\pi * (1)^2=\pi\) The radius of the smaller circle can be found by taking the square root of \(\frac{1}{2}\). The area of the smaller circle is: \(A_2=\pi * (\sqrt(\frac{1}{2}))^2=\frac{\pi}{2}\) The area between the two circles is the difference of their areas: \(A_1-A_2=\frac{\pi}{2}\) Since the total area of the larger circle is \(\pi\), the probability that a point chosen randomly inside the circle lies in the area between the two circles is: \(\frac{\frac{\pi}{2}}{\pi}=\frac{1}{2}\).
  • Applications of Integrals

    Integration has numerous applications in mathematics, physics, and engineering. In this lesson, we will explore some of the most common applications of integrals in a high school calculus course.

    1. Area Under a Curve

    Definite integrals can be used to find the area under a curve between two points on the x-axis. If \(f(x)\) is a continuous function on the interval \([a, b]\), the area under the curve is given by:

    \(\int_a^b f(x) \, dx\)

    2. Area Between Two Curves

    To find the area between two curves, \(y = f(x)\) and \(y = g(x)\), on the interval \([a, b]\), we compute the definite integral of the difference between the functions:

    \(\int_a^b (f(x) - g(x)) \, dx\)

    This gives the area between the curves when \(f(x) \ge g(x)\) for all \(x\) in \([a, b]\).

    3. Volume of Solids of Revolution

    Integration can be used to find the volume of a solid formed by revolving a region around an axis. We will discuss two methods: the disk method and the shell method.

    3.1 Disk Method

    For a region bounded by \(y = f(x)\), the x-axis, and the vertical lines \(x = a\) and \(x = b\), the volume of the solid formed by revolving the region around the x-axis is given by:

    \(V = \pi \int_a^b (f(x))^2 \, dx\)

    3.2 Shell Method

    For a region bounded by \(y = f(x)\), the x-axis, and the vertical lines \(x = a\) and \(x = b\), the volume of the solid formed by revolving the region around the y-axis is given by:

    \(V = 2 \pi \int_a^b x \cdot f(x) \, dx\)

    4. Arc Length

    To find the arc length of a smooth curve represented by the function \(y = f(x)\) on the interval \([a, b]\), we can use the following formula:

    \(L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx\)

    5. Work Done by a Variable Force

    In physics, integration is used to compute the work done by a variable force. If a force \(F(x)\) acts on an object, moving it from position \(x = a\) to position \(x = b\), the work done by the force is given by:

    \(W = \int_a^b F(x) \, dx\)

    These are just a few examples of the many applications of integrals. As you progress through calculus, you will encounter even more advanced applications in different areas of mathematics and science.

    Differential Equations

    Back to curriculum

    Exercise 1

    . Solve the following first-order linear differential equation:

    \(y' + 3xy = 6x\)
    Click to reveal the solution

    The given differential equation is a first-order linear differential equation. To solve it, first find the integrating factor, which is given by \(IF = e^{\int P(x)dx}\), where \(P(x) = 3x\). In this case, \(IF = e^{3/2 * x^2}\). Multiply both sides of the equation by the integrating factor:

    \( e^{3/2 * x^2}(y' + 3xy) = 6x * e^{3/2 * x^2}\)

    Now, the left side of the equation is the derivative of the product of \(y\) and the integrating factor, so we can rewrite it as:

    \((y * e^{3/2 * x^2})' = 6x * e^{3/2 * x^2}\)

    Integrate both sides with respect to \(x\):

    \(y * e^{3/2 * x^2} = 2 * e^{3/2 * x^2} + C\)

    Finally, solve for \(y\):

    \(y(x) = 2 + C * e^{-3/2 * x^2}\)

    Exercise 2

    Solve the following homogeneous differential equation:

    \(y' = (x^2 + y^2) / (xy)\)
    Click to reveal the solution

    The given differential equation is homogeneous. Make the substitution \(y = vx\), where \(v\) is a function of \(x\). Then, \(y' = v + xv'\). Substituting this into the original equation, we get:

    \(v + xv' = (x^2 + x^2v^2) / (x^2v)\)

    Simplify the equation and solve for \(v'\):

    \(v' = 1/(xv)\) or \(vv'=1/x\)

    Then \((1/2)v^2=ln|x|+C\)

    \(v=±\sqrt(2ln|x|+C)\)

    Exercise 3

    Solve the following Bernoulli differential equation:

    \(y' - (1/x)y = y^2/x^2\)
    Click to reveal the solution

    The given differential equation is a Bernoulli equation with \(n = 2\). To solve it, make the substitution v = 1/y. Then, \(v' = -y'/y^2\). Substituting this into the original equation, we get:

    \(-y'/y^2 - (1/x)(1/y) = 1/x^2\)

    Rewrite the equation in terms of \(v\):

    \(v' - (1/x)v = 1/x^2\)

    This is now a first-order linear differential equation. Find the integrating factor \(IF = e^{\int(P(x)dx)})\), where \(P(x) = -1/x\). In this case, \(IF = e^{-ln(x)} = 1/x\). Multiply both sides of the equation by the integrating factor:

    \((1/x)v' - (1/x^2)v = 1/x^3\)

    Now, the left side of the equation is the derivative of the product of \(v\) and the integrating factor, so we can rewrite it as:

    \((v/x)' = 1/x^3\)

    Integrate both sides with respect to \(x\):

    \(v/x = -1/(2x^2) + C\)

    Finally, solve for \(y\):

    \( y(x) = 2x/(-1+2Cx^2)\)

    Exercise 4

    Solve the following initial value problem:

    \(y' = y^2/x, y(1) = 1\)
    Click to reveal the solution

    The given differential equation is separable. Separate the variables:

    \(\int((1/y^2)dy) = \int((1/x)dx)\)

    Integrate both sides:

    \(-1/y = ln(|x|) + C\)

    Now, solve for \(y\):

    \(y(x) = -1/(ln(|x|) + C)\)

    Use the initial condition \(y(1) = 1\) to find \(C\):

    \(1 = -1/(ln(1) + C)\)

    Since \(ln(1) = 0\), we have \(C = -1\). Thus, the solution is:

    \(y(x) = -1/(ln(|x|) - 1)\)

    Exercise 5

    Solve the following exact differential equation:

    \((2xy + e^x)dx + (x^2)dy = 0\)
    Click to reveal the solution

    The given differential equation is exact. To solve it, first identify \(M(x, y) = 2xy + e^x\) and \(N(x, y) = x^2\). Then, check that \(dM/dy = dN/dx\). In this case, both partial derivatives are equal to \(2x\). Thus, the equation is exact. Now, integrate \(M\) with respect to \(x\):

    \(\int((2xy + e^x)dx) = x^2y + e^x + g(y)\)

    Now, differentiate this expression with respect to \(y\) and compare it to \(N(x, y)\):

    \(d/dy(x^2y + e^x + g(y)) = x^2 + dg/dy\)

    Since \(N(x, y) = x^2 \), it follows that \(dg/dy = 0 \), which means \(g(y)\) is a constant. Thus, the solution to the exact differential equation is:

    \(x^2y + e^x + C = 0\)

    where \(C\) is a constant.

    Exercise 6

    Solve the following second-order linear differential equation:

    \(y'' - 2y' + y = e^x\)
    Click to reveal the solution

    First, solve the homogeneous part of the equation:

    \(y_h'' - 2y_h' + y_h = 0\)

    The characteristic equation is \(r^2 - 2r + 1 = 0\), which has a repeated root \(r = 1\). Thus, the homogeneous solution is:

    \(y_h(x) = C_1e^x + C_2xe^x\)

    Now, find a particular solution \(y_p(x)\) for the non-homogeneous part. Assume a solution of the form \(y_p(x) = Ax^2e^x\). Differentiate \(y_p(x)\) and substitute it into the non-homogeneous equation:

    \(y_p'' - 2y_p' + y_p = e^x\)

    After substituting and simplifying, we get \(A = 1/2\). Thus, the particular solution is:

    \(y_p(x) = (1/2)x^2e^x\)

    The general solution is the sum of the homogeneous and particular solutions:

    \(y(x) = C_1e^x + C_2xe^x + (1/2)x^2e^x\)

    Exercise 7

    Find the general solution to the following second-order linear differential equation:

    \(y'' + 4y' + 5y = 0\)
    Click to reveal the solution

    The characteristic equation is \(r^2 + 4r + 5 = 0\). The roots are complex: \(r = -2 ± i\). Thus, the general solution is:

    \(y(x) = e^{-2x}(C_1cos(x) + C_2sin(x))\)

    Exercise 8

    Solve the following initial value problem:

    \(y'' + y = 0, y(0) = 1, y'(0) = 2\)
    Click to reveal the solution

    The characteristic equation is \(r^2 + 1 = 0\), which has the roots \(r = ±i\). Thus, the general solution is:

    \(y(x) = C_1cos(x) + C_2sin(x)\)

    Use the initial conditions to find the constants \(C_1\) and \(C_2\). First, apply \(y(0) = 1\):

    \(1 = C_1cos(0) + C_2sin(0)\)

    Since \(cos(0) = 1\) and \(sin(0) = 0\), we have \(C_1 = 1\). Next, apply \(y'(0) = 2\). The derivative of \(y(x)\) is:

    \(y'(x) = -C_1sin(x) + C_2cos(x)\)

    Substitute \(x = 0\) and \(C_1 = 1\):

    \(2 = -1sin(0) + C_2cos(0)\)

    Since \(sin(0) = 0\) and \(cos(0) = 1\), we have \(C_2 = 2\). Therefore, the solution to the initial value problem is:

    \(y(x) = cos(x) + 2sin(x)\)

    Exercise 9

    Solve the following non-homogeneous second-order linear differential equation:

    \(y'' - y' - 2y = x + 1\)
    Click to reveal the solution

    First, solve the homogeneous part of the equation:

    \(y_h'' - y_h' - 2y_h = 0\)

    The characteristic equation is \(r^2 - r - 2 = 0\), which has the roots \(r = 2\) and \(r = -1\). Thus, the homogeneous solution is:

    \(y_h(x) = C_1e^{2x} + C_2e^{-x}\)

    Now, find a particular solution \(y_p(x)\) for the non-homogeneous part. Assume a solution of the form \(y_p(x) = Ax + B\). Differentiate \(y_p(x)\) and substitute it into the non-homogeneous equation:

    \(y_p'' - y_p' - 2y_p = x + 1\)

    After substituting and simplifying, we get \(A = 1\) and \(B = 2\). Thus, the particular solution is:

    \(y_p(x) = x + 2\)

    The general solution is the sum of the homogeneous and particular solutions:

    \(y(x) = C_1e^{2x} + C_2e^{-x} + x + 2\)

    Exercise 10

    Solve the following initial value problem for a damped harmonic oscillator:

    \(y'' + 4y' + 5y = 0, y(0) = 2, y'(0) = -1\)
    Click to reveal the solution

    The characteristic equation is \(r^2 + 4r + 5 = 0\), which has the roots \(r = -2 ± i\). Thus, the general solution is:

    \(y(x) = e^{-2x}(C_1cos(x) + C_2sin(x))\)

    Use the initial conditions to find the constants \(C_1\) and \(C_2\). First, apply \(y(0) = 2\):

    \(2 = e^{-2*0}(C_1cos(0) + C_2sin(0))\)

    Since \(cos(0) = 1\) and \(sin(0) = 0\), we have \(C_1 = 2\). Next, apply \(y'(0) = -1\). The derivative of \(y(x)\) is:

    \(y'(x) = -2e^{-2x}(C_1cos(x) + C_2sin(x)) + e^{-2x}(-C_1sin(x) + C_2cos(x))\)

    Substitute \(x = 0\) and \(C_1 = 2\):

    \(-1 = -2*e^{-2*0}(2*cos(0) + C_2*sin(0)) + e^{-2*0}(-2*sin(0) + C_2*cos(0))\)

    Since \(cos(0) = 1\) and \(sin((0) = 0, we have C_2 = 1\). Therefore, the solution to the initial value problem is:

    \(y(x) = e^{-2x}(2cos(x) + sin(x))\)

    Differential Equations

    Differential equations are equations that involve an unknown function and its derivatives. They are used to model a wide range of real-world problems in various fields such as physics, engineering, and biology. In this lesson, we will focus on first-order differential equations.

    1. Definition

    A first-order differential equation is an equation that can be written in the form:

    \(F(x, y, y') = 0\)

    where \(y' = \frac{dy}{dx}\) is the first derivative of the function \(y(x)\), and \(F\) is a function of three variables.

    2. Separable Equations

    A separable first-order differential equation can be written in the form:

    \(\frac{dy}{dx} = \frac{g(y)}{h(x)}\)

    To solve a separable equation, we can rewrite it as:

    \(h(x) \, dx = g(y) \, dy\)

    Integrating both sides of the equation, we obtain:

    \(\int h(x) \, dx = \int g(y) \, dy + C\)

    where \(C\) is the constant of integration.

    3. Linear First-Order Differential Equations

    A linear first-order differential equation has the form:

    \(y' + P(x) y = Q(x)\)

    To solve this equation, we can use an integrating factor, which is defined as:

    \(\mu(x) = e^{\int P(x) \, dx}\)

    Multiplying both sides of the differential equation by the integrating factor, we obtain:

    \((\mu(x) y)' = \mu(x) Q(x)\)

    Integrating both sides with respect to \(x\), we get:

    \(\mu(x) y = \int \mu(x) Q(x) \, dx + C\)

    Finally, we can solve for \(y(x)\) by dividing by the integrating factor:

    \(y(x) = \frac{1}{\mu(x)} \left( \int \mu(x) Q(x) \, dx + C \right)\)

    4. Initial Value Problems

    An initial value problem (IVP) consists of a differential equation and an initial condition of the form:

    \(y(x_0) = y_0\)

    To solve an IVP, we first find the general solution of the differential equation and then use the initial condition to determine the constant of integration.

    These are the basics of differential equations at the high school level. As you progress in your mathematical studies, you will encounter more advanced types of differential equations and techniques for solving them.

    Sequences and Series

    Back to curriculum

    Exercise 1.

    Determine if the following sequence converges or diverges: $$a_n = \frac{n^3 + 5n}{3n^3 - n^2 + 2}$$. If it converges, find its limit. Explain your reasoning.

    Click to reveal the solution The sequence converges. Limit: $$\lim_{n \to \infty} \frac{n^3 + 5n}{3n^3 - n^2 + 2}$$. To determine the limit, we can use the fact that the highest power of $$n$$ in both the numerator and denominator is $$n^3$$. Divide both the numerator and denominator by $$n^3$$: $$\lim_{n \to \infty} \frac{1 + \frac{5}{n^2}}{3 - \frac{1}{n} + \frac{2}{n^3}} = \frac{1}{3}$$ The limit is $$\frac{1}{3}$$.

    Exercise 2.

    Consider the geometric series: $$\sum_{n=1}^{\infty} ar^n$$. If the sum of the series converges to 6 and the first term is 2, find the value of the common ratio $$r$$. Explain your reasoning.

    Click to reveal the solution For a convergent geometric series, the sum is given by the formula $$S = \frac{a}{1 - r}$$. We are given that the sum converges to 6 and the first term is 2, so we have the equation: $$6 = \frac{2}{1 - r}$$ Solving for $$r$$, we get $$r = \frac{2}{3}$$.

    Exercise 3.

    Determine if the following series converges or diverges: $$\sum_{n=1}^{\infty} \frac{n^2 + 3n + 2}{n^3 + 1}$$. If it converges, find the sum. Explain your reasoning.

    Click to reveal the solution We can apply the Limit Comparison Test to determine convergence. Let $$b_n = \frac{n^2}{n^3} = \frac{1}{n}$$. Since $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{n^2 + 3n + 2}{n^3 + 1}}{\frac{1}{n}} = 1$$, both series either converge or diverge together. The series $$\sum_{n=1}^{\infty} \frac{1}{n}$$ is the harmonic series, which diverges. Therefore, the given series also diverges.

    Exercise 4.

    Determine the radius of convergence for the power series $$\sum_{n=0}^{\infty} \frac{(x-3)^n}{n!}$$. Explain your reasoning.

    Click to reveal the solution To find the radius of convergence, we can use the Ratio Test: $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty} \left|\frac{\frac{(x-3)^{n+1}}{(n+1)!}}{\frac{(x-3)^n}{n!}}\right| = \lim_{n \to \infty} \frac{(x-3)}{n+1} = 0$$ Since the limit is 0 for all values of $$x$$, the radius of convergence is infinite.

    Exercise 5.

    Find the Maclaurin series for the function $$f(x) = e^{2x}$$ up to and including the $$x^4$$ term. Explain your reasoning.

    Click to reveal the solution To find the Maclaurin series, we need to determine the first few derivatives of the function $$f(x) = e^{2x}$$: $$f(x) = e^{2x}$$ $$f'(x) = 2e^{2x}$$ $$f''(x) = 4e^{2x}$$ $$f'''(x) = 8e^{2x}$$ $$f^{(4)}(x) = 16e^{2x}$$ Now, evaluate these derivatives at $$x = 0$$: $$f(0) = 1$$ $$f'(0) = 2$$ $$f''(0) = 4$$ $$f'''(0) = 8$$ $$f^{(4)}(0) = 16$$ The Maclaurin series up to the $$x^4$$ term is given by: $$f(x) \approx 1 + 2x + \frac{4x^2}{2!} + \frac{8x^3}{3!} + \frac{16x^4}{4!}$$ Simplifying, we get: $$f(x) \approx $$ So, the Maclaurin series for $$f(x) = e^{2x}$$ up to and including the $$x^4$$ term is $$1 + 2x + 2x^2 + \frac{4}{3}x^3 + \frac{2}{3}x^4$$

    Exercise 6.

    Given the sequence $$b_n = \frac{(-1)^n}{\sqrt{n}}$$, determine if it converges or diverges using the Alternating Series Test. Explain your reasoning.

    Click to reveal the solution The sequence converges by the Alternating Series Test. For the test to hold, two conditions must be met: 1. $$|b_{n+1}| \leq |b_n|$$ for all $$n$$ 2. $$\lim_{n \to \infty} b_n = 0$$ First, we check the decreasing condition. Since the sequence $$c_n = \frac{1}{\sqrt{n}}$$ is decreasing, and $$b_n = (-1)^n c_n$$, it follows that $$|b_{n+1}| \leq |b_n|$$ for all $$n$$. Next, we check the limit: $$\lim_{n \to \infty} \frac{(-1)^n}{\sqrt{n}} = 0$$, as the exponential term $$(-1)^n$$ does not affect the limit. Both conditions are met, so the sequence converges by the Alternating Series Test.

    Exercise 7.

    Determine if the series \(\sum_{n=1}^{\infty} (3^n/n!)\) converges or diverges. If it converges, find the sum. Explain your reasoning.
    Click to reveal the solution Solution: The series converges by the Ratio Test. For the Ratio Test, we compute the limit: $$ \lim(n->\infty) |((a_{n+1})/a_n)|$$ $$ = \lim(n->\infty) ((3^{n+1})/((n+1)!)) / ((3^n)/(n!)) = \lim(n->\infty) (3n!)/((n+1)!)$$ $$ = \lim(n->\infty) (3/(n+1)) = 0.$$ Since the limit is less than 1, the series converges. To find the sum, we can recognize the series as a variant of the Taylor series for \(e^x\) $$ e^x = \sum_{n=0}^{\infty} (x^n/n!).$$ Plugging in x = 3, we get: $$e^3 = \sum_{n=0}^{\infty} (3^n/n!).$$ The given series starts at n = 1, so we must subtract the n = 0 term from the sum: $$\sum_{n=1}^{\infty} (3^n/n!) = e^3 - (3^0/0!) = e^3 - 1.$$ Thus, the sum of the series is $$ e^3 - 1.$$

    Exercise 8.

    Find the interval of convergence for the power series $$\sum_{n=1}^{\infty} \frac{(x+4)^n}{n3^n}$$. Explain your reasoning.

    Click to reveal the solution To find the interval of convergence, we can use the Ratio Test: $$\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$ $$ = \lim_{n \to \infty} \left|\frac{\frac{(x+4)^{n+1}}{(n+1)3^{n+1}}}{\frac{(x+4)^n}{n3^n}}\right| $$ $$= \lim_{n \to \infty} \left|\frac{n(x+4)}{3(n+1)}\right| = \frac{|x+4|}{3}$$ The series converges when this limit is less than 1: $$\frac{|x+4|}{3} < 1$$ Solving for $$x$$, we get: $$-3 < x+4 < 3$$ $$-7 < x < -1$$ So, the interval of convergence is $$(-7, -1)$$. With divergence at x=-1 and convergence at x=-7.

    Exercise 9.

    Use the Integral Test to determine if the series $$\sum_{n=1}^{\infty} (1/(n^2 + 1))$$ converges or diverges. Explain your reasoning.
    Click to reveal the solution Solution: To apply the Integral Test, consider the function $$f(x) = 1 / (x^2 + 1).$$ We need to verify the conditions for the Integral Test: f(x) is continuous for x ≥ 1, f(x) is positive for x ≥ 1, and f(x) is decreasing for x ≥ 1. All three conditions hold for $$f(x) = 1 / (x^2 + 1).$$ Therefore, we can apply the Integral Test. We need to evaluate the following integral: $$\int_{1}^{\infty} (1/(x^2 + 1)) dx.$$ The integral converges, so the series also converges.

    Exercise 10.

    Evaluate the limit $$\lim(n->\infty) (1/n) * \sum_{k=1}^{n} (k^2/n^2).$$ Explain your reasoning.
    Click to reveal the solution Solution: To evaluate the limit, we can rewrite the expression as a Riemann sum: $$lim(n->\infty) (1/n^3) * \sum_{k=1}^{n} k^2.$$ Now, we use the formula for the sum of the first n squares: $$\sum_{k=1}^{n}k^2 = n(n + 1)(2n + 1)/6.$$ Substituting this into the expression and taking the limit as n goes to infinity, we get the limit to be 1/3.

    Sequences and Series

    In this lesson, we will discuss sequences, series, and some of their properties. Sequences and series are essential concepts in calculus and have various applications in mathematics and real-world problems.

    1. Sequences

    A sequence is an ordered list of numbers. We can define a sequence \(\{a_n\}\) as a function that maps each positive integer \(n\) to a real number \(a_n\). In other words, a sequence is a function with the domain being the set of positive integers.

    Convergence and Divergence of Sequences

    A sequence \(\{a_n\}\) is said to converge to a limit \(L\) if, for every \(\epsilon > 0\), there exists a positive integer \(N\) such that for all \(n > N\), we have \(|a_n - L| < \epsilon\). If a sequence does not converge, it is said to diverge.

    \(\lim_{n \to \infty} a_n = L\)

    2. Series

    A series is the sum of the terms of a sequence. Given a sequence \(\{a_n\}\), the corresponding series is denoted by:

    \(\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots\)

    Convergence and Divergence of Series

    A series is said to converge if the sequence of its partial sums converges. The partial sum \(S_n\) of a series is defined as:

    \(S_n = \sum_{k=1}^{n} a_k\)

    If the sequence \(\{S_n\}\) converges, the series converges, and if the sequence \(\{S_n\}\) diverges, the series diverges.

    3. Tests for Convergence

    There are several tests that can be used to determine the convergence or divergence of a series. Some of these tests are:

    a) The nth-Term Test

    If \(\lim_{n \to \infty} a_n \neq 0\), then the series \(\sum_{n=1}^{\infty} a_n\) diverges.

    b) The Integral Test

    If \(f(x)\) is a positive, continuous, and decreasing function on the interval \([1, \infty)\), and \(a_n = f(n)\), then the series \(\sum_{n=1}^{\infty} a_n\) converges if and only if the integral \(\int_1^{\infty} f(x) \, dx\) converges.

    c) The Comparison Test

    If \(0 \leq a_n \leq b_n\) for all \(n\), and the series \(\sum_{n=1}^{\infty} b_n\) converges, then the series \(\sum_{n=1}^{\infty} a_n\) also converges. If the series \(\sum_{n=1}^{\infty} b_n\) diverges, then the series \(\sum_{n=1}^{\infty} a_n

    also diverges.

    d) The Ratio Test

    If \(\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L\), then:

    1. If L < 1, the series \(\sum_{n=1}^{\infty} a_n\) converges absolutely. 2. If L > 1, the series \(\sum_{n=1}^{\infty} a_n\) diverges. 3. If L = 1, the test is inconclusive.

    e) The Root Test

    If \(\lim_{n \to \infty} \sqrt[n]{|a_n|} = L\), then:

    1. If L < 1, the series \(\sum_{n=1}^{\infty} a_n\) converges absolutely. 2. If L > 1, the series \(\sum_{n=1}^{\infty} a_n\) diverges. 3. If L = 1, the test is inconclusive.

    4. Power Series

    A power series is a series of the form:

    \(\sum_{n=0}^{\infty} c_n (x - a)^n\)

    where \(c_n\) is a sequence of constants, and \(a\) is a fixed number. The interval of convergence of a power series is the set of all \(x\) for which the series converges.

    By using various tests for convergence, we can determine the interval of convergence and the radius of convergence (half the length of the interval of convergence) for a given power series.

    Power series are essential in calculus and have applications in various areas of mathematics, including solving differential equations, approximating functions, and computing infinite sums.

    5. Taylor and Maclaurin Series

    A Taylor series is a power series that represents a function \(f(x)\) near a point \(a\). If a function \(f(x)\) has continuous derivatives up to order \(n\), the Taylor series for \(f(x)\) about the point \(a\) is given by:

    \(f(x) \approx \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!} (x - a)^n\)

    A Maclaurin series is a special case of the Taylor series, where the point \(a\) is equal to 0. The Maclaurin series for a function \(f(x)\) is given by:

    \(f(x) \approx \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\)

    Taylor and Maclaurin series are used to approximate functions, especially when working with complex functions or when the exact function is difficult to compute.

    6. Parametric and Polar Equations

    In some cases, it is more convenient to describe a curve in the plane using parametric equations. A parametric representation of a curve is given by:

    x = f(t) y = g(t)

    where \(f(t)\) and \(g(t)\) are functions of a parameter \(t\), usually with \(t\) in a closed interval \([a, b]\).

    To find the derivative of a parametric function, we can use the chain rule:

    \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)

    provided that \(dx/dt\) is not equal to zero.

    Polar coordinates \((r, \theta)\) are another way of representing points in the plane. A polar equation defines a curve in terms of the radial distance \(r\) and the angle \(\theta\). To find the derivative of a polar equation, we first convert the polar equation to parametric form:

    x = r cos(\(\theta\)) y = r sin(\(\theta\))

    Then, we find the derivative using the chain rule as before:

    \(\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}\)

    provided that \(dx/d\theta\) is not equal to zero.

    7. Vector Functions

    A vector function is a function that assigns a vector to each point in its domain. A vector function can be represented in component form as:

    \(\vec{r}(t) = \langle f(t), g(t), h(t) \rangle\)

    where \(f(t)\), \(g(t)\), and \(h(t)\) are scalar functions of the parameter \(t\).

    The derivative of a vector function is found by taking the derivative of each component function:

    \(\frac{d\vec{r}}{dt} = \langle \frac{df}{dt}, \frac{dg}{dt}, \frac{dh}{dt} \rangle\)

    Vector functions have applications in physics, engineering, and other fields where quantities have both magnitude and direction.

    8. Multivariable Functions

    A multivariable function is a function of two or more variables. For example, a function of two variables \(x\) and \(y\) can be written as:

    \(f(x, y)\\)

    Partial derivatives are used to find the rate of change of a multivariable function with respect to one variable while keeping the other variables constant. The partial derivative of a function \(f(x, y)\) with respect to \(x\) is denoted by:

    \(\frac{\partial f}{\partial x}\)

    Similarly, the partial derivative with respect to \(y\) is denoted by:

    \(\frac{\partial f}{\partial y}\)

    Higher-order partial derivatives can also be computed, such as the second partial derivatives:

    \(\frac{\partial^2 f}{\partial x^2}\), \(\frac{\partial^2 f}{\partial y^2}\), and \(\frac{\partial^2 f}{\partial x \partial y}\)

    These higher-order partial derivatives are used in various applications, such as optimization problems and the study of surfaces

    9. Infinite Series

    An infinite series is the sum of the terms of an infinite sequence. It can be written in the form:

    \(\sum_{n=1}^{\infty} a_n\)

    where \(a_n\) represents the terms of the sequence. Infinite series have applications in many areas of mathematics and science, including the study of functions, calculus, and numerical analysis.

    Convergence is an important concept in the study of infinite series. A series is said to converge if the sum of its terms approaches a finite value as the number of terms goes to infinity. Divergence occurs when the sum of the terms does not approach a finite value.

    Several tests can be used to determine whether an infinite series converges or diverges, including:

    1. The nth term test 2. The geometric series test 3. The integral test 4. The comparison test 5. The limit comparison test 6. The alternating series test 7. The ratio test 8. The root test

    These tests can help us determine the convergence or divergence of a given series, as well as find the sum of a convergent series in some cases.

    10. Power Series

    A power series is a series of the form:

    \(\sum_{n=0}^{\infty} a_n (x - c)^n\)

    where \(a_n\) are constants, \(c\) is a fixed number, and \(x\) is a variable. Power series can be used to represent functions and can be differentiated and integrated term-by-term within their interval of convergence.

    The interval of convergence is the set of all \(x\) values for which the power series converges. The radius of convergence is the distance from the center \(c\) to the endpoints of the interval of convergence.

    Power series have many applications in mathematics, including approximating functions, solving differential equations, and computing limits.

    11. Taylor and Maclaurin Series

    Taylor and Maclaurin series are special types of power series that can be used to approximate functions. A Taylor series is centered at a point \(c\) and can be written as:

    \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x - c)^n\)

    where \(f^{(n)}(c)\) is the nth derivative of \(f\) evaluated at \(c\). A Maclaurin series is a Taylor series centered at \(c = 0\):

    \(f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\)

    Taylor and Maclaurin series can be used to approximate functions and compute limits, derivatives, and integrals. They also have applications in physics, engineering, and other fields.

    12. Convergence of Series

    Convergence tests are methods used to determine if an infinite series converges or diverges. Here are some of the most common convergence tests:

    1. The nth term test: If \(\lim_{n \to \infty} a_n \neq 0\), the series diverges. 2. The geometric series test: If \(|r| < 1\), the geometric series converges. If \(|r| > 1\), it diverges. If \(|r| = 1\), the test is inconclusive. 3. The integral test: If \(\int_{1}^{\infty} f(x) \, dx\) converges, the series converges, and if the integral diverges, the series diverges. 4. The comparison test: If \(0 \le a_n \le b_n\) for all \(n\) and \(\sum_{n=1}^{\infty} b_n\) converges, then \(\sum_{n=1}^{\infty} a_n\) also converges. If \(\sum_{n=1}^{\infty} b_n\) diverges, then \(\sum_{n=1}^{\infty} a_n\) also diverges. 5. The limit comparison test: If \(\lim_{n \to \infty} \frac{a_n}{b_n} = c > 0\), then either both series converge or both series diverge. 6. The alternating series test: If the terms of an alternating series are decreasing and \(\lim_{n \to \infty} a_n = 0\), the series converges. 7. The ratio test: If \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} < 1\), the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, the test is inconclusive. 8. The root test: If \(\lim_{n \to \infty} \sqrt[n]{|a_n|} < 1\), the series converges. If the limit is greater than 1, the series diverges. If the limit is equal to 1, the test is inconclusive.

    These tests help us determine whether a given series converges or diverges and, in some cases, find the sum of a convergent series.

    13. Series Approximations and Error Bounds

    Sometimes, it is not possible or practical to find the exact sum of an infinite series. Instead, we can approximate the sum using a finite number of terms. The error bound is a measure of how accurate the approximation is.

    For an alternating series, the error bound can be determined using the alternating series remainder theorem:

    \(R_n = |S - S_n| \le a_{n+1}\)

    where \(S\) is the exact sum of the series, \(S_n\) is the sum of the first \(n\) terms, and \(a_{n+1}\) is the absolute value of the \((n+1)\)th term.

    For other types of series, error bounds can be determined using methods such as the integral test remainder estimate, the Taylor series remainder theorem, and the Cauchy criterion for convergence.

    These methods can help us estimate the sum of a series and determine the accuracy of the approximation

    14. Power Series

    A power series is an infinite series of the form:

    \(\sum_{n=0}^{\infty} c_n(x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \cdots\)

    where \(c_n\) are the coefficients, \(a\) is the center of the series, and \(x\) is the variable.

    Power series have a radius of convergence, which is the interval around the center where the series converges. The radius of convergence can be found using the ratio test:

    \(\lim_{n \to \infty} \frac{c_{n+1}(x-a)^{n+1}}{c_n(x-a)^n} = \lim_{n \to \infty} \frac{c_{n+1}}{c_n}(x-a) \le 1\)

    If this limit is less than 1, the power series converges. If the limit is greater than 1, the power series diverges. If the limit is equal to 1, the test is inconclusive.

    15. Taylor and Maclaurin Series

    Taylor series are a type of power series that can be used to approximate functions. A Taylor series for a function \(f(x)\) centered at \(a\) is given by:

    \(\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots\)

    where \(f^{(n)}(a)\) denotes the \(n\)th derivative of the function evaluated at \(a\).

    A Maclaurin series is a special case of a Taylor series centered at \(a=0\):

    \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \cdots\)

    Common functions can be represented as Maclaurin series:

    1. \(e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}\) 2. \(\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}\) 3. \(\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}\) 4. \(\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}x^n\) 5. \((1+x)^k = \sum_{n=0}^{\infty} \binom{k}{n}x^n\)

    Taylor and Maclaurin series can be used to approximate functions and calculate limits, derivatives, and integrals.

    16. Convergence of Series

    The convergence of a series can be determined using various tests. Some of the common tests are:

    1. Geometric Series Test 2. p-Series Test 3. Divergence Test 4. Integral Test 5. Comparison Test 6. Limit Comparison Test 7. Alternating Series Test 8. Ratio Test 9. Root Test

    Each test has its own specific conditions and is applicable to different types of series. It is important to choose the appropriate test based on the properties of the given series.

    17. Convergence of Improper Integrals

    Improper integrals are integrals where either the interval of integration is infinite or the integrand has an infinite discontinuity. They are of two types:

    1. Type 1: \(\int_{a}^{\infty} f(x) \, dx\) or \(\int_{-\infty}^{b} f(x) \, dx\) 2. Type 2: \(\int_{a}^{b} f(x) \, dx\), where \(f(x)\) has an infinite discontinuity at one or both endpoints.

    To evaluate a Type 1 improper integral, we use the limit:

    \(\int_{a}^{\infty} f(x) \, dx = \lim_{t \to \infty} \int_{a}^{t} f(x) \, dx\)

    To evaluate a Type 2 improper integral, we use the limit:

    \(\int_{a}^{b} f(x) \, dx = \lim_{t \to a^+} \int_{t}^{b} f(x) \, dx\) or \(\lim_{t \to b^-} \int_{a}^{t} f(x) \, dx\)

    An improper integral converges if the limit exists, and diverges otherwise.

    18. Parametric Equations

    Parametric equations are a pair of equations that define the coordinates of a point in terms of a third variable, usually denoted as \(t\). The equations are of the form:

    \(x = f(t)\) \(y = g(t)\)

    Parametric equations can be used to represent curves that cannot be easily expressed using a single equation in \(x\) and \(y\). They are particularly useful for representing curves in higher dimensions.

    To find the slope of the tangent line to a curve defined by parametric equations, we can use the chain rule:

    \(\frac{dy}{dx} = \frac{dy/dt}{dx/dt}\)

    To find the arc length of a curve defined by parametric equations, we can use the formula:

    \(L = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\)

    Parametric equations can also be used to find the area enclosed by a curve and to compute surface area and volume of revolution.

    19. Polar Coordinates

    Polar coordinates are an alternative coordinate system to the Cartesian coordinates. In this system, a point in the plane is represented by a distance \(r\) from the origin and an angle \(\theta\) measured counterclockwise from the positive \(x\)-axis. The relationship between Cartesian and polar coordinates is given by:

    \(x = r \cos \theta\) \(y = r \sin \theta\)

    Polar coordinates are particularly useful for representing curves with radial symmetry or those that are more easily described using angles and distances.

    20. Polar Equations

    A polar equation is an equation that relates the polar coordinates \(r\) and \(\theta\) of a point. Some common types of polar equations are:

    1. Circles: \(r = a\) 2. Line: \(r \cos(\theta - \alpha) = a\) 3. Archimedean Spiral: \(r = a\theta\) 4. Lemniscate: \(r^2 = a^2 \cos 2\theta\) 5. Rose Curves: \(r = a \cos n\theta\) or \(r = a \sin n\theta\)

    To find the slope of the tangent line to a curve defined by a polar equation, we can use the chain rule and the relationship between Cartesian and polar coordinates:

    \(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}\)

    To find the arc length of a curve defined by a polar equation, we can use the formula:

    \(L = \int_{\theta_1}^{\theta_2} \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \, d\theta\)

    Polar equations can also be used to find the area enclosed by a curve and to compute surface area and volume of revolution.

    21. Sequences

    A sequence is a set of numbers in a specific order, typically generated by a formula. A sequence can be defined recursively, where each term depends on the previous terms, or explicitly, where each term is determined directly by the formula. Examples of sequences include arithmetic sequences, geometric sequences, and Fibonacci sequences.