Calculus-I

Functions and Limits

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Exercise 1.

Find the limit as x approaches 0 of \( (sin(2x) - 2sin(x))/(x^2).\)

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Exercise 2.

Find the limit as x approaches 1 of \((\sqrt(x) - 1)/(x - 1).\)

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Exercise 3.

Find the limit as x approaches 0 of \((e^{3x} - 1 - 3x)/(x^2).\)

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Exercise 4.

Find the limit as x approaches 0 of \((1 - cos(2x))/(x^2).\)

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Exercise 5.

Find the limit as x approaches\(\frac{\pi}{2}\) of \((tan(x))^2 / (1 - cos(x)).\)

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Exercise 6.

Find the limit as x approaches 0 of \((x - sin(x))/(x^3).\)

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Exercise 7.

Find the limit as x approaches 0 of \((1 - cos(x))/(xsin(x)).\)

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Exercise 8.

Find the limit as x approaches infinity of \((ln(x))^2/x.\)

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Exercise 9.

Find the limit as x approaches infinity of \((x^2 + 2x + 1)^{1/3} - x^{1/3}.\)

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Exercise 10.

Find the limit as x approaches 0 of \((arctan(x) - arcsin(x))/x^3.\)

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Calculus I: Functions and Limits

1. Introduction

In calculus, the study of functions and their limits is fundamental to understanding the behavior of functions as their input values approach certain points. This lesson will introduce the concept of limits, provide examples of different types of limits, and demonstrate how to compute limits using various techniques.

2. Functions

A function is a relation between a set of inputs and a set of possible outputs. In calculus, we generally deal with real-valued functions, which are functions that take real numbers as input and produce real numbers as output. Functions can be represented algebraically, graphically, or numerically, and they play a crucial role in understanding the behavior of various mathematical models.

3.a Limits

A limit is a value that a function approaches as the input approaches a particular value. In other words, a limit describes the behavior of a function near a given point. Limits are essential for defining continuity, differentiability, and integrability of functions, and they are the foundation of calculus.

3.b Formal Definition of Limits (Epsilon-Delta Definition)

The formal definition of a limit, known as the epsilon-delta definition, provides a rigorous way to describe the behavior of a function near a given point. It states:

Let f(x) be a function defined on some open interval containing a point a (except possibly at a) and let L be a real number. We say that the limit of f(x) as x approaches a is equal to L, written as:

lim(x→a) f(x) = L

If for every number ε > 0, there exists a number δ > 0 such that for all x in the domain of f, if 0 < |x - a| < δ, then |f(x) - L| < ε.

This definition states that as x gets arbitrarily close to a (but not equal to a), f(x) gets arbitrarily close to L. The values ε and δ represent the tolerance for how close x must be to a and how close f(x) must be to L, respectively.

4. Properties of Limits

There are several important properties of limits that can help us compute them more easily. Some of these properties include:

  • Constant and Identity Rules
  • Sum and Difference Rules
  • Product and Quotient Rules
  • Composition Rule
  • Squeeze Theorem

5. Techniques for Finding Limits

There are several techniques for finding limits, including:

  • Direct Substitution
  • Factoring
  • Rationalizing
  • L'Hopital's Rule
  • Special Limits

6. Continuity

A function is continuous at a point if the limit of the function as the input approaches that point exists and is equal to the function's value at that point. Continuity is an important property of functions that allows us to use calculus techniques, such as differentiation and integration, to analyze their behavior.

7. Conclusion

Understanding functions and their limits is essential for studying calculus. In this lesson, we've introduced the concept of limits, explored their properties, and discussed various techniques for finding limits. We've also discussed the importance of continuity and its relation to limits. With a solid grasp of these concepts, you'll be well-prepared to tackle more advanced topics in calculus, such as derivatives and integrals.

Derivatives

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  1. Exercise 1

    Find the derivative of \(f(x) = x^4 - 7x^3 + 9x^2 - 3x + 1\).
    Click to show the solution

    \(f'(x) = 4x^3 - 21x^2 + 18x - 3\)

  2. Exercise 2

    Find the derivative of \(g(x) = \sin(2x^2 + 3x)\).
    Click to show the solution

    \(g'(x) = (4x + 3)\cos(2x^2 + 3x)\)

  3. Exercise 3

    Find the derivative of \(h(x) = \frac{x^2 + 1}{x^3 - 2}\).
    Click to show the solution

    \(h'(x) = \frac{-3x^4 + 4x^3 + 6x^2 - 2}{(x^3 - 2)^2}\)

  4. Exercise 4

    Find the derivative of \(k(x) = e^{3x} \cos(2x)\).
    Click to show the solution

    \(k'(x) = e^{3x}(3\cos(2x) - 2\sin(2x))\)

  5. Exercise 5

    Find the derivative of \(n(x) = x^2 \ln(x)\).
    Click to show the solution

    Solution: \(n'(x) = x(2\ln(x) + 1)\)

  6. Exercise 6

    Find the derivative of \(p(x) = \frac{1}{\sqrt{1-x^2}}\).
    Click to show the solution

    Solution: \(p'(x) = \frac{x}{(1-x^2)^{3/2}}\)

  7. Exercise 7

    Find the derivative of \(q(x) = \tan^3(x)\).
    Click to show the solution

    Solution: \(q'(x) = 3\tan^2(x)\sec^2(x)\)

  8. Exercise 8

    Find the derivative of \(r(x) = x^5 \arctan(x)\).
    Click to show the solution

    Solution: \(r'(x) = 5x^4\arctan(x) + x^5\frac{1}{x^2 + 1}\)

  9. Exercise 9

    Find the derivative of \(s(x) = \frac{\cos(x)}{\sin^2(x)}\).
    Click to show the solution

    Solution: \(s'(x) = \frac{-\sin(x)\sin^2(x) - 2\sin(x)\cos^2(x)}{\sin^4(x)}\)

    Solution: \(s'(x) = \frac{-\sin^2(x) - 2\cos^2(x)}{\sin^3(x)}\)

Calculus I: Derivatives Lesson

Lesson Overview

In this lesson, we will cover the following topics:

  • Introduction to derivatives
  • Derivative as a rate of change
  • Derivative as a tangent line
  • Basic rules of differentiation
  • Derivative of trigonometric functions
  • Derivative of exponential and logarithmic functions
  • Chain rule
  • Implicit differentiation
  • Higher-order derivatives

1. Introduction to Derivatives

Derivatives are a fundamental concept in calculus, which is a branch of mathematics that deals with the study of change. The derivative represents the rate at which a function is changing at a particular point.

2. Derivative as a Rate of Change

Consider a function f(x) that represents the position of an object at time x. The rate of change of the position, or the velocity, can be found by taking the derivative of f(x) with respect to x.

3. Derivative as a Tangent Line

The derivative of a function at a point can be interpreted as the slope of the tangent line to the graph of the function at that point. The tangent line represents the best linear approximation of the function at that point.

4. Basic Rules of Differentiation

There are several basic rules for finding derivatives, which include:

  • Constant rule: The derivative of a constant function is zero.
  • Power rule: The derivative of x^n, where n is a constant, is nx^(n-1).
  • Sum and difference rule: The derivative of the sum or difference of two functions is the sum or difference of their derivatives.
  • Product rule: The derivative of the product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.
  • Quotient rule: The derivative of the quotient of two functions is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

5. Derivative of Trigonometric Functions

Some common derivatives of trigonometric functions are:

  • d(sin(x))/dx = cos(x)
  • d(cos(x))/dx = -sin(x)
  • d(tan(x))/dx = sec^2(x)
  • d(csc(x))/dx = -csc(x)cot(x)
  • d(sec(x))/dx = sec(x)tan(x)
  • d(cot(x))/dx = -csc^2(x)

6. 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] and differentiable on the open interval (a, b), and if f(a) = f(b), then there exists at least one point c in the open interval (a, b) such that the derivative f'(c) = 0. In other words, there must be a horizontal tangent line at some point between a and b.

Example of Rolle's Theorem

Let f(x) = x^3 - 3x^2 + 2x be a polynomial function defined on the interval [0, 3]. We can see that f(0) = 0 and f(3) = 0. Now, let's verify if f(x) satisfies the conditions of Rolle's theorem:

1. f(x) is continuous on the closed interval [0, 3]. (Polynomial functions are always continuous.)

2. f(x) is differentiable on the open interval (0, 3). (Polynomial functions are always differentiable.)

3. f(0) = f(3) = 0.

Since all three conditions are met, Rolle's theorem applies, and there must be at least one point c in the open interval (0, 3) such that f'(c) = 0.

Now, let's find f'(x) and solve for c:

f'(x) = 3x^2 - 6x + 2

Set f'(x) to 0 and solve for x:

3x^2 - 6x + 2 = 0

Divide the equation by 3:

x^2 - 2x + 2/3 = 0

Using the quadratic formula to solve for x:

x = [2 ± sqrt(2^2 - 4 * 1 * 2/3)] / 2

x ≈ 1.15 or x ≈ 0.55

Both values of x are within the open interval (0, 3), so there are two points c where f'(c) = 0, confirming Rolle's theorem.

7. L'Hôpital's Rule

L'Hôpital's Rule is a method for evaluating the limit of a quotient of functions when the limit of the numerator and denominator is either 0 or infinity. If f(x) and g(x) are differentiable functions, and the limit of f'(x)/g'(x) exists or is infinite as x approaches a, then:

lim (x -> a) [f(x) / g(x)] = lim (x -> a) [f'(x) / g'(x)]

provided that lim (x -> a) [f(x)] = lim (x -> a) [g(x)] = 0 or ±∞.

8. Chain Rule

The chain rule is used to find the derivative of a composite function. If a function is defined as the composition of two functions, say, h(x) = f(g(x)), then the derivative of h(x) with respect to x is given by:

h'(x) = f'(g(x)) * g'(x)

9. Implicit Differentiation

Implicit differentiation is a technique for finding the derivative of a function defined implicitly, i.e., when the function is not explicitly solved for one variable in terms of the other. To differentiate implicitly, differentiate both sides of the equation with respect to the independent variable, and then solve for the derivative of the dependent variable.

Example of Implicit Differentiation

Consider the equation of a circle with radius r centered at the origin:

x^2 + y^2 = r^2

We want to find the derivative dy/dx. Differentiate both sides of the equation with respect to x:

2x + 2y(dy/dx) = 0

Now, solve for dy/dx:

2y(dy/dx) = -2x

dy/dx = -2x / 2y

dy/dx = -x / y

The derivative dy/dx represents the slope of the tangent line to the circle at any point (x, y) on the circle.

10. Higher-Order Derivatives

Higher-order derivatives are the derivatives of a function taken multiple times with respect to the same variable. The second derivative, denoted as f''(x) or d^2f/dx^2, represents the rate of change of the first derivative. Similarly, the third derivative, denoted as f'''(x) or d^3f/dx^3, represents the rate of change of the second derivative, and so on.

Taylor Formula and Taylor Series

The Taylor formula is used to approximate a function by a polynomial. The Taylor series is an infinite series of terms that represent a function as a sum of its derivatives evaluated at a specific point, multiplied by the appropriate power of the independent variable.

The Taylor series of a function f(x) around the point x = a is given by:

f(x) ≈ f(a) + f'(a)(x - a) + f''(a)(x - a)^2 / 2! + f'''(a)(x - a)^3 / 3! + ...

More generally, the Taylor series can be expressed as:

f(x) ≈ ∑[f^(n)(a)(x - a)^n / n!] for n = 0 to ∞

where f^(n)(a) represents the nth derivative of f evaluated at x = a, and n! is the factorial of n (the product of all positive integers up to n).

A special case of the Taylor series is the Maclaurin series, where the expansion is around the point x = 0:

f(x) ≈ ∑[f^(n)(0)(x^n / n!] for n = 0 to ∞

Taylor series are useful for approximating functions and solving problems in areas such as physics, engineering, and economics. Note that not all functions can be represented by a Taylor series, and the convergence of the series depends on the properties of the function and the interval of interest.

Example: Taylor Series of e^x

Let's find the Taylor series of the function f(x) = e^x around the point x = 0 (i.e., a Maclaurin series).

First, we need to find the derivatives of f(x) at x = 0:

  • f(0) = e^0 = 1
  • f'(0) = e^0 = 1
  • f''(0) = e^0 = 1
  • f'''(0) = e^0 = 1
  • ...

Since all the derivatives of e^x evaluated at x = 0 are 1, the Maclaurin series becomes:

e^x ≈ 1 + x + x^2 / 2! + x^3 / 3! + x^4 / 4! + ...

Or in summation notation:

e^x ≈ ∑[(x^n) / n!] for n = 0 to ∞

Integrals

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  1. Exercise 1

    Evaluate the following definite integral:

    01 (x3 + x2 + x + 1) dx

    Click to show the solution

    Using the power rule, we get:

    ∫ (x3 + x2 + x + 1) dx = (1/4) x4 + (1/3) x3 + (1/2) x2 + x + C

    Substituting the limits of integration, we get:

    01 (x3 + x2 + x + 1) dx = (1/4) 14 + (1/3) 13 + (1/2) 12 + 1 - (1/4) 04 - (1/3) 03 - (1/2) 02 - 0

    =25/12

    Therefore, the value of the definite integral is 25/12.

  2. Exercise 2

    Evaluate the following definite integral:

    0π/3 x cos(3x) dx

    Click to show the solution

    We can use integration by parts with u = x and dv = cos(3x) dx to get:

    ∫ x cos(3x) dx = (1/3) x sin(3x) + (1/9) cos(3x) + C

    Substituting the limits of integration, we get:

    0π/3 x cos(3x) dx = (1/3) (π/3) sin(π) + (1/9) cos(π) - (1/3) (0) sin(0) - (1/9) cos(0)

    = -2/9

    Therefore, the value of the definite integral is -2/9.

  3. Exercise 3

    Evaluate the following definite integral:

    0π/2 sin3(x) cos(x) dx

    Click to show the solution

    We can use substitution with u = sin(x) to get:

    ∫ sin3(x) cos(x) dx = ∫ u3 du

    Using the power rule, we get:

    ∫ sin3(x) cos(x) dx = (1/4) sin4(x) + C

    Substituting the limits of integration, we get:

    0π/2 sin3(x) cos(x) dx = (1/4) sin4(π/2) - (1/4) sin4(0)

    = 1/4

    Therefore, the value of the definite integral is 1/4.

  4. Exercise 4

    Evaluate the following definite integral:

    01 ex cos(ex) dx

    Click to show the solution

    We can notice that if u = sin(ex) then du = excos(ex) dx to get:

    ∫ ex cos(ex) dx = sin(ex) + C

    Substituting the limits of integration, we get:

    01 ex cos(ex) dx = (sin(e) - sin(1))

  5. Exercise 5

    Evaluate the following definite integral:

    14 ln(x)/x dx

    Click to show the solution

    Using u = (ln(x))2 and du = 2ln(x)/x dx, we get:

    ∫ ln(x)/x dx = (ln(x))2/2 + c

    Substituting the limits of integration, we get:

    14 ln(x)/x dx = (ln(4))2/2 - (ln(1))2/2

    = (ln(4))2/2

    Therefore, the value of the definite integral is approximately 0.96.

  6. Exercise 6

    Evaluate the following definite integral:

    0π x2 sin(x) dx

    Click to show the solution

    We can use integration by parts with u = x2 and dv = sin(x) dx to get:

    ∫ x2 sin(x) dx = -x2 cos(x) + 2 ∫ x cos(x) dx

    Using integration by parts again with u = x and dv = cos(x) dx, we get:

    ∫ x2 sin(x) dx = -x2 cos(x) + 2 (x sin(x)) - 2 ∫ sin(x) dx

    = -x2 cos(x) + 2 x sin(x) + 2 cos(x)

    Substituting the limits of integration, we get:

    0π x2 sin(x) dx = -2 + (π)2

    Therefore, the value of the definite integral is -2 + (π)2.

  7. Exercise 7

    Evaluate the following definite integral:

    0π/2 x2 cos(x) dx

    Click to show the solution

    We can use integration by parts with u = x2 and dv = cos(x) dx to get:

    ∫ x2 cos(x) dx = x2 sin(x) - 2 ∫ x sin(x) dx

    Using integration by parts again with u = x and dv = sin(x) dx, we get:

    ∫ x2 cos(x) dx = x2 sin(x) - 2 (-x cos(x) + ∫ cos(x) dx)

    Using the power rule, we get:

    = x2 sin(x) + 2 x cos(x) - 2 sin(x)

    Substituting the limits of integration, we get:

    0π/2 x2 cos(x) dx = (π)2/4 - 2

    Therefore, the value of the definite integral is (π)2/4 - 2.

  8. Exercise 8

    Evaluate the following definite integral:

    01 ex2 dx

    Click to show the solution

    This integral does not have an elementary antiderivative, so we must use numerical methods to approximate the value of the integral. We can use Simpson's rule with n = 4 subintervals

    To use Simpson's rule, we need to divide the interval [0, 1] into n = 4 subintervals of equal width. The width of each subinterval is: h = (1-0)/4 = 0.25 The endpoints of the subintervals are: x0 = 0 x1 = h = 0.25 x2 = 2h = 0.5 x3 = 3h = 0.75 x4 = 4h = 1 Using Simpson's rule, the integral can be approximated as: ∫₀¹ e^(x^2) dx ≈ (h/3) [f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + f(x4)] where f(x) = e^(x^2) is the integrand. Substituting the values, we get: ∫₀¹ e^(x^2) dx ≈ (0.25/3) [f(0) + 4f(0.25) + 2f(0.5) + 4f(0.75) + f(1)] = (0.25/3) [1 + 4e^(0.25^2) + 2e^(0.5^2) + 4e^(0.75^2) + e^(1^2)] ≈ 1.4636 Therefore, the approximated value of the integral using Simpson's rule with n = 4 subintervals is approximately 1.4636.
  9. Exercise 9

    Evaluate the following definite integral:

    0π/2 sin(x)cos(x) esin(x) dx

    Click to show the solution

    We can use integration by parts with u = sin(x) and dv = cos(x)esin(x) dx to get:

    = ∫ sin(x) cos(x) esin(x) dx = sin(x) esin(x) - ∫ cos(x)esin(x) dx

    sin(x) esin(x) - esin(x) + c

    Substituting the limits of integration, we get:

    0π/2 sin(x) cos(x) esin(x) dx = 1

    Therefore, the value of the definite integralis approximately 1.

  10. Exercise 10

    Evaluate the following definite integral:

    0π/4 tan(x) dx

    Click to show the solution

    We can use substitution with u = cos(x) to get:

    ∫ tan(x) dx = ∫ (sin(x)/cos(x)) dx = -∫ (1/u) du = -ln|cos(x)| + C

    Substituting the limits of integration, we get:

    0π/4 tan(x) dx = -ln|cos(π/4)| + ln|cos(0)|

    = ln(2)/2

    Therefore, the value of the definite integral is ln(2)/2.

Lesson on Integrals: Finite and Undefinite

Introduction to Integrals

In calculus, Integrals are used to find the area under the curve of a function, as well as to evaluate other quantities such as volume, displacement, and work. There are two types of integrals: Definite Integrals and Indefinite Integrals.

Definite Integrals

Definite Integrals are used to find the area under the curve of a function between two limits, a and b. The notation used for a definite integral is:

ab f(x) dx

where f(x) is the function being integrated and dx is the differential of the variable of integration. The symbol ∫ is called the integral sign.

To evaluate a definite integral, we use the following formula:

ab f(x) dx = F(b) - F(a)

where F(x) is the antiderivative or indefinite integral of f(x).

Indefinite Integrals

Indefinite Integrals are used to find a function whose derivative is equal to the function being integrated. The notation used for an indefinite integral is:

∫ f(x) dx

where f(x) is the function being integrated and dx is the differential of the variable of integration.

To evaluate an indefinite integral, we use the following formula:

∫ f(x) dx = F(x) + C

where F(x) is the antiderivative of f(x) and C is the constant of integration.

Properties of Integrals

There are several properties of Integrals that we can use to simplify calculations. These include:

Linearity

The integral of a sum of functions is equal to the sum of the integrals of the individual functions. That is,

∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx

Power Rule

The integral of a power function is given by:

∫ xn dx = (xn+1)/(n+1) + C

where n is any real number except for -1.

Constant Multiple Rule

The integral of a constant multiple of a function is equal to the constant multiplied by the integral of the function. That is,

∫ k f(x) dx = k ∫ f(x) dx

where k is a constant.

Sum Rule

The integral of a sum of functions is equal to the sum of the integrals of the individual functions. That is,

∫ [f(x) + g(x)] dx = ∫ f(x) dx + ∫ g(x) dx

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus establishes the relationship between differentiation and integration. It states that:

The First Fundamental Theorem of Calculus

If$f(x)$ is a continuous function on the interval [a,b] and F(x) is an antiderivative of $f(x)$, then the definite integral of $f(x)$ from $a$ to $b$ is given by:

ab f(x) dx = F(b) - F(a)

The Second Fundamental Theorem of Calculus

If $f(x)$ is a continuous function on the interval [a,b], and F(x) is an antiderivative of $f(x)$, then the derivative of the definite integral of $f(x)$ with respect to $x$ is equal to $f(x)$. That is,

d/dx ∫ax f(t) dt = f(x)

Techniques of Integration

There are several techniques of integration that we can use to evaluate integrals. Some of the common techniques are:

Substitution

Substitution is a technique of integration that involves substituting a function in the integral with another function. This technique is useful when the integral involves a composite function.

Integration by Parts

Integration by Parts is a technique of integration that involves breaking down an integral into two parts and integrating each part separately.

Partial Fractions

Partial Fractions is a technique of integration that involves breaking down a rational function into simpler fractions and integrating each fraction separately.

Trigonometric Substitution

Trigonometric Substitution is a technique of integration that involves substituting a trigonometric function in the integral with another function. This technique is useful when the integral involves a radical function.

Applications of Integrals

Integrals have several applications in calculus, including:

Area Under a Curve

Integrals can be used to find the area under the curve of a function between two limits, a and b.

Volumes of Solids with Known Cross-Sections

Integrals can be used to find the volume of a solid that has a known cross-section.

Length of Curves

Integrals can be used to find the length of a curve between two points.

Conclusion

In conclusion, Integrals are important tools used in Calculus to find areas, volumes, and other quantities. There are two types of Integrals: Definite and Indefinite. Properties of Integrals, such as Linearity, Power Rule, Constant Multiple Rule, and Sum Rule can be used to simplify calculations. The Fundamental Theorem of Calculus relates differentiation and integration, while Techniques of Integration such as Substitution, Integration by Parts, Partial Fractions, and Trigonometric Substitution can be used to evaluate integrals. Finally, Integrals have several applications in calculus, including finding the area under a curve, volumes of solids with known cross-sections, and the length of curves.

Advanced Integration Techniques

In addition to the above topics, students may also learn advanced integration techniques, such as Improper Integrals and Differential Equations, in higher-level calculus courses.

Improper Integrals

Improper Integrals are integrals where one or both of the limits of integration are infinite or the function being integrated is unbounded at some point in the interval of integration. To evaluate improper integrals, we take the limit of the definite integral as one or both of the limits of integration go to infinity or approach the point of unboundedness.

Differential Equations

A Differential Equation is an equation that involves the derivative of a function. The solution to a differential equation involves finding a function that satisfies the equation. Integrals are used in the process of solving certain types of differential equations, such as Separable Differential Equations and Homogeneous Differential Equations.

Examples

Let's look at some examples to apply the concepts we have learned:

Example 1:

Find the definite integral of the function f(x) = x3 + 2x2 - 5x + 2 between the limits x = 1 and x = 4.

Solution:

We can find the antiderivative of f(x) as follows:

F(x) = ∫ f(x) dx = (x4/4) + (2x3/3) - (5x2/2) + 2x + C

where C is the constant of integration.

Substituting the limits of integration, we get:

14 (x3 + 2x2 - 5x + 2) dx = F(4) - F(1)

= [(44/4) + (2(43)/3) - (5(42)/2) + 2(4)] - [(14/4) + (2(13)/3) - (5(12)/2) + 2(1)]

= 67/3

Therefore, the definite integral of f(x) between x = 1 and x = 4 is 67/3.

Example 2:

Find the indefinite integral of the function f(x) = 3x2 + 2x + 1.

Solution:

We can find the antiderivative of f(x) as follows:

F(x) = ∫ f(x) dx = x3 + x2 + x + C

where C is the constant of integration.

Therefore, the indefinite integral of f(x) is x3 + x2 + x + C.

Conclusion

In this lesson, we learned about the two types of Integrals: Definite and Indefinite, and how to evaluate them. We also looked at some of the properties of Integrals, such as Linearity, Power Rule, Constant Multiple Rule, and Sum Rule, and how they can be used to simplify calculations. We learned about the Fundamental Theorem of Calculus, which relates differentiation and integration, and Techniques of Integration, such as Substitution, Integration by Parts, Partial Fractions, and Trigonometric Substitution, which can be used to evaluate integrals. Finally, we looked at some of the applications of Integrals, including finding the area under a curve, volumes of solids with known cross-sections, and the length of curves. We also briefly discussed some of the advanced integration techniques that students may learn in higher-level calculus courses, such as Improper Integrals and Differential Equations. By applying these concepts and techniques to examples, we were able to see how they can be used in practice to evaluate Integrals and solve problems in calculus.

Differential Equations

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  1. Exercise 1

    Solve the separable differential equation: $\frac{dy}{dx} = \frac{y}{x}$
    Click to show the solution $\int \frac{1}{y} dy = \int \frac{1}{x} dx$
    $\ln |y| = \ln |x| + C = \ln |x| + \ln(\lambda) $
    $y = \lambda x$
  2. Exercise 2

    Solve the first-order linear differential equation: $\frac{dy}{dx} - 3y = e^{2x}$
    Click to show the solution Integrating factor: $I(x) = e^{-3x}$
    $e^{-3x} \frac{dy}{dx} - 3e^{-3x} y = e^{-x}$
    $\frac{d}{dx} (ye^{-3x}) = e^{-x}$
    $ye^{-3x} = -e^{-x} + C$
    $y = -e^{2x} + Ce^{3x}$
  3. Exercise 3

    Solve the homogeneous differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$
    Click to show the solution Using the substitution $y = vx$:
    $v + x \frac{dv}{dx} = \frac{x^2 + x^2 v^2}{x^2 v}$
    $v + x \frac{dv}{dx} = \frac{1}{v} + v$
    $x \frac{dv}{dx} = \frac{1}{v}$
    $\int v dv = \int \frac{1}{x} dx$
    $\frac{1}{2}v^2 = \ln |x| + C$
    $v^2 = 2\ln |x| + 2C$
    $y^2 = 2x^2 \ln |x| + 2Cx^2$
  4. Exercise 4

    Solve the differential equation: $(2x + y^2) dx + (2y^2 + 2yx) dy = 0$
    Click to show the solution To solve the given differential equation, $(2x + y^2) dx + (2y^2 + 2yx) dy = 0$, we'll first check if it's an exact differential equation.
    A first-order differential equation of the form $M(x, y) dx + N(x, y) dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$.
    Here, $M(x, y) = 2x + y^2$ and $N(x, y) = 2y^2 + 2yx$. Let's compute the partial derivatives:
    $\frac{\partial M}{\partial y} = \frac{\partial (2x + y^2)}{\partial y} = 2y$
    $\frac{\partial N}{\partial x} = \frac{\partial (2y^2 + 2yx)}{\partial x} = 2y$
    Since $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$, the given differential equation is exact.
    Now we need to find a function $F(x, y)$ such that $\frac{\partial F}{\partial x} = M(x, y)$ and $\frac{\partial F}{\partial y} = N(x, y)$.
    We'll first integrate $M(x, y)$ with respect to $x$:
    $F(x, y) = \int (2x + y^2) dx = x^2 + xy^2 + g(y)$ Here, $g(y)$ is an unknown function of $y$. To find $g(y)$, we differentiate $F(x, y)$ with respect to $y$ and compare it with $N(x, y)$:
    $\frac{\partial F}{\partial y} = \frac{\partial (x^2 + xy^2 + g(y))}{\partial y} = 2xy + g'(y)$
    Comparing this with $N(x, y) = 2y^2 + 2yx$, we have:
    $2xy + g'(y) = 2y^2 + 2yx$
    Thus, $g'(y) = 2y^2$. Integrating $g'(y)$ with respect to $y$, we get: $g(y) = \int 2y^2 dy = \frac{2}{3}y^3 + C$
    Now we can write the complete solution for $F(x, y)$: $F(x, y) = x^2 + xy^2 + \frac{2}{3}y^3 + C$
    Since the exact differential equation is given by $dF = 0$, the general solution is:
    $F(x, y) = x^2 + xy^2 + \frac{2}{3}y^3 + C$
  5. Exercise 5

    Solve the differential equation: $\frac{dy}{dx} + 2y = e^x$
    Click to show the solution Integrating factor: $I(x) = e^{\int 2 dx} = e^{2x}$
    $e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x + x}$
    $\frac{d}{dx} (ye^{2x}) = e^{3x}$
    $ye^{2x} = \int e^{3x} dx + C$
    $y = e^{-2x}[(1/3)e^{3x} + C] = (1/3)e^x + Ce^{-2x}$
  6. Exercise 6

    Solve the separable differential equation: $\frac{dy}{dx} = \frac{x^2}{y^2}$
    Click to show the solution $\int y^2 dy = \int x^2 dx$
    $\frac{1}{ 3}y^3 = \frac{1}{3}x^3 + C$
    $y^3 = x^3 + 3C$
    $y = \sqrt[3]{x^3 + 3C}$
  7. Exercise 7

    Solve the first-order linear differential equation: $\frac{dy}{dx} + y \cot x = \sin x$
    Click to show the solution Integrating factor: $I(x) = e^{\int \cot x dx} = e^{\ln |\sin x|} = \sin x$
    $\sin x \frac{dy}{dx} + y \cos x = \sin^2 x$
    $\frac{d}{dx} (y \sin x) = \sin^2 x = (1/2)[1 - cos(2x)]$
    $y \sin x = (1/2)[x - (1/2)sin(2x)]+ C$
    $y = \frac{(1/2)[x - (1/2)sin(2x)]+ C}{\sin x}$
  8. Exercise 8

    Solve the homogeneous differential equation: $\frac{dy}{dx} = \frac{2y^2 - 2yx}{xy - x^2 }$
    Click to show the solution Using the substitution $y = vx$:
    $v + x \frac{dv}{dx} = \frac{2v^2 x^2 - 2vx^2}{v x^2 - x^2}$
    $v + x \frac{dv}{dx} = \frac{2v^2 - 2v}{v - 1}$
    $x \frac{dv}{dx} = \frac{2v^2 - 2v}{v - 1} - v = v$
    $\int \frac{1}{v}dv = \int \frac{1}{x} dx$
    $ v = x + C$ and $ y = x(x + C)$ or $ v = -x + C$ and $ y = x(-x + C)$
  9. Exercise 9

    Solve the exact differential equation: $(ye^{xy} + 1) dx + xe^{xy} dy = 0$
    Click to show the solution To solve the given exact differential equation $(ye^{xy} + 1) dx + xe^{xy} dy = 0$, we first need to check if it is indeed exact. An exact differential equation has the form $M(x, y) dx + N(x, y) dy = 0$, where $M$ and $N$ are functions of $x$ and $y$. For it to be exact, it must satisfy the following condition: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ In our case, $M(x, y) = ye^{xy} + 1$ and $N(x, y) = xe^{xy}$. Let's compute the partial derivatives: $\frac{\partial M}{\partial y} = e^{xy} + xye^{xy}$ $\frac{\partial N}{\partial x} = e^{xy} + xye^{xy}$ We can see that the condition is satisfied: $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$ Since the equation is exact, we can find a potential function $F(x, y)$ such that: $\frac{\partial F}{\partial x} = M(x, y) = ye^{xy} + 1$ $\frac{\partial F}{\partial y} = N(x, y) = xe^{xy}$ Let's first integrate the first equation with respect to $x$: $F(x, y) = \int (ye^{xy} + 1) dx = e^{xy} + x + g(y)$ The function $g(y)$ is an unknown function of $y$ that may arise due to the integration. Now, let's differentiate $F(x, y)$ with respect to $y$: $\frac{\partial F}{\partial y} = xe^{xy} + g'(y)$ Since we already know that $\frac{\partial F}{\partial y} = N(x, y) = xe^{xy}$, we can equate the two expressions: $xe^{xy} + g'(y) = xe^{xy}$ This implies that $g'(y) = 0$. Integrating $g'(y)$ with respect to $y$ gives us: $g(y) = C$, where $C$ is a constant. So, the potential function $F(x, y)$ is given by: $F(x, y) = e^{xy} + x + C$ The general solution to the exact differential equation is: $F(x, y) = e^{xy} + x + C = K$, where $K$ is an arbitrary constant.
  10. Exercise 10

    Solve the differential equation: $\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$
    Click to show the solution Using the substitution $y = vx$:
    $v + x \frac{dv}{dx} = \frac{x^2 + v^2 x^2}{vx^2}$
    $v + x \frac{dv}{dx} = \frac{1 + v^2}{v}$
    $x \frac{dv}{dx} = \frac{1 + v^2}{v} - v = \frac{1}{v}$
    $\int v dv = \int \frac{1}{x} dx$
    $(1/2)v^2 = ln|x| + C$ and $v = \sqrt{2ln|x| + 2C}$ so $y = x \sqrt{2ln|x| + 2C}$
    or $v = -\sqrt{2ln|x| + 2C}$ so $y = -x \sqrt{2ln|x| + 2C}$
Lesson: Differential Equations in College Calculus I
Introduction
Definition of a Differential Equation
Separable Differential Equations
3.1. Solving Separable Differential Equations
3.2. Example
First-Order Linear Differential Equations
4.1. Solving First-Order Linear Differential Equations
4.2. Example
Homogeneous Differential Equations
5.1. Solving Homogeneous Differential Equations
5.2. Example
Exact Differential Equations
6.1. Conditions for Exactness
6.2. Solving Exact Differential Equations
6.3. Example
Integrating Factors
7.1. Finding Integrating Factors
7.2. Example
Applications of Differential Equations
8.1. Population Growth and Decay
8.2. Mixing Problems
8.3. Newton's Law of Cooling
Conclusion
Introduction
Differential equations are mathematical equations that relate a function to its derivatives. They are used to describe various phenomena in physics, engineering, biology, and other disciplines. In College Calculus I, we mainly deal with first-order differential equations, which involve the first derivative of the function.
Definition of a Differential Equation
A first-order differential equation is an equation of the form: dy/dx = f(x, y) where x and y are variables, and f(x, y) is a given function.
Separable Differential Equations
A first-order differential equation is called separable if it can be written in the form: dy/dx = g(x)h(y)
To solve a separable differential equation, we can rewrite it as: (1/h(y))dy/dx = g(x)
3.1. Solving Separable Differential Equations
Integrating both sides with respect to x yields: ∫(1/h(y))dy/dx dx = ∫g(x) dx
Using substitution, we get: ∫(1/h(y)) dy = ∫g(x) dx
After integrating both sides and solving for y, we can find the solution to the differential equation.
3.2. Example
Consider the separable differential equation: dy/dx = 2xy
Solution:
The given equation can be written as (1/y)dy/dx = 2x. Integrating both sides, we get: ∫(1/y) dy = ∫2x dx
ln|y| = x^2 + C
y = Ce^(x^2), where C is the integration constant.
First-Order Linear Differential Equations
A first-order linear differential equation is an equation of the form: dy/dx + P(x)y = Q(x) where P(x) and Q(x) are continuous functions of x.
4.1. Solving First-Order Linear Differential Equations
To solve this type of equation, we first compute an integrating factor, I(x), given by: I(x) = e^(∫ P(x) dx)
Next, we multiply both sides of the equation by I(x): I(x)dy/dx + I(x)P(x)y = I(x)Q(x)
This transformed equation is now exact, and we can integrate both sides with respect to x and solve for y.
4.2. Example
Consider the first-order linear differential equation: dy/dx + 2xy = x
Solution:
Here, P(x) = 2x and Q(x) = x. The integrating factor, I(x), is given by: I(x) = e^(∫ 2x dx) = e^(x^2)
Multiplying the equation by I(x): e^(x^2) dy/dx + 2x e^(x^2) y = x e^(x^2)
Now, integrating both sides with respect to x: ∫(e^(x^2) dy/dx + 2x e^(x^2) y) dx = ∫(x e^(x^2)) dx
Using substitution, let u = x^2, then du = 2x dx: ∫(e^(x^2) dy/dx + 2x e^(x^2) y) dx = ∫(x e^(u)) (1/2) du
Integrating the right-hand side, we get: y e^(x^2) = (1/2) e^(x^2) + C y = (1/2) + C e^(-x^2)
Homogeneous Differential Equations
A first-order differential equation is called homogeneous if it can be written in the form: dy/dx = F(y/x)
To solve a homogeneous differential equation, we can use the substitution y = vx, where v = y/x.
5.1. Solving Homogeneous Differential Equations
Substituting y = vx and differentiating y with respect to x, we get: dy/dx = v + x dv/dx
Replacing dy/dx and y in the original equation: v + x dv/dx = F(v)
Now, separate the variables and integrate both sides to find the solution.
5.2. Example
Consider the homogeneous differential equation: dy/dx = (x + y)/(x - y)
Solution:
Rewrite the given equation in the form dy/dx = F(y/x): dy/dx = (1 + y/x)/(1 - y/x)
Using the substitution y = vx: v + x dv/dx = (1 + v)/(1 - v)
Separating the variables and integrating:
∫(1 - v) dv = ∫(1/x) dx
(1/2) v^2 - v = ln|x| + C
v^2 - 2v = 2 ln|x| + 2C
Now, substitute back y/x for v: (y/x)^2 - 2(y/x) = 2 ln|x| + 2C
Exact Differential Equations
A first-order differential equation is called exact if it can be written in the form: M(x, y) dx + N(x, y) dy = 0
where M and N are functions of x and y, and the partial derivatives satisfy: ∂M/∂y = ∂N/∂x
6.1. Conditions for Exactness
If the given differential equation satisfies the condition ∂M/∂y = ∂N/∂x, then it's an exact differential equation.
6.2. Solving Exact Differential Equations
To solve an exact differential equation, find a function ψ(x, y) such that: ∂ψ/∂x = M(x, y) and ∂ψ/∂y = N(x, y)
Integrating the first equation with respect to x, and the second equation with respect to y, and comparing the results, we getthe solution ψ(x, y) = C, where C is a constant.
6.3. Example
Consider the exact differential equation: (2xy + y^2) dx + (x^2 + 2xy) dy = 0
Solution:
Here, M(x, y) = 2xy + y^2 and N(x, y) = x^2 + 2xy. We need to check if the equation is exact:
∂M/∂y = 2x + 2y and ∂N/∂x = 2x + 2y
Since ∂M/∂y = ∂N/∂x, the equation is exact. Now, we need to find a function ψ(x, y) such that:
∂ψ/∂x = 2xy + y^2 and ∂ψ/∂y = x^2 + 2xy
Integrating the first equation with respect to x, we get: ψ(x, y) = x^2y + xy^2 + h(y)
Now, differentiating ψ(x, y) with respect to y, and comparing it with N(x, y): x^2 + 2xy + h'(y) = x^2 + 2xy
From the comparison, we can deduce that h'(y) = 0, so h(y) = constant. Therefore, the solution is: ψ(x, y) = x^2y + xy^2 = C
Integrating Factors
In some cases, a first-order differential equation may not be exact, but can be made exact by multiplying it by an integrating factor µ(x, y).
7.1. Finding Integrating Factors
To find an integrating factor for a non-exact differential equation, we need to satisfy the condition:
(∂(Mµ)/∂y) - (∂(Nµ)/∂x) = 0
Solving for µ(x, y) and multiplying the original equation by µ(x, y) will make it exact, and then it can be solved using the method for exact differential equations.
7.2. Example
Consider the differential equation: (3x^2 + 2xy) dx + (2x^2 - y^2) dy = 0
Solution:
Here, M(x, y) = 3x^2 + 2xy and N(x, y) = 2x^2 - y^2. Since ∂M/∂y ≠ ∂N/∂x, the equation is not exact. We will attempt to find an integrating factor µ(y) dependent only on y: (∂(Mµ)/∂y) - (∂(Nµ)/∂x) = 0
Solving for µ(y), we find that µ(y) = 1/y. Multiplying the original equation by µ(y), we get: (3x^2/y + 2x) dx + (2x^2/y - y) dy = 0
Now, the new equation is exact, and we can solve it using the method for exact differential equations.
Applications of Differential Equations
Differential equations have many applications in various fields, including:
8.1. Population Growth and Decay
The logistic differential equation is used to model population growth and decay: dP/dt = kP(1 - P/M)
where P(t) is the population at time t, k is the growth rate, and M is the carrying capacity.
8.2. Mixing Problems
Differential equations can be used to model theconcentration of a substance in a mixing tank, taking into account the flow rates of the incoming and outgoing solutions.
8.3. Newton's Law of Cooling
Newton's Law of Cooling states that the rate of cooling of an object is proportional to the difference in temperature between the object and its surroundings. This can be modeled using a first-order differential equation: dT/dt = -k(T - T_s)
where T(t) is the temperature of the object at time t, T_s is the surrounding temperature, and k is a constant representing the cooling rate.
Conclusion
In College Calculus I, we focus on solving first-order differential equations, specifically separable, first-order linear, homogeneous, and exact differential equations, as well as using integrating factors. Understanding these methods provides a foundation for solving more complex differential equations in advanced courses and applying them to real-world problems in various fields.