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Notes on

Calculus II

Integral Calculus

Miguel A. Lerma

• November 22,
• Introduction
• Chapter 1. Integrals
• 1.1. Areas and Distances. The Definite Integral
• 1.2. The Evaluation Theorem
• 1.3. The Fundamental Theorem of Calculus
• 1.4. The Substitution Rule
• 1.5. Integration by Parts
• 1.6. Trigonometric Integrals and Trigonometric Substitutions
• 1.7. Partial Fractions
• 1.8. Integration using Tables and CAS
• 1.9. Numerical Integration
• 1.10. Improper Integrals
• Chapter 2. Applications of Integration
• 2.1. More about Areas
• 2.2. Volumes
• 2.3. Arc Length, Parametric Curves
• 2.4. Average Value of a Function (Mean Value Theorem)
• 2.5. Applications to Physics and Engineering
• 2.6. Probability
• Chapter 3. Differential Equations
• 3.1. Differential Equations and Separable Equations
• 3.2. Directional Fields and Euler’s Method
• 3.3. Exponential Growth and Decay
• Chapter 4. Infinite Sequences and Series
• 4.1. Sequences
• 4.2. Series
• 4.3. The Integral and Comparison Tests
• 4.4. Other Convergence Tests
• 4.5. Power Series
• 4.6. Representation of Functions as Power Series
• 4.7. Taylor and MacLaurin Series
• CONTENTS
• 4.8. Applications of Taylor Polynomials
• Appendix A. Hyperbolic Functions
• A.1. Hyperbolic Functions
• Appendix B. Various Formulas
• B.1. Summation Formulas
• Appendix C. Table of Integrals

Introduction

These notes are intended to be a summary of the main ideas in course MATH 214-2: Integral Calculus. I may keep working on this document as the course goes on, so these notes will not be completely finished until the end of the quarter.

The textbook for this course is Stewart: Calculus, Concepts and Contexts (2th ed.), Brooks/Cole. With few exceptions I will follow the notation in the book.

If you find any typos or errors, or you have any suggestions, please, do not hesitate to let me know. You may email me, or use the web form for feedback on the web pages for the course:

http://www.math.northwestern.edu/~mlerma/courses/math214-2-02f/

Miguel A. Lerma [email protected] Northwestern University Fall 2002

5

1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 7

The area Si of the strip between xi− 1 and xi can be approximated as the area of the rectangle of width ∆x and height f (x∗ i ), where x∗ i is a sample point in the interval [xi, xi+1]. So the total area under the curve is approximately the sum

∑^ n

i=

f (x∗ i ) ∆x = f (x∗ 1 ) ∆x + f (x∗ 2 ) ∆x + · · · + f (x∗ n) ∆x.

This expression is called a Riemann Sum.

The estimation is better the thiner the strips are, and we can iden- tify the exact area under the graph of f with the limit:

A = lim n→∞

∑^ n

i=

f (x∗ i ) ∆x

As long as f is continuous the value of the limit is independent of the sample points x∗ i used.

That limit is represented

∫ (^) b a f^ (x)^ dx, and is called^ definite integral of f from a to b: ∫ (^) b

a

f (x) dx = lim n→∞

∑^ n

i=

f (x∗ i ) ∆x

The symbols at the left historically were intended to mean an infinite sum, represented by a long “S” (the integral symbol

), of infinitely small amounts f (x) dx. The symbol dx was interpreted as the length of an “infinitesimal” interval, sort of what ∆x becomes for infinite n. This interpretation was later abandoned due to the difficulty of reasoning with infinitesimals, but we keep the notation.

Remark : Note that in intervals where f (x) is negative the graph of y = f (x) lies below the x-axis and the definite integral takes a negative value. In general a definite integral gives the net area between the graph of y = f (x) and the x-axis, i.e., the sum of the areas of the regions where y = f (x) is above the x-axis minus the sum of the areas of the regions where y = f (x) is below the x-axis.

1.1.2. Evaluating Integrals. We will soon study simple and ef- ficient methods to evaluate integrals, but here we will look at how to evaluate integrals directly from the definition.

Example: Find the value of the definite integral

0 x

(^2) dx from its

definition in terms of Riemann sums.

1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 8

Answer : We divide the interval [0, 1] into n equal parts, so xi = i/n and ∆x = 1/n. Next we must choose some point x∗ i in each subinterval [xi− 1 , xi]. Here we will use the right endpoint of the interval x∗ i = i/n. Hence the Riemann sum associated to this partition is:

∑^ n

i=

i n

1 /n =

n^3

∑^ n

i=

i^2 =

n^3

2 n^3 + 3n^2 + n 6

2 + 3/n + 2/n^2 6

So: ∫ (^1)

0

x^2 dx = lim n→∞

2 + 3/n + 2/n^2 6

In order to check that the result does not depend on the sample points used, let’s redo the computation using now the left endpoint of each subinterval:

∑^ n

i=

i − 1 n

1 /n =

n^3

∑^ n

i=

(i−1)^2 =

n^3

2 n^3 − 3 n^2 + n 6

2 − 3 /n + 2/n^2 6

So: ∫ (^1)

0

x^2 dx = lim n→∞

2 − 3 /n + 2/n^2 6

1.1.3. The Midpoint Rule. The Midpoint Rule consists of com- puting Riemann sums using xi = (xi− 1 + xi)/2 = midpoint of each interval as sample point. This yields the following approximation for the value of a definite integral:

∫ (^) b

a

f (x) dx ≈

∑^ n

i=

f (xi) ∆x = ∆x [f (x 1 ) + f (x 2 ) + · · · + f (xn)].

Example: Use the Midpoint Rule with n = 5 to approximate

0 x

(^2) x.

Answer : The subintervals are [0, 0 .2], [0. 2 , 0 .4], [0. 4 , 0 .6], [0. 6 , 0 .8], [0. 8 , 1], the midpoints are 0. 1 , 0. 3 , 0. 5 , 0. 7 , 0 .9, and ∆x = 1/5, so

∫ (^1)

0

x^2 dx ≈

[

]

which agrees up to the second decimal place with the actual value 1/3.

1.1. AREAS AND DISTANCES. THE DEFINITE INTEGRAL 10

(a)

∫ (^) c

a

f (x) dx +

∫ (^) b

c

f (x) dx =

∫ (^) b

a

f (x) dx.

(b)

∫ (^) a

b

f (x) dx = −

∫ (^) b

a

f (x) dx,

(c)

∫ (^) a

a

f (x) dx = 0.

(4) Comparison:

(a) f (x) ≥ 0 ⇒

∫ (^) b

a

f (x) dx ≥ 0.

(b) f (x) ≥ g(x) ⇒

∫ (^) b

a

f (x) dx ≥

∫ (^) b

a

g(x) dx.

(c) m ≤ f (x) ≤ M ⇒ m (b − a) ≤

∫ (^) b

a

f (x) dx ≤ M (b − a).

1.2. THE EVALUATION THEOREM 11

1.2. The Evaluation Theorem

1.2.1. The Evaluation Theorem. If f is a continuous function and F is an antiderivative of f , i.e., F ′(x) = f (x), then ∫ (^) b

a

f (x) dx = F (b) − F (a).

Example: Find

0 x

(^2) dx using the evaluation theorem.

Answer : An antiderivative of x^2 is x^3 /3, hence: ∫ (^1)

0

x^2 dx =

[

x^3 3

] 1

0

1.2.2. Indefinite Integrals. If F is an antiderivative of a function f , i.e., F ′(x) = f (x), then for any constant C, F (x) + C is another antiderivative of f (x). The family of all antiderivatives of f is called indefinite integral of f and represented: ∫ f (x) dx = F (x) + C.

Example:

x^2 dx =

x^3 3

+ C.

1.2.3. Table of Indefinite Integrals. We can make an integral table just by reversing a table of derivatives.

xn^ dx =

xn+ n + 1

• C (n 6 = −1).

(2)

x

dx = ln |x| + C.

(3)

ex^ dx = ex^ + C.

(4)

ax^ dx =

ax ln a

+ C.

sin x dx = − cos x + C.

(6)

cos x dx = sin x + C.

(7)

sec^2 x dx = tan x + C.

1.2. THE EVALUATION THEOREM 13

Answer : The displacement is ∫ (^4)

1

(t^2 − t − 6) dx =

[

t^3 3

t^2 2

− 6 t

] 4

1

=

In order to find the total distance traveled we need to separate the intervals in which the velocity takes values of different signs. Those intervals are separated by points at which v(t) = 0, i.e., t^2 −t−6 = 0 ⇒ t = −2 and t = 3. Since we are interested only in what happens in [1, 4] we only need to look at the intervals [1, 3] and [3, 4]. Since v(1) = −6, the velocity is negative in [1, 3], and since v(4) = 6, the velocity is positive in [3, 4]. Hence: ∫ (^4)

1

|v(t)| dt =

1

[−v(t)] dt +

3

v(t) dt

1

(t^2 − t − 6) dt +

3

(t^2 − t − 6) dt

[

t^3 3

t^2 2

• 6t

] 3

1

[

t^3 3

t^2 2

− 6 t

] 4

3

1.3. THE FUNDAMENTAL THEOREM OF CALCULUS 14

1.3. The Fundamental Theorem of Calculus

1.3.1. The Fundamental Theorem of Calculus. The Funda- mental Theorem of Calculus (FTC) connects the two branches of cal- culus: differential calculus and integral calculus. It says the following:

Suppose f is continuous on [a, b]. Then:

(1) The function

g(x) =

∫ (^) x

a

f (t) dt

is an antiderivative of f , i.e., g′(x) = f (x). (2) (Evaluation Theorem) If F is an antiderivative of f , i.e. F ′(x) = f (x), then ∫ (^) b

a

f (x) dx = F (b) − F (a).

The two parts of the theorem can be rewritten like this:

d dx

∫ (^) x

a

f (t) dt = f (x).

∫ (^) b

a

F ′(x) dx = F (b) − F (a).

So the theorem states that integration and differentiation are in- verse operations, i.e., the derivative of an integral of a function yields the original function, and the integral of a derivative also yields the function originally differentiated (up to a constant).

Example: Find

d dx

∫ (^) x 2

0

t^3 dt.

Answer : We solve this problem in two ways. First directly:

g(x) =

∫ (^) x 2

0

t^3 dt =

[

t^4 4

]x^2

0

(x^2 )^4 4

x^8 4

hence

g′(x) =

8 x^7 4

= 2 x^7.

1.4. THE SUBSTITUTION RULE 16

1.4. The Substitution Rule

1.4.1. The Substitution Rule. The substitution rule is a trick for evaluating integrals. It is based on the following identity between differentials (where u is a function of x):

du = u′^ dx.

Hence we can write:

∫ f (u) u′^ dx =

f (u) du

or using a slightly different notation:

∫ f (g(x)) g′(x) dx =

f (u) du

where u = g(x).

Example: Find

1 + x^2 2 x dx.

Answer : Using the substitution u = 1 + x^2 we get ∫ (^) √ 1 + x^2 2 x dx =

u u′^ dx

u du = 23 u^3 /^2 + C

= 23 (1 + x^2 )^3 /^2 + C.

Most of the time the only problem in using this method of integra- tion is finding the right substitution.

Example: Find

cos 2x dx.

Answer : We want to write the integral as

cos u du, so cos u = cos 2x ⇒ u = 2x, u′^ = 2. Since we do not see any factor 2 inside the

1.4. THE SUBSTITUTION RULE 17

integral we write it, taking care of dividing by 2 outside the integral:

∫ cos 2x dx =

cos 2x 2 dx

cos u u′^ dx

cos u du

sin u + C

(always remember to undo the substitution)

sin 2x + C.

In general we need to look at the integrand as a function of some expression (which we will later identify with u) multiplied by the de- rivative of that expression.

Example: Find

e−x

2 x dx.

Answer : We see that x is “almost”, the derivative of −x^2 , so we use the substitution u = −x^2 , u′^ = − 2 x, hence in order to get u′^ inside the integral we do the following:

∫ e−x

2 x dx = −

e−x

2

︸︷︷︸ eu

(− 2 x) dx ︸ ︷︷ ︸ du

= −

eu^ du = −^12 eu^ + C = −^12 e−x

2

• C.

Sometimes the substitution is hard to see until we make some in- genious transformation in the integrand.

Example: Find

tan x dx.

1.4. THE SUBSTITUTION RULE 19

Since we do not know yet how to integrate cos^2 t we leave it like this and will be back to it later (after we study integrals of trigonometric functions).

1.4.3. The Substitution Rule for Definite Integrals. When computing a definite integral using the substitution rule there are two possibilities:

(1) Compute the definite integral first, then use the evaluation theorem: ∫ f (u) u′^ dx = F (x) ; ∫ (^) b

a

f (u) u′^ dx = F (b) − F (a).

(2) Use the substitution rule for definite integrals: ∫ (^) b

a

f (u) u′^ dx =

∫ (^) u(b)

u(a)

f (u) du.

The advantage of the second method is that we do not need to undo the substitution.

Example: Find

0

2 x + 1 dx.

Answer : Using the first method first we compute the indefinite in- tegral: ∫ √ 2 x + 1 dx =

2 x + 1 2x dx (u = 2x + 1)

u du

u^3 /^2 + C

(2x + 1)^3 /^2 + C.

Then we use it for computing the definite integral:

∫ (^4)

0

2 x + 1 dx =

[

(2x + 1)^3 /^2

] 4

0

93 /^2 −

13 /^2 =

1.4. THE SUBSTITUTION RULE 20

In the second method we compute the definite integral directly ad- justing the limits of integration after the substitution: ∫ (^4)

0

2 x + 1 dx =

0

2 x + 1 2x dx (u = 2x + 1; u′^ = 2)

1

u du

(note the change in the limits of integration to u(0) = 1 and u(4) = 9)

[

u^3 /^2

] 9

1

93 /^2 −

13 /^2