CHAPTER 6Integration

SECTION 6.1

Antiderivatives and Indefinite Integration

Definition of an Antiderivative

A function F is an antiderivative of f on an interval I if F’(x) = f(x) for all x in I.

Theorem – Representation of Antiderivtives

If F is an antiderivative of f on an interval I, then G is an antiderivative of f on the interval I if and only if G is on the form:

G(x) = F(x) + C, for all x in I where C is a constant

Terminology

C - is called the constant of integration

G(x) = x2 + c is the general solution of the differential equation G’(x) = 2x

Integration is the “inverse” of Differentiation

Differential Equation

Is an equation that involves x, y and derivatives of y.

EXAMPLESy’ = 3x and y’ = x2 + 1

Notation for Antiderivatives

When solving a differential equation of the form

dy/dx = f(x) you can writedy = f(x)dx

and is called antidifferentiation and is denoted by an integral

sign ∫

Integral Notation

y = ∫ f(x)dx = F(x) + C

f(x) – integrand

dx – variable of integration

C – constant of integration

Basic Integration Rules

y = ∫ F’(x)dx = F(x) + C

f(x) – integrand

dx – variable of integration

C – constant of integration

Basic Integration Rules

DIFFERENTIATION FORMULA

INTEGRATION FORMULA

d/dx [c] = 0 d/dx [kx] = k d/dx [kf(x)] = kf’(x) d/dx [f(x) ± g(x)] =

f’(x) ± g’(x) d/dx [xn] = nxn-1

∫ 0 dx = c ∫ kdx = kx + c ∫ kf(x)dx = k ∫ f(x)

dx ∫ f(x) ± g(x)]dx =

∫ f(x) dx ± ∫ g(x) dx

∫ xn dx = (xn+1)/(n+1) + c, n≠ - 1

(Power Rule)

Examples

1. ∫ 3x dx2. ∫ 1/x3 dx3. ∫ (x + 2) dx4. ∫ (x + 1)/√x dx

Finding a Particular Solution

EXAMPLEF’(x) = 1/x2, x > 0 and find the

particular solution that satisfies the initial condition F(1) = 0

Finding a Particular Solution

1. Find the general solution by integrating, - 1/x + c. x > 0

2. Use initial condition F(1) = 0 and solve for c, F(1) = -1/1 + c, so c = 1

3. Write the particular solution F(x) = - 1/x + 1, x > 0

Solving a Vertical Motion Problem

A ball is thrown upward with an initial velocity of 64 ft/sec from an initial height of 80 ft.

a) Find the position function giving the height s as a function of the time t

b) When does the ball hit the ground?

Solutiona) Let t = 0 represent the initial time; s(0) =

80 and s’(0) = 64b) Use -32 ft/sec as the acceleration due to

gravity, then s”(t) = - 32c) ∫ s”(t) dt = ∫ -32 dt = -32 t + cd) s’(0) =64 = -32(0) = c, so c = 64e) s(t) = ∫ s’(t) = ∫ (-32t + 64) dt = -16t2 +

64t + Cf) s(0) = 80 = -16(02) + 64(0) + C, hence C =

80g) s(t) = -16t2 + 64t + 80 = 0, solve and t = 5

SECTION 6.2

Area

Sigma Notation

The sum of n terms a1, a2, a3…,an is written as

n

∑ ai = a1 + a2 + a3+ …+ ani = 1

Where i the index of summation, a1 is the ith term of the sum, and the upper and lower bounds of summation are n and 1, respectively.

EXAMPLES

6

∑ i= 1 + 2 + 3 + 4 + 5 + 6i = 1

5

∑ (i + 1)= 1 + 2 + 3 + 4 + 5 + 6i = 0

7

∑ j2 = 9 + 16 + 25 + 36 + 49 j = 3

SUMMATION FORMULAS n

1. ∑ c = cni = 1

n

2. ∑ i = n(n + 1)/2 i = 1

n

3. ∑ i2 = n(n + 1)(n + 2)(2n + 1)/6 i= 1

n

4. ∑ i3 = n2(n + 1)2/4 i = 1

PROPERTIES OF SUMMATION

n n

1. ∑ kai = k ∑ aii = 1 i = 1

n n n

2. ∑ (ai ± bi ) = ∑ ai ± ∑ bi i = 1 i = 1 i = 1

EXAMPLE

Find the sum for n = 10 and n = 100

n

1. ∑ ( i + 1)/n2

i = 1

AREA OF A PLANE REGION

Find the area of the region lying between the graph of f(x) = - x2 + 5 and the x-axis between x = 0 and x = 2 using five rectangles to find an approximation of the area. You should use both inscribed rectangles and circumscribed rectangles. In doing so you will be able to find a lower and upper sum.

Limits of the Lower and Upper Sums

Let f be continuous and nonnegative on the interval [a,b]. The limits as n→∞ of both the lower and upper sums exist and are equal to each other. That is,

n

lim s(n) = lim ∑ f(mi)x and n→∞ n→∞ i = 1

Limits of the Lower and Upper Sums

n

lim s(n) = lim ∑ f(Mi)x n→∞ n→∞ i = 1

n

lim S(n) = lim ∑ f(Mi)x n→∞ n→∞ i = 1

Definition of the Area of a Region in the Plane

Let f be continuous and nonnegative on the interval [a,b]. The area of the region bounded by the graph of f, the x-axis, and the vertical lines x= a and x = b is

n

Area = lim ∑ f(ci)x, xi -1 ci xi n→∞ i = 1

Where x = (b-a)/n

EXAMPLE

Fine the area of the region bounded by the graph of f(x) =x3, the x-axis, and the vertical lines x=0 and x =1.

1. Partition the interval [0,1] into n subintervals each of width 1/n = x

2. Simplify using the formula below and A = 1/4

n

Area = lim ∑ f(ci)x, xi -1 ci xi n→∞ i = 1

Where x = (b-a)/n

SECTION 6.3

Riemann Sums and Definite Integrals

Definition of Riemann Sum

Let f be defined on the closed interval [a, b] and let be a partition of [a,b] given by

a = xo < x1 < x2 < …<xn-1<xn =b

Where xi is the width of the ith subinterval. If ci is any point in the ith subinterval, then the sum

n

f(ci) xi , xi-1 ci xi is called a Riemann i =

1 sum of f for the partition

Definition of the Norm

The width of the largest subinterval of a partition is the norm of the partition

and is denoted by . If every subinterval is of equal width, the partition is regular and the norm is denoted by

= x = (b – a)/n

Definite Integrals

If f is defined on the closed interval [a, b] and the limit n

lim ∑ f(ci)x → 0 i = 1

exists, then f is integrable on [a,b] and the limit is b

∫ f(x) dx a

The number a is the lower limit of integration, and the number b is the upper limit of integration

Continuity Implies Integrability

If a function f is continuous on the closed interval [a,b], then f is integrable on ]a,b]

Evaluating a Definite Integral

Evaluate the definite integral

1

∫ 2xdx -2

The Definite Integral as the Area of a Region

If f is continuous and nonnegative closed interval [a, b] the the area of the region bounded by the graph of f, the x-axis, and the vertical lines x =a and x = b is given by

b

∫ f(x) dx a

Examples

Evaluate the Definite Integral

0

∫ (x + 2)dx

3Sketch the region and use formula for

trapezoid

Additive Interval Property

If f is integrable on the three closed intervals determined by a, b and c, then ,

b c b

∫ f(x) dx = ∫ f(x) dx = ∫ f(x) dx

a a c

Properties of Definite Integrals

If f and g are integrable on [a,b] and k is a constant, then the functions of kf and f ± g are integrable on [a,b], and

b b

1. ∫kf dx = k ∫ f(x) dx

a a

Properties of Definite Integrals

If f and g are integrable on [a,b] and k is a constant, then the functions of kf and f ± g are integrable on [a,b], and

b b b

1. ∫[f(x) ± g(x)] dx = ∫ f(x) dx ± ∫ f(x) dx

a a a

SECTION 6.4THE FUNDAMENTAL

THEOREM OF CALCULUS

The Fundamental Theorem of Calculus

If a function f is continuous on the closed interval [a,b] and F is an antiderivative of f on the interval [a,b], then

b

∫ f(x) dx = F(b) – F(a)

a

Using the Fundamental Theorem of Calculus

1. Find the antiderivative of f if possible 2. Evaluate the definite integral

Example: ∫ x3dx on the interval [1,3]

Using the Fundamental Theorem of Calculus to Find Area

Find the area of the region bounded by the graph of y = 2x2 – 3x +2, the x-axis, and the vertical lines x=0 and x= 2.

1. Graph2. Find the antiderivative3. Evaluate on your interval

Mean Value Theorem for Integrals

If f is continuous on the closed interval [a,b], then there exists a number c in the closed interval [a,b] such that

∫ f(x)dx = f(c)(b-a)

Average Value of a Function on an Interval

If f is integrable on the closed interval [a,b], then the average value of f on the interval is

b

1/(b-a) ∫ f(x)dx a

The Second Fundamental Theorem of Calculus

If f is continuous on an open interval I containing a, then, for every x in the interval

x

d/dx[ ∫ f(t) dt] = f(x)

a

SECTION 6.5

INTEGRATION BY SUBSTITUTION

Antidifferentiation of a Composite Function

Let g be a function whose range is an interval I, and let f be a function that is continuous on I. If g is differentiable on its domain and F is an antiderivative of f on I, then

∫ f(g(x))g’(x)dx = F(g(x)) + CIf u = g(x), then du = g’(x)dx and

∫ f(u)du = F(u) + C

Change in Variable

You completely rewrite the integral in terms of u and du. This is useful technique for complicated intergrands.

∫ f(g(x))g’(x)dx = ∫ f(u) du = F(u)+ C

Example

Find

∫ (2x – 1).5 dxLet u = 2x - 1, then du/dx = 2dx/dxSolve for dx and substitute back to

obtain the antiderivative.

Check your answer.

Power Rule for Integration

If g is a differentiable function of x, then,

∫ ((g(x))ng’(x)dx = ∫ (g(x))n+1/(n+1) + C

Change of Variables for Definite Integrals

If the function u = g(x) has a continuous derivative on the closed interval [a,b] and f is continuous on the range of g, then,

b g(b)

∫ (g(x)g’(x)dx = ∫ f(u)du a g(a)

Integration of Even and Odd Functions

Let f be integrable on the closed interval [ -a,a].

If f is an even function, then

a a

∫ f(x) dx=2 ∫ f(x) dx

-a 0

Integration of Even and Odd Functions

Let f be integrable on the closed interval [ -a,a].

If f is an odd function, then

a

∫ f(x) dx= ∫ f(x) dx = 0

-a

SECTION 6.6

NUMERICAL INTEGRATION

Trapezoidal Rule

Let f be continuous on [a,b]. The trapezoidal Rule for approximating

∫ f(x) dx

(b-a)/2n [f(x0) = 2(f(x1) +…..+2f(xn-1) + f(xn)]

Simpson’s Rule

If p(x) = Ax2 + Bx + c, then b

∫ p(x) dx = a(b-a)/6 [p(a) + 4p[(a+b)]/2) + p(b)]

END