Post on 11-Jan-2016
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