1. Chapter 6
Integration

2. Antiderivatives and Indefinite Integration
Section 6.1
Antiderivatives and Indefinite Integration

3. 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.

4. 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

5. 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

6. Differential Equation
Is an equation that involves x, y and derivatives of y. EXAMPLES y’ = 3x and y’ = x2 + 1

7. Notation for Antiderivatives
When solving a differential equation of the form dy/dx = f(x) you can write dy = f(x)dx and is called antidifferentiation and is denoted by an integral sign ∫

8. Integral Notation
y = ∫ f(x)dx = F(x) + C f(x) – integrand dx – variable of integration C – constant of integration

9. Basic Integration Rules
y = ∫ F’(x)dx = F(x) + C f(x) – integrand dx – variable of integration C – constant of integration

10. 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)

11. Examples
∫ 3x dx
∫ 1/x3 dx
∫ (x + 2) dx
∫ √x dx
∫ (x + 1)/√x dx

12. Finding a Particular Solution
EXAMPLE F’(x) = 1/x2, x > 0 and find the particular solution that satisfies the initial condition F(1) = 0

13. Finding a Particular Solution
Find the general solution by integrating, - 1/x + c. x > 0
Use initial condition F(1) = 0 and solve for c, F(1) = -1/1 + c, so c = 1
Write the particular solution F(x) = - 1/x + 1, x > 0

14. 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.
Find the position function giving the height s as a function of the time t
When does the ball hit the ground?

15. Solution
Let t = 0 represent the initial time; s(0) = 80 and s’(0) = 64
Use -32 ft/sec as the acceleration due to gravity, then s”(t) = - 32
∫ s”(t) dt = ∫ -32 dt = -32 t + c
s’(0) =64 = -32(0) = c, so c = 64
s(t) = ∫ s’(t) = ∫ (-32t + 64) dt = -16t2 + 64t + C
s(0) = 80 = -16(02) + 64(0) + C, hence C = 80
s(t) = -16t2 + 64t + 80 = 0, solve and t = 5

16. Section 6.2
Area

17. Sigma Notation The sum of n terms a1, a2, a3…,an is written as
∑ ai = a1 + a2 + a3+ …+ an
i = 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.

18. EXAMPLES ∑ i= 1 + 2 + 3 + 4 + 5 + 6 ∑ (i + 1)= 1 + 2 + 3 + 4 + 5 + 67
∑ j2 =
j = 3

19. SUMMATION FORMULAS ∑ i3 = n2(n + 1)2/4 ∑ c = cn ∑ i = n(n + 1)/2
∑ i2 = n(n + 1)(n + 2)(2n + 1)/6
i= 1
∑ i3 = n2(n + 1)2/4

20. PROPERTIES OF SUMMATION
n n
∑ kai = k ∑ ai
i = 1 i = 1
n n n
∑ (ai ± bi ) = ∑ ai ± ∑ bi
i = i = i = 1

21. EXAMPLE
Find the sum for n = 10 and n = 100 n
∑ ( i + 1)/n2
i = 1

22. 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.

23. 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

24. Limits of the Lower and Upper Sums
lim s(n) = lim ∑ f(Mi)x
n→∞ n→∞ i = 1
lim S(n) = lim ∑ f(Mi)x
n→∞ n→∞ i = 1

25. 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

26. EXAMPLE Area = lim ∑ f(ci)x, xi -1  ci xi
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.
Partition the interval [0,1] into n subintervals each of width 1/n = x
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

27. Riemann Sums and Definite Integrals
Section 6.3
Riemann Sums and Definite Integrals

28. 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 

29. 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

30. Definite Integrals lim ∑ f(ci)x    → 0 i = 1
If f is defined on the closed interval [a, b] and the limit n
lim ∑ f(ci)x
   → 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

31. Continuity Implies Integrability
If a function f is continuous on the closed interval [a,b], then f is integrable on ]a,b]

32. Evaluating a Definite Integral
Evaluate the definite integral 1
∫ 2xdx -2

33. 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

34. Examples Evaluate the Definite Integral ∫ (x + 2)dx 3
∫ (x + 2)dx 3
Sketch the region and use formula for trapezoid

35. Additive Interval Property
If f is integratable 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

36. 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

37. 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

38. THE FUNDAMENTAL THEOREM OF CALCULUS
Section 6.4
THE FUNDAMENTAL THEOREM OF CALCULUS

39. 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

40. Using the Fundamental Theorem of Calculus
Find the antiderivative of f if possible
Evaluate the definite integral
Example: ∫ x3dx on the interval [1,3]

41. 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.
Graph
Find the antiderivative
Evaluate on your interval

42. 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)

43. 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

44. 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

45. INTEGRATION BY SUBSTITUTION
Section 6.5
INTEGRATION BY SUBSTITUTION

46. 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)) + C
If u = g(x), then du = g’(x)dx and
∫ f(u)du = F(u) + C

47. Change in Variable ∫ f(g(x))g’(x)dx = ∫ f(u) du = F(u)+ C
You completely rewrite the integral in terms of u and du. This is a useful technique for complicated intergrands.
∫ f(g(x))g’(x)dx = ∫ f(u) du = F(u)+ C

48. Example ∫ (2x – 1).5 dx Find Let u = 2x - 1, then du/dx = 2dx/dx
Solve for dx and substitute back to obtain the antiderivative.
Check your answer.

49. 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

50. 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)

51. 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

52. 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

53. NUMERICAL INTEGRATION
Section 6.6
NUMERICAL INTEGRATION

54. Trapezoidal Rule ∫ f(x) dx 
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)]

55. Simpson’s Rule ∫ p(x) dx = If p(x) = Ax2 + Bx + c, then b a
(b-a)/6 [p(a) + 4p[(a+b)]/2) + p(b)]

56. END