1. P.S : I definitely deserve
Integration
By : Wonhee Lee
Edited by : Jiwoo Lee
P.S : I definitely deserve
an A for this

2. Index 1. An Overview of the Area Problem
2. The Indefinite Integral; Integral Curves and Direction Fields
3. Integration by Substitution
4. Sigma Notation; Area as a Limit
5. The Definite Integral
6. The Fundamental Theorem of Calculus
7. Rectilinear Motion Revisited; Average Value
8. Evaluating Definite Integrals by Substitution
9. Logarithmic Functions from the Integral Point of View

3. An Overview of the Area Problem
The Antiderivative Method
Finding the area function by “undoing” a differentiation

4. The Rectangle Method

5. The Rectangle Method (Cont.)
Divide the Interval [a,b] into n subintervals
The Area of all the rectangles (Right)
= ((b-a)/n)[(f(a+(b-a)/n) + (f(a+2(b-a)/n) (f(b-(b-a)/n) + f(b)]
As more and more subintervals are created, the calculation becomes increasingly accurate

6. Example Problem
Use both the rectangle method and the antiderivate method to solve for the area under the curve y=x2 [0,2] (n=4,REA)
(2/4)[f(2/4) + f(4/4) + f(6/4) + f(8/4)]
= 15/4
Y(x) = X3/3 + C
∴ Y(2)-Y(0) = 8/3

7. The Indefinite Integral
Function F is the antiderivative of function f on interval I if F’(x) = f(x)
(d/dx)[∫f(x)dx] = f(x)
The integral of f(x) with respect to x is equal to F(x) plus a constant, C.
∫f(x)dx = F(x) + C

8. Integration Formulas ∫cos x dx = sin x + C ∫sin x dx = -cos x + C
∫sec2x dx = tan x + C
∫csc2x dx = -cot x + C
∫sec x tan x dx = sec x + C
∫csc x cot x dx = -csc x + C

9. Integration Formulas (Cont.)
∫ex dx = ex + C
∫(1/x) dx = ln |x| + C
∫(1/(1+x2)) dx = tan-1 x + C
∫(1/(1-x2)^(1/2)) dx = sin-1 x + C
∫(1/(x(x2-1)^(1/2))) dx = sec-1 |x| + C

10. General Integration Theorems
∫cf(x) dx = c ∫ f(x) dx
∫[f(x) + g(x)]dx = ∫ f(x) dx + ∫ g(x) dx
∫[f(x) – g(x)]dx = ∫ f(x) dx - ∫ g(x) dx
Try it out!
∫5550x4 dx
= 5550∫x4 dx = 1110x5 + C
∫5x4 + 4x3 + 3x2 + 2x + 1 dx
= 5∫x4 dx + 4∫x3 dx + 3∫x2 dx + 2∫x dx + ∫1 dx
= x5 + x4 + x3 + x2 + x + C

11. Initial-Value Problem
dy/dx = f(x), y(x0) = y0 => Initial Condition
Ex) dy/dx = sinx , y(0) = 0
Y = ∫ sinx dx = -cos x + C
0 = -cos(0) + C , C = 1
∴ Y = -cosx + 1

12. Direction Fields (Slope Fields)
y’ = e-y
Choose a rectangular gird of points
Calculate the slopes of the tangent lines
Draw small portions of the tangent lines

13. Example Problems Evaluate the following integrals
(a) ∫x2(1+x4) dx (b) ∫(sinx/cos2x) dx
(a) = ∫x2 + x6 dx = x3/3 + x7/7 + C
(b) = ∫tanxsecx dx = secx + C

14. Integration by Substitution
(d/dx)[F(g(x))] = F’(g(x))g’(x)
∫ F’(g(x))g’(x)dx = F(g(x)) + C
∫ f(g(x))g’(x)dx = F(g(x)) + C
u = g(x), du = g’(x) dx
∫ f(u)du = F(u) + C

15. Integration by Substitution(Cont)
Find a function u = g(x) so that the equation ∫ f(g(x))g’(x) dx can be transformed to ∫ f(u) du
Evaluate the intgeral ∫ f(u) du in terms of u
Replace u = g(x) so that the final answer is in terms of x

16. Some more integration formulas
∫1/(1+x2) dx = tan-1x + C
∫1/(1-x2)1/2 dx = sin-1x + C
∫1/(x(x2-1)1/2) dx = sec-1|x| + C

17. Example Problems ∫(x2+10)20 • 2x dx u = x2+ 10
∴ ∫(u)20 • du = u21/21 + C
= ((x2+10)21)/21 + C
∫sin7x dx
u = 7x, du = 7 dx, dx = du/7
∴ ∫sinu (1/7)du = -(1/7)cosu + C
= -(1/7)cos7x + C

18. Example Problems (Cont)
∫ dx/(1+4x2)
u = 2x, du = 2dx
∴(1/2)∫du/(1+u2) = (1/2)tan-1u + C
= (1/2)tan-1(2x) + C
∫ dy/(4-y2)1/2
= sin-1(y/2) + C

19. Sigma Notation; Area as a Limit
The summation of the function f(k) from k = a to k = b
All numbers from a to b must be 0 or natural numbers
Also known as the summation notation

20. Sigma Notation Theorems
A constant factor can be moved through a sigma sign.
Sigma distributes across sums.
Sigma distributes across differences.

21. Common evaulations ∑ k = 1 + 2 + … + n = n(n+1)/2 k=1
∑ k2 = … + n2 = n(n+1)(2n+1)/6
Open form : Closed form

22. Area under a curve f must be continuous on [a,b]
f(x) ≥ 0 for all x in [a,b]
If f(x) is not always greater than or equal to 0, the equation above becomes the Net Signed Area.

23. Methods of approximation

24. Example Problems Write the following expression in sigma notation.
4•1 + 4•2 + 4•3 + … + 4•30 30
= ∑ 4k
k=1
Evaluate the sigma notation.
200
∑ k = 200(200+1)/2 = 20100

25. The Definite Integral b ∫ f(x) dx = a
The Definite Integral of f from a to b
If f is continuous on [a,b], f is integrable on [a,b]

26. Duh! a
∫ f(x) dx = 0
b a
-∫ f(x) dx = ∫ f(x) dx
a b

27. Sooooo Obvious
b c b
∫ f(x) dx = ∫ f(x) dx + ∫ f(x) dx
a a c

28. Example Problems Find the area beneath the curve y = sinx from [0,∏]
∏ ∏
∫ sinx dx = [-cosx] = -cos ∏ + cos 0 = = 2
Evaluate 4
∫ 1 – (1/3)x2 dx 2
= [x-(1/9)x3] = 4-(64/9)-2+(8/9) = -(38/9)

29. The Fundamental Theorem of Calculus (1)
If f is continuous on [a,b] and F is any antiderivative of f on [a,b], then b
∫ f(x) dx = F(b) – F(a) a

30. The Mean-Value Theorem for Integralsb
∫ f(x) dx = f(x*)(b-a) a
There exists at least one number x* in [a,b] such that it satisfies the equation above
The two shaded regions are equal in area

31. The Fundamental Theorem of Calculus (2)
x x
∫ f(t) dt = F(x) , (d/dx)[∫ f(t) dt] = f(x)
a a
If f is continuous on interval I, then f has an antiderivative on I. If a is any number in I, then the function F is an antiderivative of f on I, that is, F’(x) = f(x) for each x in I, or in an alternative notation

32. Example Problems 1 ∫ (3x-2)3 dx = F(1) – F(0) = [(3x-2)4/12]01 = -5/4
= F(1) – F(0)
= [(3x-2)4/12]01
= -5/4 1
∫ xex dx -1
= F(1) – F(-1)
= [(xex – ex)]-11
= 2/e

33. Rectilinear Motion Revisited; Average Value
s(t) = ∫ v(t)dt
v(t) = ∫ a(t)dt
s(t) = s0 + v0t + (1/2)at2
v(t) = v0 + at

34. Distance vs Displacement
Distance : The length of the path that an object takes to get from one point to another
Displacement: The shortest distance possible to get from one point to another

35. Distance vs Displacement (Cont.)
Displacement = ∫ v(t) dt t0
|Distance| = ∫ |v(t)| dt

36. Average Value b fave = (1/(b-a))∫ f(x) dx a t1
vave = (1/(t1-t0))∫ v(t) dt t0

37. Example Problems
v(t) = -2t2 – t + 10, find both the displacement and the distance traveled between t=0 and t= 4
∫v(t)dt = s(t) = -(2/3)t3 – t2/2 + 10t + C
Displacement = s(4)-s(0) = -32/3
|Distance| = (s(2) - s(0)) – (s(4) – s(2))
= -36 , ∴ Distance = 36

38. Evaluating Definite Integrals by Substitution
Recall the process from Indefinite Integrals
d/dx[F(g(x))] = F’(g(x))g’(x)
∫ F’(g(x))g’(x)dx = F(g(x)) + C
∫ f(g(x))g’(x)dx = F(g(x)) + C
u = g(x), du = g’(x) dx
∫ f(u)du = F(u) + C
b g(b)
∫ f(g(x))g’(x) dx = ∫ f(u) du
a g(a)

39. Example Problems Evaluate ∫ sinx cos2x dx∏
Evaluate ∫ sinx cos2x dx
g(x) = u = cosx, du = -sinx dx
g(∏)
∫ -u2 du = -[u3/3] = 2/3
g(0)

40. Logarithmic Functions from the Integral Point of Viewx
ln x = ∫ (1/t) dt , x > 0 1
(d/dx)[ln x] = (1/x) (x > 0)
Fundamental Theorem of Calculus, Part 2, Remember?

41. The Value of ln(x) in graphs

42. Euler’s number Recall that ∫ex dx = ex + C ∴ (d/dx)[ex] = ex
lim (1 + x)1/x = e
X-> 0

43. Example Problem Represent ln 10 on a graph and approximate it. 10
ln 10 = ∫ (1/t) dt ≈ 2.303 1

44. Bibliography