INSTRUCTORS Raffi Hovasapian John Zhu

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• ## Related Books

 4 answersLast reply by: Chonglin XuWed Jan 7, 2015 10:46 PMPost by James Xie on July 23, 2012For ex. 3, i got 15/8 + ln4 - ln1 as an answer. should i just leave the natural logs (... ln4 - ln1) as they are? I'm really not sure what else to do with them.

### Fundamental Theorem of Calculus

• Simply evaluating integral at 2 bounds
• Area under a curve
• Accumulated value of anti-derivative function

### Fundamental Theorem of Calculus

15 dx
• 15 dx = x |15
• 15 dx = 5 − 1
15 dx = 4
0π cosx (sinx)3 dx
• u = sinx
• du = cosx  dx
• 0π cosx (sinx)3 dx = ∫0π u3 du
• 0π cosx (sinx)3 dx = [(u4)/4] |0π
• Remember, the limits of the integral are dependent on x. We must plug back in for x before using the original limits
• 0π cosx (sinx)3 dx = [((sinx)4)/4] |0π
• 0π cosx (sinx)3 dx = [((sinπ)4)/4] − [((sin0)4)/4]
• 0π cosx (sinx)3 dx = [0/4] − [0/4]
0π cosx (sinx)3 dx = 0
−[(π)/2][(π)/2] 5 cosx  dx
• −[(π)/2][(π)/2] 5 cosx  dx = 5 sinx |−[(π)/2][(π)/2]
• −[(π)/2][(π)/2] 5 cosx  dx = 5 sin[(π)/2] − 5 sin[(−π)/2]
• −[(π)/2][(π)/2] 5 cosx  dx = 5 − (−5)
−[(π)/2][(π)/2] 5 cosx  dx = 10
Confirm that ∫−22 x2 dx = 2 ∫02 x2 dx
• −22 x2 dx = [(x3)/3] |−22
• −22 x2 dx = [(23)/3] − [((−2)3)/3]
• −22 x2 dx = [8/3] + [8/3]
• −22 x2 dx = [16/3]
• 2 ∫02 x2 dx = 2 [(x3)/3] |02
• 2 ∫02 x2 dx = 2 [(23)/3] − 2[(03)/3]
• 2 ∫02 x2 dx = [16/3]
• This is a property of integrals involving even functions
• −aa f(x)  dx = 2 ∫0a f(x)  dx if f(x) is even
Yes, ∫−22 x2 dx = 2 ∫02 x2 dx
0ln5 e2x dx
• u = 2x
• du = 2  dx
• 0ln5 e5x dx = [1/2] ∫0ln5 eu du
• 0ln5 e5x dx = [1/2] eu |0ln5
• 0ln5 e5x dx = [1/2] e2x |0ln5
• 0ln5 e5x dx = [1/2] e2 ln5 − [1/2] e2(0)
• 0ln5 e5x dx = [1/2] eln52 − [1/2] (1)
• 0ln5 e5x dx = [25/2] − [1/2]
0ln5 e5x dx = 12
−11 x2 + 3x + 1  dx
• −11 x2 + 3x + 1  dx = ∫−11 x2 dx + ∫−11 3x  dx + ∫−11  dx
• −11 x2 + 3x + 1  dx = [(x3)/3] |−11 + [(3x2)/2] |−11 + x |−11
• −11 x2 + 3x + 1  dx = [1/3] + [1/3] + [3/2] − [3/2] + 1 + 1
−11 x2 + 3x + 1  dx = [8/3]
−11 x3 dx
• −11 x3 dx = [(x4)/4] |−11
• −11 x3 dx = [1/4] − [1/4]
• Property of odd functions
• −aa f(x) = 0 if f(x) is odd
−11 x3 dx = 0
02 [1/(√{4 − x2})] dx
• 02 [1/(√{4 − x2})] dx = ∫02 [1/(√{22 − x2})] dx
• 02 [1/(√{4 − x2})] dx = sin−1 [x/2] |02
• 02 [1/(√{4 − x2})] dx = sin−1 1 − sin−1 0
• 02 [1/(√{4 − x2})] dx = [(π)/2] − 0
02 [1/(√{4 − x2})] dx = [(π)/2]
01 [1/(√{4 − x2})] dx + ∫12 [1/(√{4 − x2})] dx
• 01 [1/(√{4 − x2})] dx + ∫12 [1/(√{4 − x2})] dx = sin−1 [x/2] |01 + sin−1 [x/2] |12
• 01 [1/(√{4 − x2})] dx + ∫12 [1/(√{4 − x2})] dx = sin−1 [1/2] − sin−1 0 + sin−1 1 − sin−1 [1/2]
• 01 [1/(√{4 − x2})] dx + ∫12 [1/(√{4 − x2})] dx = sin−1 1 − sin−1 0
• Another property of integrals.
• ab f(x)  dx + ∫bc f(x)  dx = ∫ac f(x)  dx
01 [1/(√{4 − x2})] dx + ∫12 [1/(√{4 − x2})] dx = [(π)/2]
Show that ∫12 ex dx = − ∫21 ex dx
• 12 ex dx = ex |12
• 12 ex dx = e2 − e1
• − ∫21 ex dx = − ex |21
• − ∫21 ex dx = −e1 − (−e2)
• − ∫21 ex dx = e2 − e1
12 ex dx = − ∫21 ex dx

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Fundamental Theorem of Calculus

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Fundamental Theorem of Calculus: Properties 0:10
• Definition of Integral
• Example 1
• Fundamental Theorem of Calculus: Properties 2:40
• Rule 1
• Rule 2
• Rule 3
• Rule 4
• Example 2 4:07
• Example 3 6:17
• Example 4 9:31
• Example 5 10:52
• Example 6 13:34