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For more information, please see full course syllabus of Calculus AB

For more information, please see full course syllabus of Calculus AB

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### Fundamental Theorem of Calculus

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

### Fundamental Theorem of Calculus

∫

_{1}^{5}dx- ∫
_{1}^{5}dx = x |_{1}^{5} - ∫
_{1}^{5}dx = 5 − 1

∫

_{1}^{5}dx = 4∫

_{0}^{π}cosx (sinx)^{3}dx- u = sinx
- du = cosx dx
- ∫
_{0}^{π}cosx (sinx)^{3}dx = ∫_{0}^{π}u^{3}du - ∫
_{0}^{π}cosx (sinx)^{3}dx = [(u^{4})/4] |_{0}^{π} - Remember, the limits of the integral are dependent on x. We must plug back in for x before using the original limits
- ∫
_{0}^{π}cosx (sinx)^{3}dx = [((sinx)^{4})/4] |_{0}^{π} - ∫
_{0}^{π}cosx (sinx)^{3}dx = [((sinπ)^{4})/4] − [((sin0)^{4})/4] - ∫
_{0}^{π}cosx (sinx)^{3}dx = [0/4] − [0/4]

∫

_{0}^{π}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 = 10Confirm that ∫

_{−2}^{2}x^{2}dx = 2 ∫_{0}^{2}x^{2}dx- ∫
_{−2}^{2}x^{2}dx = [(x^{3})/3] |_{−2}^{2} - ∫
_{−2}^{2}x^{2}dx = [(2^{3})/3] − [((−2)^{3})/3] - ∫
_{−2}^{2}x^{2}dx = [8/3] + [8/3] - ∫
_{−2}^{2}x^{2}dx = [16/3] - 2 ∫
_{0}^{2}x^{2}dx = 2 [(x^{3})/3] |_{0}^{2} - 2 ∫
_{0}^{2}x^{2}dx = 2 [(2^{3})/3] − 2[(0^{3})/3] - 2 ∫
_{0}^{2}x^{2}dx = [16/3] - This is a property of integrals involving even functions
- ∫
_{−a}^{a}f(x) dx = 2 ∫_{0}^{a}f(x) dx if f(x) is even

Yes, ∫

_{−2}^{2}x^{2}dx = 2 ∫_{0}^{2}x^{2}dx∫

_{0}^{ln5}e^{2x}dx- u = 2x
- du = 2 dx
- ∫
_{0}^{ln5}e^{5x}dx = [1/2] ∫_{0}^{ln5}e^{u}du - ∫
_{0}^{ln5}e^{5x}dx = [1/2] e^{u}|_{0}^{ln5} - ∫
_{0}^{ln5}e^{5x}dx = [1/2] e^{2x}|_{0}^{ln5} - ∫
_{0}^{ln5}e^{5x}dx = [1/2] e^{2 ln5}− [1/2] e^{2(0)} - ∫
_{0}^{ln5}e^{5x}dx = [1/2] e^{ln52}− [1/2] (1) - ∫
_{0}^{ln5}e^{5x}dx = [25/2] − [1/2]

∫

_{0}^{ln5}e^{5x}dx = 12∫

_{−1}^{1}x^{2}+ 3x + 1 dx- ∫
_{−1}^{1}x^{2}+ 3x + 1 dx = ∫_{−1}^{1}x^{2}dx + ∫_{−1}^{1}3x dx + ∫_{−1}^{1}dx - ∫
_{−1}^{1}x^{2}+ 3x + 1 dx = [(x^{3})/3] |_{−1}^{1}+ [(3x^{2})/2] |_{−1}^{1}+ x |_{−1}^{1} - ∫
_{−1}^{1}x^{2}+ 3x + 1 dx = [1/3] + [1/3] + [3/2] − [3/2] + 1 + 1

∫

_{−1}^{1}x^{2}+ 3x + 1 dx = [8/3]∫

_{−1}^{1}x^{3}dx- ∫
_{−1}^{1}x^{3}dx = [(x^{4})/4] |_{−1}^{1} - ∫
_{−1}^{1}x^{3}dx = [1/4] − [1/4] - Property of odd functions
- ∫
_{−a}^{a}f(x) = 0 if f(x) is odd

∫

_{−1}^{1}x^{3}dx = 0∫

_{0}^{2}[1/(√{4 − x^{2}})] dx- ∫
_{0}^{2}[1/(√{4 − x^{2}})] dx = ∫_{0}^{2}[1/(√{2^{2}− x^{2}})] dx - ∫
_{0}^{2}[1/(√{4 − x^{2}})] dx = sin^{−1}[x/2] |_{0}^{2} - ∫
_{0}^{2}[1/(√{4 − x^{2}})] dx = sin^{−1}1 − sin^{−1}0 - ∫
_{0}^{2}[1/(√{4 − x^{2}})] dx = [(π)/2] − 0

∫

_{0}^{2}[1/(√{4 − x^{2}})] dx = [(π)/2]∫

_{0}^{1}[1/(√{4 − x^{2}})] dx + ∫_{1}^{2}[1/(√{4 − x^{2}})] dx- ∫
_{0}^{1}[1/(√{4 − x^{2}})] dx + ∫_{1}^{2}[1/(√{4 − x^{2}})] dx = sin^{−1}[x/2] |_{0}^{1}+ sin^{−1}[x/2] |_{1}^{2} - ∫
_{0}^{1}[1/(√{4 − x^{2}})] dx + ∫_{1}^{2}[1/(√{4 − x^{2}})] dx = sin^{−1}[1/2] − sin^{−1}0 + sin^{−1}1 − sin^{−1}[1/2] - ∫
_{0}^{1}[1/(√{4 − x^{2}})] dx + ∫_{1}^{2}[1/(√{4 − x^{2}})] dx = sin^{−1}1 − sin^{−1}0 - Another property of integrals.
- ∫
_{a}^{b}f(x) dx + ∫_{b}^{c}f(x) dx = ∫_{a}^{c}f(x) dx

∫

_{0}^{1}[1/(√{4 − x^{2}})] dx + ∫_{1}^{2}[1/(√{4 − x^{2}})] dx = [(π)/2]Show that ∫

_{1}^{2}e^{x}dx = − ∫_{2}^{1}e^{x}dx- ∫
_{1}^{2}e^{x}dx = e^{x}|_{1}^{2} - ∫
_{1}^{2}e^{x}dx = e^{2}− e^{1} - − ∫
_{2}^{1}e^{x}dx = − e^{x}|_{2}^{1} - − ∫
_{2}^{1}e^{x}dx = −e^{1}− (−e^{2}) - − ∫
_{2}^{1}e^{x}dx = e^{2}− e^{1}

∫

_{1}^{2}e^{x}dx = − ∫_{2}^{1}e^{x}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.

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

4 answers

Last reply by: Chonglin Xu

Wed Jan 7, 2015 10:46 PM

Post by James Xie on July 23, 2012

For 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.