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For more information, please see full course syllabus of College Calculus: Level I

For more information, please see full course syllabus of College Calculus: Level I

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### Riemann Sums, Definite Integrals, Fundamental Theorem of Calculus

- Geometrically, Riemann sums represent sums of rectangle approximations.
- The definite integral is a limit of Riemann sums.
- For very simple functions, it is possible to directly compute Riemann sums and then take the limit.
- Though this topic is very important theoretically, in practice we will compute integrals by using the First Fundamental Theorem of Calculus.

### Riemann Sums, Definite Integrals, 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}dxGiven the values below, use right-hand Riemann Sum with 4 intervals to approximate ∫

f(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

_{0}^{10}f(x) dxf(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

- ∫
_{0}^{10}f(x) dx ≈ 5(3 − 0) + 5.5 (4 − 3) + 3(7 − 4) + 7(10 − 7) - ∫
_{0}^{10}f(x) dx ≈ 15 + 5.5 + 9 + 21

∫

_{0}^{10}f(x) dx ≈ 50.5Given the values below, use left-hand Riemann Sum with 4 intervals to approximate ∫

f(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

_{0}^{10}f(x) dxf(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

- ∫
_{0}^{10}f(x) dx ≈ 2(3 − 0) + 5(4 − 3) + 5.5(7 − 4) + 3(10 − 7) - ∫
_{0}^{10}f(x) dx ≈ 6 + 5 + 16.5 + 9

∫

_{0}^{10}f(x) dx ≈ 36.5Given the values below, use right-hand Riemann Sum with 2 intervals to approximate ∫

f(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

_{0}^{4}f(x) dxf(0) = 2 f(3) = 5 f(4) = 5.5 f(7) = 3 f(10) = 7

- ∫
_{0}^{4}f(x) dx ≈ 5(3 − 0) + 5.5(4 − 3) - ∫
_{0}^{4}f(x) dx ≈ 15 + 5.5

∫

_{0}^{4}f(x) dx ≈ 20.5Given the values below, use right-hand Riemann Sum with 5 intervals to approximate ∫

f(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

_{−5}^{5}f(x) dxf(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

- ∫
_{−5}^{5}f(x) dx ≈ 1(−1 − (−5)) + 0(0 − (−1)) + 4(2 − 0) + 16(4 − 2) + 25(5 − 4) - ∫
_{−5}^{5}f(x) dx ≈ 4 + 0 + 8 + 32 + 25

∫

_{−5}^{5}f(x) dx ≈ 69Given the values below, use right-hand Riemann Sum with 3 intervals to approximate ∫

f(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

_{0}^{5}f(x) dxf(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

- ∫
_{0}^{5}f(x) dx ≈ 4(2 − 0) + 16(4 − 2) + 25(5 − 4) - ∫
_{0}^{5}f(x) dx ≈ 8 + 32 + 25

∫

_{0}^{5}f(x) dx ≈ 65Given the values below, use left-hand Riemann Sum with 2 intervals to approximate ∫

f(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

_{−5}^{0}f(x) dxf(−5) = 25 f(−1) = 1 f(0) = 0 f(2) = 4 f(4) = 16 f(5) = 25

- ∫
_{−5}^{0}f(x) dx ≈ 25(−1 − (−5)) + 1(0 − (−1)) - ∫
_{−5}^{0}f(x) dx ≈ 100 + 1

∫

_{−5}^{0}f(x) dx ≈ 101Given the values below, use right-hand Riemann Sum with 4 intervals to approximate ∫

f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8

_{−2}^{2}f(x) dxf(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8

- ∫
_{−2}^{2}f(x) dx ≈ −1(−1 −(−2)) + 0(0 − (−1)) + 1(1 − 0) 8(2 − 1) - ∫
_{−2}^{2}f(x) dx ≈ −1 + 0 + 1 + 8

∫

_{−2}^{2}f(x) dx ≈ 8Given the values below, use left-hand Riemann Sum with 4 intervals to approximate ∫

f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8

_{−2}^{2}f(x) dxf(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8

- ∫
_{−2}^{2}f(x) dx ≈ −8(−1 − (−2)) + −1(0 − (−1)) + 0(1 − 0) + 1(2 − 1) - ∫
_{−2}^{2}f(x) dx ≈ −8 − 1 + 0 + 1

∫

_{−2}^{2}f(x) dx ≈ −8Given the values below, use right-hand Riemann Sum with 6 intervals to approximate ∫

f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8 f(4) = 64 f(8) = 512 f(10) = 1000 f(12) = 1728

_{0}^{12}f(x) dxf(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8 f(4) = 64 f(8) = 512 f(10) = 1000 f(12) = 1728

- ∫
_{0}^{12}f(x) dx ≈ 1(1 − 0) + 8(2 − 1) + 64(4 − 2) + 512(8 − 4) + 1000(10 − 8) + 1728(12 − 10) - ∫
_{0}^{12}f(x) dx ≈ 1 + 8 + 128 + 2048 + 2000 + 3456

∫

_{0}^{12}f(x) dx ≈ 7641Given the values below, use right-hand Riemann Sum with 6 intervals to approximate ∫

f(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8 f(4) = 64 f(8) = 512 f(10) = 1000 f(12) = 1728

_{0}^{8}f(x) dxf(−2) = −8 f(−1) = −1 f(0) = 0 f(1) = 1 f(2) = 8 f(4) = 64 f(8) = 512 f(10) = 1000 f(12) = 1728

- ∫
_{0}^{8}f(x) dx ≈ 1(1 − 0) + 8(2 − 1) + 64(4 − 2) + 512(8 − 4) - ∫
_{0}^{8}f(x) dx ≈ 1 + 8 + 128 + 2048

∫

_{0}^{8}f(x) dx ≈ 2185*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

### Riemann Sums, Definite Integrals, 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
- Important Equations 0:22
- Riemann Sum
- Integral
- Integrand
- Limits of Integration (Upper Limit, Lower Limit)
- Other Equations
- Fundamental Theorem of Calculus
- Lecture Example 1 5:04
- Lecture Example 2 10:43
- Lecture Example 3 13:52
- Additional Example 4
- Additional Example 5

0 answers

Post by Israel Haile on October 23, 2015

I loved your methodology ,I wanna encourage you, I am 99.99% satisfied with your class,Stay Blessed !!!

0 answers

Post by Valeriya Pinkhasova on February 5, 2015

I am unable to watch additional example. Everytime I press on it ,lecture starts from the beginning .

0 answers

Post by Jason Kim on November 5, 2014

wait for the antideriviative of the lecture example 2 how do you exactly know that anti deriviative of 2x-5 is just x^2-5X. It could also be x^2-5x plus some number because the number just disappears when we derive it. So it could be x^2-5x+9 and the deriviative would be also 2x-5.

How can you assume that the antideriviative is just x^2-5x?

0 answers

Post by Johnny Zamora on January 14, 2014

She is Amazing

0 answers

Post by Jose Gonzalez-Gigato on November 10, 2013

'Additional Example 5' was excellent!

1 answer

Last reply by: Martina Alvarez

Fri Dec 9, 2011 4:28 AM

Post by Martina Alvarez on December 9, 2011

Ex: 2

13 min.

=(36-30) - (1-5)

=6 - (-4)

=10 not 8

0 answers

Post by David Bascom on September 4, 2011

In lecture example 3 where did 3xsquared come from?

1 answer

Last reply by: Jose Gonzalez-Gigato

Sun Nov 10, 2013 11:30 AM

Post by Okwudili Ezeh on July 21, 2011

The graph was supposed to be a sine graph not a cosine graph.