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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

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Related Books

### Area Under A Curve

- Approximating area under curve using rectangles
- Setup appropriate intervals and evaluate
*f(x*_{i}*)* - Sum all of
*f(x*_{i}*)* - To avoid confusion, think physical, geometric area, rather than calculus

### Area Under A Curve

Find the area under the curve f(x) = x on interval [0,5] using 5 rectangles and the right-endpoint

- f(1) = 1
- f(2) = 2
- f(3) = 3
- f(4) = 4
- f(5) = 5
- Area under the curve is approximately equal to [(5 − 0)/5](1 + 2 + 3 + 4 + 5)

Area is approximately equal to 15

Find the area under the curve f(x) = x on interval [0,5] using 5 rectangles and the left-endpoint

- f(0) = 0
- f(1) = 1
- f(2) = 2
- f(3) = 3
- f(4) = 4
- Area under the curve is approximately equal to [(5 − 0)/5](0 + 1 + 2 + 3 + 4)

Area under the curve is approximately equal to 10

Find the area under the curve f(x) = x

^{2}on interval [−2,2] using 4 rectangles and the right-endpoint- f(−1) = 1
- f(0) = 0
- f(1) = 1
- f(2) = 4
- Area under the curve is approximately equal to [(2 − (−2))/4](1 + 0 + 1 + 4)

Area under the curve is approximately equal to 6

Find the area under the curve f(x) = 3cosx on interval [−[(π)/2],[(π)/2]] using 4 rectangles and the right-endpoint

- f(−[(π)/4]) = 3 cos−[(π)/4] = [3/(√2)]
- f(0) = 3 cos0 = 3
- f([(π)/4]) = 3 cos[(π)/4] = [3/(√2)]
- f([(π)/2]) = 3 cos[(π)/2] = 0
- Area under the curve is approximately equal to [([(π)/2] − (−[(π)/2]))/4]([3/(√2)] + 3 + [3/(√2)] + 0)

A ≈ [(π)/4](3 + [6/(√2)])

Find the area under the curve f(x) = x + 3 on interval [0,1] using 4 rectangles and the right-endpoint

- Find the width of each rectangle
- [(1 − 0)/4] = .25
- f(.25) = 3.25
- f(.5) = 3.5
- f(.75) = 3.75
- f(1) = 4
- Area under the curve is approximately equal to .25(3.25 + 3.5 + 3.75 + 4)

A ≈ 3.625

Find the area under the curve f(x) = x + 3 on interval [0,1] using 10 rectangles and the right-endpoint

- Find the width of each rectangle
- [(1 − 0)/10] = .1
- f(.1) = 3.1
- f(.2) = 3.2
- f(.3) = 3.3
- f(.4) = 3.4
- f(.5) = 3.5
- f(.6) = 3.6
- f(.7) = 3.7
- f(.8) = 3.8
- f(.9) = 3.9
- f(1) = 4
- Area under the curve is approximately equal to [(1 − 0)/10](3.1 + 3.2 + 3.3 + 3.4 + 3.5 + 3.6 + 3.7 + 3.8 + 3.9 + 4)

A ≈ 3.55

Find the area under the curve f(x) = x

^{2}+ 3x + 1 on interval [−1,1] using 2 rectangles and the right-endpoint- [(1 − (−1))/2] = 1
- f(0) = 1
- f(1) = 1 + 3 + 1 = 5
- Area under the curve is approximately 1(1 + 5)

A ≈ 6

Find the area under the curve f(x) = sin2x on interval [0, [(π)/2]] using 2 rectangles and the left-endpoint

- [([(π)/2] − 0)/2] = [(π)/4]
- f(0) = sin0 = 0
- f([(π)/4]) = [1/(√2)]
- Area under the curve is approximately equal to [(π)/4](0 + [1/(√2)])

A ≈ [(π)/(4√2)]

Find the area under the curve f(x) = sinx on interval [−[(π)/2], [(π)/2]] using 4 rectangles and the left-endpoint

- [([(π)/2] − (−[(π)/2]))/4] = [(π)/4]
- f(−[(π)/2]) = sin[(−π)/2] = −1
- f(−[(π)/4]) = sin[(−π)/4] = −[1/(√2)]
- f(0) = sin0 = 0
- f([(π)/4]) = sin[(π)/4] = [1/(√2)]
- Area under the curve is approximately equal to [(π)/4](−1 − [1/(√2)] + 0 + [1/(√2)])

A ≈ −[(π)/4]

Find the area under the curve f(x) = sinx on interval [−[(π)/2], [(π)/2]] using 4 rectangles and the right-endpoint

- [([(π)/2] − (−[(π)/2]))/4] = [(π)/4]
- f(−[(π)/4]) = sin[(−π)/4] = −[1/(√2)]
- f(0) = sin0 = 0
- f([(π)/4]) = sin[(π)/4] = [1/(√2)]
- f([(π)/2]) = sin[(π)/2] = 1
- Area under the curve is approximately equal to [(π)/4](− [1/(√2)] + 0 + [1/(√2)] + 1)

A ≈ [(π)/4]

*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

### Area Under A Curve

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
- Area Under Curve 0:07
- Definition of Integral
- Left Endpoint
- Right Endpoint
- Midpoints
- Example 1 2:40
- Example 2 4:59
- Example 3 8:48
- Example 4 10:23
- Example 5 12:30
- Example 6 15:32

0 answers

Post by mohammad salem on March 29, 2013

example 5 is wrong, the width should be pi/4.

0 answers

Post by GUOJI LIU on January 29, 2013

he did worry.

1 answer

Last reply by: Arshin Jain

Sun Feb 9, 2014 5:16 AM

Post by James Xie on July 23, 2012

for example 5, shouldn't the answer have pi/4 instead of pi/2? The x-value of each rectangle equals pi/4 not pi/2.