INSTRUCTORS Raffi Hovasapian John Zhu

Start learning today, and be successful in your academic & professional career. Start Today!

• Related Books

 0 answersPost by mohammad salem on March 29, 2013example 5 is wrong, the width should be pi/4. 0 answersPost by GUOJI LIU on January 29, 2013he did worry. 1 answerLast reply by: Arshin JainSun Feb 9, 2014 5:16 AMPost by James Xie on July 23, 2012for 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.

Area Under A Curve

• Approximating area under curve using rectangles
• Setup appropriate intervals and evaluate f(xi)
• Sum all of f(xi)
• 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) = x2 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) = x2 + 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.

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