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

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

Loading video...

This is a quick preview of the lesson. For full access, please Log In or Sign up.

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

## Discussion

## Study Guides

## Practice Questions

## Download Lecture Slides

## Table of Contents

## Related Books

### Area Between Curves

- I recommend always starting with a sketch and drawing in a “sample rectangle.”
- If an interval is not given, you may need to set the two functions equal in order to determine the interval involved.
- If two curves cross, then you will need to break up the integral into more than one integral.
- Sometimes, geometrical considerations can help you to check your results or even provide a non-calculus way to finish a problem.

### Area Between Curves

Find the area enclosed by y = 4 and y = x

^{2}- First, we find the bounds. To do so, we set the equations equal to each other
- x
^{2}= 4 - x = ±2
- A = ∫
_{−2}^{2}4 − x^{2}dx - A = (4x − [(x
^{3})/3]) |_{−2}^{2} - A = 8 − [8/3] − (−8 − [(−8)/3])
- A = 16 − [16/3]

A = [32/3]

Find the area under the velocity curve v(t) = 3t

^{2}+ t from t = 0 to t = 2- A = ∫
_{0}^{2}3t^{2}+ t − 0 dt - The subtraction by zero has no effect. It is just to illustrate that we're taking the x-axis as the lower-bound.
- A = (t
^{3}+ [(t^{2})/2]) |_{0}^{2} - A = 8 + 2 − 0 − 0
- The area under a velocity curve also happens to be the displacement!
- ∫
_{t1}^{t2}v(t) dt = x(t_{2}) − x(t_{1})

A = 10

Find the area under y = e

^{x}from x = 0 to x = ln10- A = ∫
_{0}^{ln10}e^{x}dx - A = e
^{x}|_{0}^{ln10} - A = e
^{ln10}− e^{0} - A = 10 − 1
- A = 9

A = 9

Find the area under y = e

^{[t/3]}from t = 0 to t = 6- A = ∫
_{0}^{6}e^{[t/3]}dt - u = [t/3]
- du = [1/3] dt
- A = 3 ∫
_{0}^{6}e^{u}du - A = 3 e
^{[t/3]}|_{0}^{6} - A = 3 (e
^{2}− e^{0})

A = 3(e

^{2}− 1)Find the area bounded by y = x

^{2}and y = |x|- Find the bounds of the integral
- x
^{2}= |x| - x
^{2}= √{x^{2}} - x
^{4}= x^{2} - x
^{2}= 1 - x = ±1
- y = |x| has different behavior on either side of zero. We can treat this as a piecewise function with the area as the sum of two different integrals
- A = ∫
_{−1}^{0}−x − x^{2}dx + ∫_{0}^{1}x − x^{2}dx - A = (−[(x
^{2})/2] − [(x^{3})/3]) |_{−1}^{0}+ ([(x^{2})/2] − [(x^{3})/3]) |_{0}^{1} - A = (−0 − 0) − (−[((−1)
^{2})/2] − [((−1)^{3})/3]) + ([1/2] − [1/3]) − 0 - A = [1/2] − [1/3] + [1/2] − [1/3]
- A = 1 − [2/3]

A = [1/3]

Find the area bound by x = y

^{3}and the y-axis from y = 0 to y = 2- A = ∫
_{0}^{2}y^{3}dy - A = [(y
^{4})/4] |_{0}^{2} - A = [(2
^{4})/4] − 0

A = 4

Find the area bound by y = sinx and the x-axis from x = 0 to x = 2π

- A = ∫
_{0}^{2π}sinx dx - A = − cosx |
_{0}^{2π} - A = −cos2π− (− cos0)
- A = −1 + 1 = 0
- An area of zero doesn't quite make sense here, so using the straight integral is insufficient. What we can do is treat this as two separate integrals, one where the area is above the x-axis and one where it is below and add their effective area.
- sinx = 0, x = 0, π, 2π in this interval.
- A = ∫
_{0}^{π}sinx dx + (−∫_{π}^{2π}sinx dx) - The negative sign is there to compensate for the "negative area"
- A = −cosx |
_{0}^{π}+ cosx |_{π}^{2π} - A = − cosπ+ cos0 + cos2π− cosπ
- A = 1 + 1 + 1 + 1

A = 4

Find the area bound by x = y

^{2}and y = x − 2- We could find the integral in terms of dx, but the we would have to split it up into at least two separate integrals. Besides, x = y
^{2}is not a function of x (there are two outputs for every x input). - x = y
^{2} - x = y + 2
- Find the limits of the integral
- y
^{2}= y + 2 - y
^{2}− y − 2 = 0 - (y + 1)(y − 2) = 0
- y = 2, −1
- A = ∫
_{−1}^{2}(y + 2) − y^{2}dy - A = ([(y
^{2})/2] + 2x − [(y^{3})/3]) |_{−1}^{2} - A = (2 + 4 − [8/3]) − ([1/2] − 2 + [1/3])
- A = 8 − [1/2] − [9/3]
- The area above the

A = [9/2]

Find the area bound by y = x

^{2}+ 1 and y = −x^{2}+ 3- Find the limits of the integral
- x
^{2}+ 1 = −x^{2}+ 3 - 2x
^{2}= 2 - x
^{2}= 1 - x = ±1
- A = ∫
_{−1}^{1}(−x^{2}+ 3) − (x^{2}+ 1) dx - A = ∫
_{−1}^{1}−2x^{2}+ 2 dx - A = (−[(2x
^{3})/3] + 2x) |_{−1}^{1} - A = −[2/3] + 2 − (−[(−2)/3] − 2)
- A = −[2/3] + 2 − [2/3] + 2
- A = 4 − [4/3]

A = [8/3]

Find the area bound by y = x

^{2}and y = 2x- Find the limits
- x
^{2}= 2x - x
^{2}− 2x = 0 - x(x − 2) = 0
- x = 0, 2
- A = ∫
_{0}^{2}2x − x^{2}dx - A = x
^{2}− [(x^{3})/3] |_{0}^{2} - A = 4 − [8/3] − 0

A = [4/3]

*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 Between Curves

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 Between Two Curves 0:12
- Graphic Description
- Lecture Example 1 1:44
- Lecture Example 2 5:39
- Lecture Example 3 8:45
- Additional Example 4
- Additional Example 5

0 answers

Post by Eti Sinha on September 8, 2014

For additional example #5, is it possible to find the area underneath the curve by: Integral of ( 1/y ) dy, from 1/3 to 3? Using dy as opposed to dx?

0 answers

Post by Rufino Hernandez on July 5, 2014

Asides from getting the area of this triangle using this method, you can also get it using another method. The Area of a triangle is square root of (s(s-a)(s-b)(s-c)). Where s=(1/2)(a+b+c). Where a is one side of the triangle, b is another side of the triangle, and c is the third side of the triangle. Since the Triangle's lower angle is the same as the angle made by the function y=|x|, it can be deduced that the angle made is 90 degrees. If it is a 90 degree triangle, then we can simply use the Area of a triangle to come to the conclusion of the area; A=(1/2)b*h, where b=base and h=height. Since we have the end points of the sides of the triangle, we can simply derive sides. For example, the right side has an endpoint of (2,2); y=2, x=2. Using this we can make a second triangle where the sides are 2 and 2 which can lead us to find the third side which is square root of (8) = 2 * square root of(2). We now know the length of the right side of the triangle. Using this same same method, we can derive the length of the base of the triangle (bottom left). The endpoint of this side is (-2/3, -2/3). So we can make a third triangle that has these sides and derive the length of the second side of the original triangle. This second side will be (-2/3)^2+(-2/3)^2=c^2 ==> (4/9)+(4/9)= c^2 ==> 8/9=c^2 ==> c= (2 * square root of (2))/3. Now since we have the height and base, we can simply put this into the Area of a triangle; A=(1/2)(base)(height) ==> A=(1/2)((2square root of (2))/2)(2 * square root of (2) ==> 4/3. Or we can solve for the third side of the triangle and input them into the (The Area of a triangle is square root of (s(s-a)(s-b)(s-c)). Where s=(1/2)(a+b+c). Where a is one side of the triangle, b is another side of the triangle, and c is the third side of the triangle.) equation and get the same results. This method will yield the same answer as the method stated in this video.

0 answers

Post by Fadel Hanoun on December 17, 2013

What about when x = 5pi/4, where Tan = 1?

0 answers

Post by Troy Ling on February 4, 2011

k= -15/16

1 answer

Last reply by: Nopparat SAntisathaporn

Wed Oct 6, 2010 2:36 AM

Post by Nopparat SAntisathaporn on October 6, 2010

Hi there, I don't know how to do number 7 in this test http://www.nzqa.govt.nz/nqfdocs/ncea-resource/exams/2007/90286-exm-07.pdf

Can you please explain how to solve for K. Thank You.