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You CAN Ace Calculus

17calculus > integrals > area under/between curves

### Calculus Main Topics

Integrals

Integral Applications

Single Variable Calculus

Multi-Variable Calculus

### Tools

math tools

general learning tools

related topics on other pages

For discussion of area under a parametric curve, see the parametric calculus page.

For discussion of area under a polar curve, see the polar calculus page.

Applied Integration - Area

on this page: ► area under a curve     ► area between curves

Area Under A Curve - Area under a curve is just a special case of area between two curves where the lower curve is the x-axis.

Area Under A Curve

To find the area under a curve that is above the x-axis, you just need to integrate the curve between two specific points. This is a natural idea from definite integrals. Here are a couple of videos explaining this.

 rootmath - Intro to Area Under a Curve [4min-28secs]
 rootmath - Exact Area Under A Curve [9min-55secs]

Okay, to really understand this idea, let's work these practice problems.
Calculate the area under these curves.

Practice 1

$$y=x^2 ~~ [0,4]$$

solution

Practice 2

$$y=x^3 ~~ [1,3]$$

solution

Practice 3

$$y=x^2-2x+8 ~~ [1,2]$$

solution

Practice 4

$$y=1/x^2 ~~ [1,4]$$

solution

Practice 5

$$y=\sin x ~~ [0,\pi]$$

solution

Practice 6

$$y=\sin x \cos x$$ $$[\pi/4, \pi/2]$$

solution

Practice 7

$$\displaystyle{ y=\frac{x^2}{\sqrt{x^3+9}} ~~ [-1,1] }$$

solution

Area Between Curves

Want a full lecture on area between curves? Check out this video.

 Prof Leonard - Finding Area Between Two Curves [1hr-33min-46secs]

Area between curves is a very natural extension of area under a curve. You know that the integral $$\int_{a}^{b}{f(x)~dx}$$ gives you the area under the curve of $$f(x)$$ between a and b. You can think of it in terms of rectangles again. You are taking rectangles of height $$f(x)$$ and width $$dx$$, multiplying them together and adding them up. Think of the height of the rectangle as being $$f(x) - 0$$ where $$f(x)$$ is the y-value above the x-axis and zero is the distance above the x-axis for the line $$y=0$$. So if your two curves are $$y=f(x)$$ and $$y=0$$, the height of the rectangle is $$f(x)-0$$ and the area between the two curves is $$\int_{a}^{b}{( f(x) - 0) ~dx} = \int_{a}^{b}{f(x)~dx}$$.

Now, if we extend this to two general curves, $$f(x)$$ and $$g(x)$$ with $$f(x) \geq g(x)$$ in the entire interval $$[a,b]$$, the area between these curves is $$\int_{a}^{b}{ f(x) - g(x) ~dx}$$. You can think of the height $$f(x)-g(x)$$ being swept from a to b, and multiplied by the width $$dx$$ along the way and summing all those areas up to give the area between the curves. It's that easy.

Okay, let's watch a video clip explaining this with a graph.

 PatrickJMT - Finding Areas Between Curves

Things To Watch For

1. Notice in the last paragraph above, we said that $$f(x) \geq g(x)$$ in the entire interval $$[a,b]$$. This is absolutely essential or the integral you calculate will not be the area between the curves. You can break the integral into sections where the functions cross and evaluate each integral separately and add the results together.

2. It doesn't matter if the functions are above or below the x-axis. As long as you have point 1 above covered, you are good.

3. It will help a lot (and I require it when I teach this material to help my students) to plot the equations, shade the area that you are asked to find and draw a rectangle(s) somewhere within the area in question. This helps you to visualize the area and the concept of a sweeping rectangle.

4. If you follow the advice in the point about drawing a rectangle and thinking about the rectangle sweeping across an area, notice carefully about what happens at each end of the rectangle. If the ends of the rectangle stay on the same curve all the way across the area, you need only one integral to calculate the area. However, if the rectangle jumps to another curve, you need to break the rectangle at the switching point and set up a new integral for the area, then add the areas together to get the total area.

Okay, let's pause and watch a video. Here is a good explanation of several of the points above.

 Firefly - Area Between Curves

5. One twist that you will see is that the integration is done in the other direction, i.e. $$\int_{c}^{d}{ p(y) - q(y) ~dy}$$. In this case, you need $$p(y) \geq q(y)$$ in the entire interval $$[c,d]$$. Using the same concept, you can think about the rectangle with height $$p(y)-q(y)$$ and width $$dy$$ sweeping from bottom to top of the area you are calculating.

Here is a quick video clip showing this idea of integrating with respect to y. He also shows why, in some cases, it is better to integrate in the y-direction instead of the x-direction.

 PatrickJMT - Area Between Curves - Integrating with Respect to y

### Search 17Calculus

Instructions - - Unless otherwise instructed, calculate the area between the curves, giving your answers in exact terms.

 Basic Problems

Practice 8

For the region bounded by the graphs $$y=9-x^2$$, $$x=2$$ and $$x=3$$ in the first quadrant, a. accurately sketch the region; b. using calculus, calculate the area of the region that you sketched in part a.

solution

Practice 9

$$y=e^x; ~ y=xe^x; ~ x=0$$

solution

Practice 10

$$x=y^2; ~~~ x=y+6$$

solution

Practice 11

$$y=x^2-4x; ~~~ y=2x$$

solution

Practice 12

$$y=8-x^2; ~ y=x^2;$$ $$x=-3; ~ x=3$$

solution

Practice 13

$$x=y^2; ~ x=2-y^2$$

solution

Practice 14

$$y=x+1; ~ y=9-x^2$$ $$x=-1; ~ x=2$$

solution

Practice 15

$$y=x;$$    $$y=x^2$$

solution

Practice 16

$$y=x^3; ~ y=3x-2$$

solution

Practice 17

$$y=2x; ~ y=5x-x^2$$

solution

Practice 18

$$y=e^x; ~ y=x^2;$$ $$x=0; ~ x=2$$

solution

Practice 19

$$y=x^2-2x;$$ $$y=-2x^2+7x$$

solution

Practice 20

$$x=y^2; ~~ y=x-2$$

solution

Practice 21

$$y=(1/4)x^2; ~~ y=x$$

solution

Practice 22

$$y=(x-2)^2+5$$; $$y=x/2-2$$; $$x=1$$; $$x=4$$

solution

Practice 23

$$y=10/(x+1)$$; $$y=\sqrt{x}$$; $$x=1$$

solution

 Intermediate Problems

Practice 24

Calculate the area of the triangle with vertices (0,0), (3,1) and (1,2).

solution

Practice 25

Find the area between two consecutive points of intersection of $$y=\sin(x)$$ and $$y=\cos(x)$$.

solution

Practice 26

$$f(x)=12x-3x^2;$$ $$x=1$$ $$g(x)=6x-24;$$ $$x=7$$

solution

Practice 27

$$x=2y^2; ~ x=4+y^2$$

solution

Practice 28

$$x=y^3-y; ~ x=1-y^4$$

solution

Practice 29

$$y=x+2; ~ y=\sqrt{x};$$ $$y=2; ~ y=0$$

solution

Practice 30

$$y=e^x;$$ $$x=-1;$$ $$y=x^2-1;$$ $$x=1$$

solution

Practice 31

$$y=-x-1$$; $$y=x-1$$; $$y=1-x^2$$

solution

Practice 32

$$y=1$$; $$y=7-x$$; $$y=\sqrt{x}+1$$

solution

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