## 17Calculus Integrals - Area Between Curves

Area between curves is a very natural extension of 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. Make sure you fully understand how to calculate area under a curve before working through this material.

If you want a full lecture on area between two curves, we recommend this video from one of our favorite lecturers.

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

video by Prof Leonard

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 [9min-50secs]

video by PatrickJMT

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.

5. The area between curves will always be positive regardless of where the area lies in the xy-plane. So, even though you can get a negative or zero area for area UNDER a curve, area BETWEEN curves will always be positive.

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

### Firefly - Area Between Curves [6min-9secs]

video by Firefly

6. 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 [7min-36secs]

video by PatrickJMT

Okay, let's work some practice problems.

Practice

Unless otherwise instructed, calculate the area bounded by these curves, giving your answer in exact form.

Basic

Set up both integrals (x-direction and y-direction) to calculate the area between the curves $$y=\sqrt{x}$$ and $$y=x/2$$ but do not evaluate the integrals.

Problem Statement

Set up both integrals (x-direction and y-direction) to calculate the area between the curves $$y=\sqrt{x}$$ and $$y=x/2$$ but do not evaluate the integrals.

Solution

### 3789 video

video by blackpenredpen

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=e^x;$$ $$y=xe^x;$$ $$x=0$$

Solution

### 1158 video

video by Krista King Math

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$x=y^2;$$ $$x=y+6$$

Solution

### 1160 video

video by Krista King Math

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x^2-4x;$$ $$y=2x$$

Solution

### 1161 video

video by PatrickJMT

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=8-x^2;$$ $$y=x^2;$$ $$x=-3;$$ $$x=3$$

Solution

### 1162 video

video by PatrickJMT

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$x=y^2;$$ $$x=2-y^2$$

Solution

### 1165 video

video by PatrickJMT

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x+1;$$ $$y=9-x^2$$ $$x=-1;$$ $$x=2$$

Solution

### 1166 video

video by Krista King Math

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$$y=x;$$ $$y=x^2$$

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x;$$ $$y=x^2$$

Solution

### 1167 video

video by Krista King Math

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x^3;$$ $$y=3x-2$$

Solution

### 1169 video

video by MIT OCW

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=2x;$$ $$y=5x-x^2$$

Solution

### 1171 video

video by MathTV

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=e^x;$$ $$y=x^2;$$ $$x=0;$$ $$x=2$$

Solution

### 2121 video

video by Michel vanBiezen

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x^2-2x;$$ $$y=-2x^2+7x$$

Solution

### 2122 video

video by Michel vanBiezen

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$x=y^2;$$ $$y=x-2$$

Solution

### 2123 video

video by Michel vanBiezen

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=(1/4)x^2;$$ $$y=x$$

Solution

### 2124 video

video by Michel vanBiezen

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=(x-2)^2+5$$; $$x=1$$; $$y=x/2-2$$; $$x=4$$

Solution

### 2125 video

video by Michel vanBiezen

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=10/(x+1)$$; $$y=\sqrt{x};$$ $$x=1$$

Solution

### 2126 video

video by Michel vanBiezen

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$$y = \sqrt{x-1};$$ $$y = x/3 + 1/3$$

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y = \sqrt{x-1};$$ $$y = x/3 + 1/3$$

1/6

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y = \sqrt{x-1};$$ $$y = x/3 + 1/3$$

Solution

### 2430 video

video by Michel vanBiezen

1/6

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$$y=x^3; y=x$$

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x^3; y=x$$

Hint

To simplify your work, use symmetry.

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x^3; y=x$$

1/2

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x^3; y=x$$

Hint

To simplify your work, use symmetry.

Solution

### 2431 video

video by Michel vanBiezen

1/2

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$$y=x+2;$$ $$y=4-x^2$$; $$x=-3$$

Problem Statement

Calculate the area between bounded by curves, giving your answer in exact form. $$y=x+2;$$ $$y=4-x^2$$; $$x=-3$$

Hint

The area that you are asked to calculate is in two sections, as shown in this plot. built with GeoGebra

Problem Statement

Calculate the area between bounded by curves, giving your answer in exact form. $$y=x+2;$$ $$y=4-x^2$$; $$x=-3$$

$$31/6$$

Problem Statement

Calculate the area between bounded by curves, giving your answer in exact form. $$y=x+2;$$ $$y=4-x^2$$; $$x=-3$$

Hint

The area that you are asked to calculate is in two sections, as shown in this plot. built with GeoGebra

Solution

### 2432 video

video by Michel vanBiezen

$$31/6$$

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Intermediate

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

Problem Statement

Calculate the area of the triangle with vertices (0,0), (3,1) and (1,2). Give your answer in exact form.

Solution

### 1157 video

video by Krista King Math

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

Problem Statement

Find the area between two consecutive points of intersection of $$y=\sin(x)$$ and $$y=\cos(x)$$. Give your answer in exact form.

Solution

### 1170 video

video by MIT OCW

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$$y=12x-3x^2;$$ $$x=1$$ $$y=6x-24;$$ $$x=7$$

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=12x-3x^2;$$ $$x=1$$ $$y=6x-24;$$ $$x=7$$

Solution

### 1168 video

video by Krista King Math

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$x=2y^2;$$ $$x=4+y^2$$

Solution

### 1159 video

video by Krista King Math

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$x=y^3-y;$$ $$x=1-y^4$$

Solution

The solution to this problem is shown in two consecutive videos. He sets up the integral in the first video and then evaluates it in the second video.

### 1163 video

video by PatrickJMT

### 1163 video

video by PatrickJMT

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=x+2;$$ $$y=\sqrt{x};$$ $$y=2;$$ $$y=0$$

Solution

### 1164 video

video by PatrickJMT

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=e^x;$$ $$x=-1;$$ $$y=x^2-1;$$ $$x=1$$

Solution

### 44 video

video by Krista King Math

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=-x-1$$; $$y=x-1$$; $$y=1-x^2$$

Solution

### 2112 video

video by PatrickJMT

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

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=1$$; $$y=7-x$$; $$y=\sqrt{x}+1$$

Solution

### 2113 video

video by PatrickJMT

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$$y=2x/\pi;$$ $$y=\sin x;$$ $$x=\pi;$$ $$x\geq 0$$

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=2x/\pi;$$ $$y=\sin x;$$ $$x=\pi;$$ $$x\geq 0$$

Hint

The area that you are asked to calculate is in two sections, as shown in this plot. built with GeoGebra

Problem Statement

Calculate the area bounded by these curves, giving your answer in exact form. $$y=2x/\pi;$$ $$y=\sin x;$$ $$x=\pi;$$ $$x\geq 0$$

Hint

The area that you are asked to calculate is in two sections, as shown in this plot. built with GeoGebra

Solution

### 2444 video

video by Michel vanBiezen

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You CAN Ace Calculus

 integration definite integrals

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.

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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