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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.

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area between curves youtube playlist

17Calculus Subjects Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

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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

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

Basic Problems

\(y=e^x; ~~ y=xe^x; ~~ x=0\)

Problem Statement

\(y=e^x; ~~ y=xe^x; ~~ x=0\)

Solution

1158 solution video

video by Krista King Math

close solution

\( x=y^2; ~~~ x=y+6 \)

Problem Statement

\( x=y^2; ~~~ x=y+6 \)

Solution

1160 solution video

video by Krista King Math

close solution

\( y=x^2-4x; ~~~ y=2x \)

Problem Statement

\( y=x^2-4x; ~~~ y=2x \)

Solution

1161 solution video

video by PatrickJMT

close solution

\(y=8-x^2; ~~~ y=x^2;\) \(x=-3; ~~~ x=3\)

Problem Statement

\(y=8-x^2; ~~~ y=x^2;\) \(x=-3; ~~~ x=3\)

Solution

1162 solution video

video by PatrickJMT

close solution

\(x=y^2; ~~~ x=2-y^2\)

Problem Statement

\(x=y^2; ~~~ x=2-y^2\)

Solution

1165 solution video

video by PatrickJMT

close solution

\(y=x+1; ~~~ y=9-x^2\) \(x=-1; ~~~ x=2\)

Problem Statement

\(y=x+1; ~~~ y=9-x^2\) \(x=-1; ~~~ x=2\)

Solution

1166 solution video

video by Krista King Math

close solution

\(y=x;\)    \(y=x^2\)

Problem Statement

\(y=x;\)    \(y=x^2\)

Solution

1167 solution video

video by Krista King Math

close solution

\(y=x^3; ~~~ y=3x-2\)

Problem Statement

\(y=x^3; ~~~ y=3x-2\)

Solution

1169 solution video

video by MIT OCW

close solution

\(y=2x; ~~~ y=5x-x^2\)

Problem Statement

\(y=2x; ~~~ y=5x-x^2\)

Solution

1171 solution video

video by MathTV

close solution

\(y=e^x; ~ y=x^2;\) \(x=0; ~ x=2\)

Problem Statement

\(y=e^x; ~ y=x^2;\) \(x=0; ~ x=2\)

Solution

2121 solution video

video by Michel vanBiezen

close solution

\( y=x^2-2x; \) \( y=-2x^2+7x \)

Problem Statement

\( y=x^2-2x; \) \( y=-2x^2+7x \)

Solution

2122 solution video

video by Michel vanBiezen

close solution

\( x=y^2; ~~~ y=x-2 \)

Problem Statement

\( x=y^2; ~~~ y=x-2 \)

Solution

2123 solution video

video by Michel vanBiezen

close solution

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

Problem Statement

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

Solution

2124 solution video

video by Michel vanBiezen

close solution

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

Problem Statement

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

Solution

2125 solution video

video by Michel vanBiezen

close solution

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

Problem Statement

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

Solution

2126 solution video

video by Michel vanBiezen

close solution

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

Problem Statement

Calculate the area bounded by the curves \( y=\sqrt{x-1}; y=x/3+ 1/3\)

Final Answer

1/6

Problem Statement

Calculate the area bounded by the curves \( y=\sqrt{x-1}; y=x/3+ 1/3\)

Solution

2430 solution video

video by Michel vanBiezen

Final Answer

1/6

close solution

\(y=x^3; y=x\)

Problem Statement

Calculate the area bounded by the curves \(y=x^3; y=x\)

Hint

To simplify your work, use symmetry.

Problem Statement

Calculate the area bounded by the curves \(y=x^3; y=x\)

Final Answer

1/2

Problem Statement

Calculate the area bounded by the curves \(y=x^3; y=x\)

Hint

To simplify your work, use symmetry.

Solution

2431 solution video

video by Michel vanBiezen

Final Answer

1/2

close solution

Given \(y=x+2\) and \(y=4-x^2\). Calculate the area between the curves and the line \(x=-3\).

Problem Statement

Given \(y=x+2\) and \(y=4-x^2\). Calculate the area between the curves and the line \(x=-3\).

Hint

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

built with GeoGebra

Problem Statement

Given \(y=x+2\) and \(y=4-x^2\). Calculate the area between the curves and the line \(x=-3\).

Final Answer

31/6

Problem Statement

Given \(y=x+2\) and \(y=4-x^2\). Calculate the area between the curves and the line \(x=-3\).

Hint

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

built with GeoGebra

Solution

2432 solution video

video by Michel vanBiezen

Final Answer

31/6

close solution

Intermediate Problems

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).

Solution

1157 solution video

video by Krista King Math

close solution

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 \(y=\cos(x)\).

Solution

1170 solution video

video by MIT OCW

close solution

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

Problem Statement

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

Solution

1168 solution video

video by Krista King Math

close solution

\( x=2y^2; ~~~ x=4+y^2 \)

Problem Statement

\( x=2y^2; ~~~ x=4+y^2 \)

Solution

1159 solution video

video by Krista King Math

close solution

\(x=y^3-y; ~~~ x=1-y^4\)

Problem Statement

\(x=y^3-y; ~~~ x=1-y^4\)

Solution

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

1163 solution video

video by PatrickJMT

1163 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

1164 solution video

video by PatrickJMT

close solution

\(y=e^x;\) \(x=-1;\) \(y=x^2-1;\) \(x=1\)

Problem Statement

\(y=e^x;\) \(x=-1;\) \(y=x^2-1;\) \(x=1\)

Solution

44 solution video

video by Krista King Math

close solution

\(y=-x-1\); \(y=x-1\); \(y=1-x^2\)

Problem Statement

\(y=-x-1\); \(y=x-1\); \(y=1-x^2\)

Solution

2112 solution video

video by PatrickJMT

close solution

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

Problem Statement

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

Solution

2113 solution video

video by PatrickJMT

close solution

Given \( y=2x/\pi \) and \( y=\sin x \). Calculate the area between the curves and the line \(x=\pi\) for \( x\geq 0 \).

Problem Statement

Given \( y=2x/\pi \) and \( y=\sin x \). Calculate the area between the curves and the line \(x=\pi\) for \( x\geq 0 \).

Hint

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

built with GeoGebra

Problem Statement

Given \( y=2x/\pi \) and \( y=\sin x \). Calculate the area between the curves and the line \(x=\pi\) for \( x\geq 0 \).

Hint

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

built with GeoGebra

Solution

2444 solution video

video by Michel vanBiezen

close solution
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