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How to Find the Area Between Two Graphs by Integration


When finding the area between two graphs, there are some things you need to take into account:

1.
The areas’ position in the coordinate system does not matter—that is, it is not important if the graphs are above or below the x-axis in this context.
2.
To get a positive area in your answer, you always have to take the top graph minus the bottom one.
3.
Due to Item 2, you must compute an integral for each area.

Then, assuming the graph of f lies above the graph of g between x = a and x = b, the area between the graphs is given by:

Formula

The Area Between Two Graphs

A =abf(x)dx abg(x)dx =ab (f(x) g(x)) dx

Note! You always want to put the top graphs first in the subtraction. If you don’t, the computation yields a negative sign. If you don’t take this into account, you may end up with a completely wrong answer.

Study the following two examples closely and read through them several times until you understand them.

Example 1

Find the area that is bounded by the functions f(x) = x3 + 3x2 13x 15 g(x) = 10x + 7

First, you draw the graphs, such that you can see the bounded areas. In this case, there are two areas.

Example of area between two graphs 1

Example of area between two graphs 1

In Area 1, g(x) is above, and in Area 2, f(x) is above. That means you have to determine where the graphs intersect—that is, where f(x) = g(x):
x3 + 3x2 13x 15 = 10x + 7

When solving this equation, you get that x 6.16, x 0.88 and x 4.04. This means that A = 6.16, B = 0.88 and C = 4.04, where g(x) lies above f(x) between 6.16 and 0.88, and f(x) lies above g(x) between 0.88 and 4.04. Therefore, the total area is:

A = A1 + A2

First, find A1:

A1 =6.160.88x3 + 3x2 13x 15 (10x + 7)dx =6.160.88x3 + 3x2 23x 22dx = (1 4x4 + x3 23 2 x2 22x)| 6.160.88 = (1 4(0.88)4 + (0.88)3 23 2 (0.88)2 22(0.88) ) (1 4(6.16)4 + (6.16)3 23 2 (6.16)2 22(6.16) ) 9.92 (174.63) 184.55

A1 =6.160.88x3 + 3x2 13x 15 (10x + 7)dx =6.160.88x3 + 3x2 23x 22dx = (1 4x4 + x3 23 2 x2 22x)| 6.160.88 = (1 4(0.88)4 + (0.88)3 23 2 (0.88)2 22(0.88)) (1 4(6.16)4 + (6.16)3 23 2 (6.16)2 22(6.16)) 9.92 (174.63) 184.55

Then you find A2:
A2 =0.884.0410x + 7 (x3 + 3x2 13x 15) dx =0.884.04 x3 3x2 + 23x + 22dx = (1 4x4 x3 + 23 2 x2 + 22x)| 0.884.04 = ( 1 4(4.04)4 (4.04)3 + 23 2 (4.04)2 + 22(4.04) ) ( 1 4(0.88)4 (0.88)3 + 23 2 (0.88)2 + 22(0.88) ) 144.04 (9.92) = 153.96

A2 =0.884.0410x + 7 (x3 + 3x2 13x 15) dx =0.884.04 x3 3x2 + 23x + 22dx = (1 4x4 x3 + 23 2 x2 + 22x)| 0.884.04 = (1 4(4.04)4 (4.04)3 + 23 2 (4.04)2 + 22(4.04)) (1 4(0.88)4 (0.88)3 + 23 2 (0.88)2 + 22(0.88)) 144.04 (9.92) = 153.96

Finally, the total area is
A = A1 + A2 184.55 + 153.96 338.5

A = A1 + A2 184.55 + 153.96 = 338.5

Example 2

Find the area that is bounded by the functions f(x) = 1 2x2 5 2x g(x) = 5cos (3 2x 3 5)

for x [ 1.23,6.18]

First, you draw the graphs so that you can see the bounded areas. In this case, it’s three areas in total:

Example of area between two graphs 2

Example of area between two graphs 2

In Area 1, f(x) lies above, in Area 2, g(x) lies above and in Area 3, f(x) is above again. You therefore have to find where the graphs intersect. That is, where f(x) = g(x):
1 2x2 5 2x = 5 cos (3 2x 3 5)

When you solve this equation, you get x 1.23, x 1.14, x 3.85 and x 6.18. (You also get two other intersections, but these lie outside the relevant interval.)

This means that A = 1.23, B = 1.14, C = 3.85 and D = 6.18, where f(x) lies above between 1.23 and 1.14 and from 3.85 to 6.18. The function g(x) lies above between 1.14 and 3.85. The total area is therefore:

A = A1 + A2 + A3

First, you find the area A1:

A1 =1.231.141 2x2 5 2x (5 cos (3 2x 3 5 )) dx =1.231.141 2x2 5 2x + 5 cos (3 2x 3 5 ) dx = (1 6x3 5 4x2 + 5 sin (3 2x 3 5 ) 2 3 )|1.231.14 = (1 6x3 5 4x2 + 10 3 sin (3 2x 3 5 )) |1.231.14 = (1 6(1.14)3 5 4(1.14)2 + 10 3 sin (3 2(1.14) 3 5 ) ) (1 6(1.23)3 5 4(1.23)2 + 10 3 sin (3 2(1.23) 3 5 ) ) 1.61 (4.34) = 5.95

A1 =1.231.141 2x2 5 2x (5 cos (3 2x 3 5)) dx =1.231.141 2x2 5 2x + 5 cos (3 2x 3 5) dx = (1 6x3 5 4x2 + 5 sin (3 2x 3 5) 2 3) |1.231.14 = (1 6x3 5 4x2 + 10 3 sin (3 2x 3 5)) |1.231.14 = (1 6(1.14)3 5 4(1.14)2 + 10 3 sin (3 2(1.14) 3 5)) (1 6(1.23)3 5 4(1.23)2 + 10 3 sin (3 2(1.23) 3 5)) 1.61 (4.34) = 5.95

Then you find the area A2:
A2 =1.143.85 5 cos (3 2x 3 5) (1 2x2 5 2x)dx =1.143.85 5 cos (3 2x 3 5) 1 2x2 + 5 2xdx = ( 5 sin (3 2x 3 5) 2 3 1 6x3 + 5 4x2 )| 1.143.85 = ( 10 3 cos (3 2x 3 5) 1 6x3 + 5 4x2 )| 1.143.85 = ( 10 3 cos (3 2(3.85) 3 5) 1 6(3.85)3 + 5 4(3.85)2 ) (10 3 cos (3 2(1.14) 3 5) 1 6(1.14)3 + 5 4(1.14)2 ) 12 (1.61) 13.61

A2 =1.143.85 5 cos (3 2x 3 5) (1 2x2 5 2x)dx =1.143.85 5 cos (3 2x 3 5) 1 2x2 + 5 2xdx = (5 sin (3 2x 3 5) 2 3 1 6x3 + 5 4x2) | 1.143.85 = (10 3 cos (3 2x 3 5) 1 6x3 + 5 4x2) | 1.143.85 = (10 3 cos (3 2(3.85) 3 5) 1 6(3.85)3 + 5 4(3.85)2) (10 3 cos (3 2(1.14) 3 5) 1 6(1.14)3 + 5 4(1.14)2) 12 (1.61) 13.61

Next, you find the area A3:
A3 =3.856.181 2x2 5 2x (5 cos (3 2x 3 5 )) dx =3.856.181 2x2 5 2x + 5 cos (3 2x 3 5 ) dx = (1 6x3 5 4x2 + 5 sin (3 2x 3 5 ) 2 3 )|3.856.18 = (1 6x3 5 4x2 + 10 3 sin (3 2x 3 5 )) |3.856.18 = (1 6(6.18)3 5 4(6.18)2 + 10 3 sin (3 2(6.18) 3 5 ) ) (1 6(3.85)3 5 4(3.85)2 + 10 3 sin (3 2(3.85) 3 5 ) ) 6.12 (12) 5.88

A3 =3.856.181 2x2 5 2x (5 cos (3 2x 3 5)) dx =3.856.181 2x2 5 2x + 5 cos (3 2x 3 5) dx = (1 6x3 5 4x2 + 5 sin (3 2x 3 5) 2 3) |3.856.18 = (1 6x3 5 4x2 + 10 3 sin (3 2x 3 5)) |3.856.18 = (1 6(6.18)3 5 4(6.18)2 + 10 3 sin (3 2(6.18) 3 5)) (1 6(3.85)3 5 4(3.85)2 + 10 3 sin (3 2(3.85) 3 5)) 6.12 (12) 5.88

Finally, the total area is
A = A1 + A2 + A3 5.95 + 13.61 + 5.88 25.44

A = A1 + A2 + A3 5.95 + 13.61 + 5.88 25.44

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