Area Between Curves Area Between Curves
Introduction
To find the accumulated area under a curve, we integrate the function.
\text{Area} = \int_{a}^{b} f(x) \, dx
If we want to find the total area under a curve, not allowing for positive and negative area to cancel out, we integrate the absolute value of the function.
\text{Absolute Area} = \int_{a}^{b} |f(x)| \, dx
To find the area between two different functions, we integrate the absolute value of their difference.
\text{Area} = \int_{a}^{b} |f(x) - g(x)| \, dx
Since this is the integral of the absolute value of a function, we must separate the integral whenever the function crosses the x-axis. In this case, whenever
f(x) = g(x)
.
Examples
Example 1: Quadratic vs. Linear
Find the area of the space enclosed by
f(x) = x^2 - 2x
and
g(x) = x
.
The enclosed space starts and ends where f
and g
intersect,x^2 - 2x = x
This can be rewritten as x^2 - 3x = 0
.
x = 0, 3
.
\begin{align*}
\text{Area} &= \int_{0}^{3} |g(x) - f(x)| \, dx \\
&= |\int_{0}^{3} (3x - x^2) \, dx| \\
&= \left[ \frac{3}{2}x^2 - \frac{1}{3}x^3 \right]_0^3 \\
&= \left( \frac{27}{2} - 9 \right) - 0 = \frac{9}{2} = 4.5 \, \text{units}^2
\end{align*}
Example 2: Sine vs. Cosine
Find the area between
y = \sin(x)
and
y = \cos(x)
from
x = 0
to
x = \pi
.
In this interval, \sin(x) = \cos(x)
at x = \frac{\pi}{4}
.
\begin{align*}
\text{Area} &= \int_{0}^{\pi} |\cos(x) - \sin(x)| \, dx \\
&= |\int_{0}^{\pi/4} [\cos(x) - \sin(x)] \, dx| + |\int_{\pi/4}^{\pi} [\cos(x) - \sin(x)] \, dx| \\
&= \left| \sin(x) + \cos(x) \right|_0^{\pi/4} + \left| \sin(x) + \cos(x) \right|_{\pi/4}^{\pi} \\
&= |\frac{2}{\sqrt{2}} - 1| + |-1 - \frac{2}{\sqrt{2}}| \\
&= 2\sqrt{2} \\
&= 2.83 \, \text{units}^2
\end{align*}
The total area between
\sin(x)
and
\cos(x)
on
[0, \pi]
is
2.83
square units.
Area Between Curves as a Function of Y
We normally integrate with respect to x, finding the area between a function and the x-axis. However, we can also integrate curves in respect to y, finding the area between the curve and the y-axis.
The area between two curves
x = f(y)
and
x = g(y)
on
y = [c,d]
is:
\text{Area} = \int_{c}^{d} |f(y) - g(y)| \, dy
Solving this integral is identical to solving integrals in respect to
x
.
Examples
Example 3: Quadratic vs. Linear
Find the area enclosed by
x = y^2
and
x = 2y
.
y^2 = 2y
when y = 0
and y = 2
.
\begin{align*}
\text{Area} &= \int_{0}^{2} |2y - y^2| \, dy \\
&= \left[ y^2 - \frac{1}{3}y^3 \right]_0^2 \\
&= \left(4 - \frac{8}{3}\right) - 0 = \frac{4}{3} \\
&= 1.33 \, \text{units}^2
\end{align*}
The area of the space enclosed by
f(y)
and
g(y)
is
1.33
square units.
Example 4: Exponential vs. Linear
Find the area between
x = e^y
and
x = y + 1
from
y = 0
to
y = 1
.
f(y)
and g(y)
intersect at y = 0
. Therefore, the integral does not need to be split.
\begin{align*}
\text{Area} &= \int_{0}^{1} |e^y - (y + 1)| \, dy \\
&= \left[ e^y - \frac{1}{2}y^2 - y \right]_0^1 \\
&= \left(e - \frac{1}{2} - 1\right) - 1 \\
&= e - 2.5 \\
&= 0.218 \, \text{units}^2
\end{align*}
The area between
f(y)
and
g(y)
on
[0,1]
is
0.218
square units.