Areas between curves
In Chapter 5, the definite integral gave the area between a curve and the $x$-axis. Now we find the area between two curves.
The formula
Suppose $f(x)\ge g(x)$ for every $x$ in $[a,b]$. So the graph of $f$ is on top, and the graph of $g$ is below it. The area of the region between them is
$\displaystyle A=\int_a^b[f(x)-g(x)]\,dx$.
In words: integrate the top curve minus the bottom curve.
Why: Cut the region into thin vertical rectangles, as in Riemann sums. A rectangle at $x$ has height $f(x)-g(x)$ and width $\Delta x$. Adding the areas and taking the limit gives the integral.
This works even when part of the region is below the $x$-axis. The height is always top minus bottom.
How to find the area between two curves
- Sketch the curves, and shade the region.
- Find where the curves meet. These are often the limits of integration.
- Decide which curve is on top.
- Integrate (top) $-$ (bottom).
Example 1: Find the area of the region enclosed by
$y=x^2$ and $y=2x-x^2$.
Solution:
Where they meet: Set the two equal:
$\begin{align*}x^2&=2x-x^2\\2x^2-2x&=0\\2x(x-1)&=0\end{align*}$
So they meet at $x=0$ and $x=1$.
Which is on top: Test $x=\tfrac{1}{2}$:
- $2x-x^2=\tfrac{3}{4}$
- $x^2=\tfrac{1}{4}$
So $2x-x^2$ is on top.
Integrate:
$\begin{align*}&A\\&=\int_0^1[(2x-x^2)-x^2]\,dx\\&=\int_0^1(2x-2x^2)\,dx\\&=\left[x^2-\dfrac{2x^3}{3}\right]_0^1\\&=1-\dfrac{2}{3}=\dfrac{1}{3}\end{align*}$
Example 2: Find the area of the region enclosed by
$y=\sqrt{x}$ and $y=\dfrac{x}{2}$.
Solution:
Where they meet: Square both sides of $\sqrt{x}=\dfrac{x}{2}$:
$\begin{align*}x&=\dfrac{x^2}{4}\\4x&=x^2\\x(x-4)&=0\end{align*}$
So $x=0$ or $x=4$. (Both check in the original equation.)
Which is on top: At $x=1$, $\sqrt{1}=1$ and $\dfrac{1}{2}$. So $\sqrt{x}$ is on top.
Integrate:
$\begin{align*}&A\\&=\int_0^4\left(\sqrt{x}-\dfrac{x}{2}\right)dx\\&=\left[\dfrac{2}{3}x^{3/2}-\dfrac{x^2}{4}\right]_0^4\\&=\dfrac{2}{3}(8)-4\\&=\dfrac{4}{3}\end{align*}$
When the curves cross
If the curves cross inside the interval, the top curve changes. Split the interval at each crossing point. On each piece, integrate top minus bottom. Then add the areas.
Example 3: Find the area between
$y=\sin x$ and $y=\cos x$
from $x=0$ to $x=\dfrac{\pi}{2}$.
Solution:
The curves cross where $\sin x=\cos x$. In this interval, that is $x=\dfrac{\pi}{4}$.
On $\left[0,\tfrac{\pi}{4}\right]$, cosine is on top:
$\begin{align*}&\int_0^{\pi/4}(\cos x-\sin x)\,dx\\&=\Big[\sin x+\cos x\Big]_0^{\pi/4}\\&=\left(\dfrac{\sqrt{2}}{2}+\dfrac{\sqrt{2}}{2}\right)-(0+1)\\&=\sqrt{2}-1\end{align*}$
On $\left[\tfrac{\pi}{4},\tfrac{\pi}{2}\right]$, sine is on top:
$\begin{align*}&\int_{\pi/4}^{\pi/2}(\sin x-\cos x)\,dx\\&=\Big[-\cos x-\sin x\Big]_{\pi/4}^{\pi/2}\\&=(0-1)-\left(-\dfrac{\sqrt{2}}{2}-\dfrac{\sqrt{2}}{2}\right)\\&=\sqrt{2}-1\end{align*}$
The total area is
$2(\sqrt{2}-1)=2\sqrt{2}-2\approx 0.828$.
Integrating with respect to $y$
Some regions are easier to describe from left to right. If the curves are $x=f(y)$ (on the right) and $x=g(y)$ (on the left), for $c\le y\le d$, then
$\displaystyle A=\int_c^d[f(y)-g(y)]\,dy$.
In words: integrate the right curve minus the left curve, with respect to $y$. The rectangles are now horizontal.
Example 4: Find the area of the region enclosed by
$x=y^2$ and $y=x-2$.
Solution:
Write the line as $x=y+2$. It is on the right, and the parabola is on the left.
Where they meet:
$\begin{align*}y^2&=y+2\\y^2-y-2&=0\\(y-2)(y+1)&=0\end{align*}$
So $y=-1$ or $y=2$.
The points are $(1,-1)$ and $(4,2)$.
Integrate:
An antiderivative is
$F(y)=\dfrac{y^2}{2}+2y-\dfrac{y^3}{3}$.
Its values at the ends:
$F(2)=2+4-\dfrac{8}{3}=\dfrac{10}{3}$
$F(-1)=\dfrac{1}{2}-2+\dfrac{1}{3}=-\dfrac{7}{6}$
So
$\begin{align*}&A\\&=\int_{-1}^{2}(y+2-y^2)\,dy\\&=F(2)-F(-1)\\&=\dfrac{10}{3}+\dfrac{7}{6}\\&=\dfrac{9}{2}\end{align*}$
With respect to $x$, this region would need two integrals, because the bottom curve changes at $x=1$. With respect to $y$, one integral is enough.
Summary
- Area between curves: $\displaystyle\int_a^b(\text{top}-\text{bottom})\,dx$.
- Find where the curves meet, and check which one is on top.
- If the curves cross, split the interval and add the pieces.
- Sometimes it is easier to use $\displaystyle\int_c^d(\text{right}-\text{left})\,dy$.