Suppose two given curves $$y=f(x)$$ and $$y=g(x)$$ intersect at $$x=a$$ and $$x=b$$ and $f(x)\geq g(x)\quad(a<x<b).$ To find the area of the region bounded between these two curves, consider a vertical rectangle as shown in Figure 1. The height of this rectangle is $$f(x)-g(x)$$ and its width is $$dx$$. The area of this infinitesimal rectangle is $dA=[f(x)-g(x)]dx,$ and the total area is $A=\int_{a}^{b}dA=\int_{a}^{b}[f(x)-g(x)]dx.$

• Notice that we integrate from the smaller limit $$a$$ to the larger one $$b$$, and $$f(x)$$ is greater than $$g(x)$$ in the entire interval of integration; otherwise the integral (and the area) becomes negative which is meaningless.
• Because the curves $$y=f(x)$$ and $$y=g(x)$$ intersect at $$x=a$$ and $$x=b$$, they have the same $$y$$-values at these points. If in a problem $$a$$ and $$b$$ are not given, we can find them from solving the equation $$f(x)=g(x)$$.
Example 1
Find the area of the region bounded by the curves $$y=\frac{1}{2}x^{2}$$ and $$y=2$$.

Solution 1
Using vertical rectangles, we realize that the height of the typical rectangle is $$2-\frac{1}{2}x^{2}$$ and its width is $$dx$$. So the element of area is $dA=\left(2-\frac{1}{2}x^{2}\right)\,dx$

Because these two curves intersect at $$x=\pm2$$: $\frac{1}{2}x^{2}=2\Rightarrow x^{2}=4\Rightarrow x=\pm2,$ the total area bounded by $$y=x^{2}/2$$ and $$y=2$$ is then \begin{aligned} A=\int_{-2}^{2}dA & =\int_{-2}^{2}\left(2-\frac{1}{2}x^{2}\right)\,dx\\ & =\left[2x-\frac{1}{6}x^{3}\right]_{x=-2}^{x=2}\\ & =2\left(4-\frac{1}{6}\cdot8\right)\\ & =\frac{16}{3}.\end{aligned} Because the area is symmetric with respect to the $$y$$-axis, we could simply integrate from $$0$$ to $$2$$ and multiply the result by $$2$$. That is, \begin{aligned} A & =2\int_{0}^{2}\left(2-\frac{1}{2}x^{2}\right)dx\\ & =2\left[2x-\frac{1}{6}x^{3}\right]_{x=-2}^{x=2}\\ & =2(4-8/6)=16/3.\end{aligned}

Sometimes, instead of vertical rectangles, we could or we should consider horizontal rectangles. To find the area of the region bounded by the curves $$x=h(y)$$ and $$x=k(y)$$ (with $$k(y)\leq h(y)$$) and the horizontal lines $$y=c$$ and $$y=d$$ ($$c\leq d$$) (Figure 4), we consider horizontal rectangles of length $$h(y)-k(y)$$ and width $$dy$$. The element of area is then $dA=[h(y)-k(y)]dy$ and the total area is $A=\int_{c}^{d}dA=\int_{c}^{d}[h(y)-k(y)]dy.$

For instance, in the above example, if we use horizontal rectangles, the length of the rectangle is the $$x$$ value (in terms of $$y$$) of the right curve ($$x=\sqrt{2y}$$) minus the $$x$$ value of the left curve ($$x=-\sqrt{2y})$$. Because the width of the rectangle is $$dy$$, the element of area is \begin{aligned} dA & =\left[\sqrt{2y}-(-\sqrt{2y})\right]\,dy\\ & =2\sqrt{2}\sqrt{y}dy.\end{aligned} As the horizontal rectangle sweeps across the region, $$y$$ varies between $$0$$ and $$2$$. Therefore, the total area is \begin{aligned} A & =\int_{0}^{2}2\sqrt{2}\sqrt{y}dy\\ & =2\sqrt{2}\int_{0}^{2}y^{1/2}dy\\ & =\frac{2\sqrt{2}}{3/2}\left.y^{3/2}\right|_{y=0}^{y=2}\\ & =\frac{4\sqrt{2}}{3}\sqrt{8}\\ & =\frac{16}{3}.\end{aligned}

Example 2
Find the area of the region bounded by the curves $$y^{2}=x$$ and $$y=x-2$$.

Solution 2
First we need to find where these curves intersect $x=y^{2},\ x=y+2\Rightarrow y^{2}=y+2$ or $y^{2}-y-2=0$ $\Rightarrow y=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{1\pm\sqrt{1+4(2)}}{2}$ $\Rightarrow y=2,-1$ $x=\left.y^{2}\right|_{y=2}=4$ $x=\left.y^{2}\right|_{y=-1}=1$ Therefore, these curves intersect at $$(2,4)$$ and $$(1,-1)$$.

To compute the area of the region, it is easier to consider horizontal thin rectangles (Figure 6). Because $$x=y+2$$ is on the right, the length of the typical rectangle is $$y+2-y^{2}$$ and its width is $$dy$$. Therefore, the element of area is $dA=(y+2-y^{2})dy.$ Because $$y$$ varies between $$-1$$ and $$2$$, the area is \begin{aligned} A & =\int_{-1}^{2}dA=\int_{-1}^{2}(y+2-y^{2})dy\\ & =\left[\frac{1}{2}y^{2}+2y-\frac{y^{3}}{3}\right]_{-1}^{2}\\ & =\frac{10}{3}-\left(-\frac{7}{6}\right)=\frac{9}{2}.\end{aligned}

To compute the area, we can also consider thin vertical rectangles (Figure 7). However, we need to divide the region into two subregions. If $$x>1$$, then the height of the rectangle is $$\sqrt{x}-(x-2)$$ and if $$0<x<1$$, the height of the rectangle is $$\sqrt{x}-(-\sqrt{x})=2\sqrt{x}$$. Therefore $dA=\begin{cases} \sqrt{x}-x+2 & (\text{if }1\leq x\leq4)\\ 2\sqrt{x} & (\text{if }0\leq x\leq1) \end{cases}$ and the total area is \begin{aligned} A & =\int_{0}^{4}dA\\ & =\int_{0}^{1}2\sqrt{x}dx+\int_{1}^{4}(\sqrt{x}-x+2)dx\\ & =\left[\frac{4}{3}x^{3/2}\right]_{0}^{1}+\left[\frac{2}{3}x^{3/2}-\frac{1}{2}x^{2}+2x\right]_{1}^{4}\\ & =\frac{4}{3}+\frac{16}{3}-\frac{13}{6}=\frac{9}{2}.\end{aligned}

Example 3
Find the area between the graphs of $$y=\cos2x$$ and $$y=\sin x$$ on the interval $$\left[0,\frac{\pi}{2}\right]$$.
Solution 3
It is a good idea to sketch these two curves. To graph $$y=\cos2x$$ we start with the graph of $$y=\cos x$$ and compress it horizontally by a factor of $$2$$ (Figure 8).

As we can see from Figure 9, the region between the curves consists of two parts. In one part, the curve $$y=\cos2x$$ is the upper curve and in the other part, the curve $$y=\sin x$$ is the upper one. To find exactly where these two curves intersect, we must solve the equation $$\cos2x=\sin x$$. We do this by writing

\begin{aligned}\cos2x & =\sin x\\ \cos^{2}x-\sin^{2}x & =\sin x\qquad\qquad{\small{(\cos2x=\cos^{2}x-\sin^{2}x)}}\\ 1-2\sin^{2}x & =\sin x\qquad\qquad{\small{(\cos^{2}x+\sin^{2}x=1)}}\\ 2\sin^{2}x+\sin x-1 & =0\end{aligned} The above equation is a quadratic equation in terms of $$\sin x$$. That is, if $$u=\sin x$$, then $2u^{2}+u-1=0$ $\Rightarrow u=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{-1\pm\sqrt{1+8}}{4}$ so $\sin x=-1\quad\text{or}\quad\sin x=\frac{1}{2}.$ $\sin x=\frac{1}{2}\Rightarrow x=\frac{\pi}{6}$ The equation $$\sin x=-1$$ has no solution in the interval $$\left[0,\frac{\pi}{2}\right]$$.

Therefore $dA=\begin{cases} (\cos2x-\sin x)dx & 0\leq x\leq\frac{\pi}{6}\\ (\sin x-\cos2x)dx & \frac{\pi}{6}\leq x\leq\frac{\pi}{2} \end{cases}$ and the desired area is: \begin{aligned} & \int_{0}^{\frac{\pi}{6}}dA+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}dA\\ & =\int_{0}^{\frac{\pi}{6}}(\cos2x-\sin x)dx+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\sin x-\cos2x)dx\\ & =\left[\frac{1}{2}\sin2x+\cos2x\right]_{0}^{\frac{\pi}{6}}+\left[-\cos x-\frac{1}{2}\sin2x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\\ & =\left(\frac{1}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)-(0+1)+\left(-(0)-\frac{1}{2}(0)\right)-\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\right)\\ & =\frac{3\sqrt{3}}{2}-1.\end{aligned}