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 \(f(x)=g(x)\).
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 \[\begin{aligned} A & =\int_{0}^{2}2\sqrt{2}\sqrt{y}dy\\ & =2\sqrt{2}\int_{0}^{2}y^{1/2}dy\\ & =\frac{2\sqrt{2}}{3/2}\left.y^{3/2}\right|_{y=0}^{y=2}\\ & =\frac{4\sqrt{2}}{3}\sqrt{8}\\ & =\frac{16}{3}.\end{aligned}\]
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 \[\begin{aligned} A & =\int_{0}^{4}dA\\ & =\int_{0}^{1}2\sqrt{x}dx+\int_{1}^{4}(\sqrt{x}-x+2)dx\\ & =\left[\frac{4}{3}x^{3/2}\right]_{0}^{1}+\left[\frac{2}{3}x^{3/2}-\frac{1}{2}x^{2}+2x\right]_{1}^{4}\\ & =\frac{4}{3}+\frac{16}{3}-\frac{13}{6}=\frac{9}{2}.\end{aligned}\]
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: \[\begin{aligned} & \int_{0}^{\frac{\pi}{6}}dA+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}dA\\ & =\int_{0}^{\frac{\pi}{6}}(\cos2x-\sin x)dx+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(\sin x-\cos2x)dx\\ & =\left[\frac{1}{2}\sin2x+\cos2x\right]_{0}^{\frac{\pi}{6}}+\left[-\cos x-\frac{1}{2}\sin2x\right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\\ & =\left(\frac{1}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)-(0+1)+\left(-(0)-\frac{1}{2}(0)\right)-\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}\cdot\frac{\sqrt{3}}{2}\right)\\ & =\frac{3\sqrt{3}}{2}-1.\end{aligned}\]