Short and Sweet Calculus

## 7.1 The Area Between Two Curves

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 7.2. 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 7.1. Find the area of the region bounded by the curves $$y=\frac{1}{2}x^{2}$$ and $$y=2$$.

Solution

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 $A=\int_{-2}^{2}dA=\int_{-2}^{2}\left(2-\frac{1}{2}x^{2}\right)\,dx=\left[2x-\frac{1}{6}x^{3}\right]_{x=-2}^{x=2}=2\left(4-\frac{1}{6}\cdot8\right)=\frac{16}{3}.$

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 7.3), 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 $dA=\left[\sqrt{2y}-(-\sqrt{2y})\right]\,dy=2\sqrt{2}\sqrt{y}dy.$ As the horizontal rectangle sweeps across the region, $$y$$ varies between $$0$$ and $$2$$. Therefore, the total area is $A=\int_{0}^{2}2\sqrt{2}\sqrt{y}dy=2\sqrt{2}\int_{0}^{2}y^{1/2}dy=\frac{2\sqrt{2}}{3/2}\left.y^{3/2}\right|_{y=0}^{y=2}=\frac{4\sqrt{2}}{3}\sqrt{8}=\frac{16}{3}.$