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## Area Between Curves

If $$f$$ and $$g$$ are continuous functions on $$[a,b]$$, then the area of the region bounded by the curves $$y=f(x)$$, $$y=g(x)$$, and the vertical lines $$x=a$$, $$x=b$$ is $\int_a^b |f(x)-g(x)|\;dx.$ In particular, if $$f(x)\geq g(x)$$ for all $$x$$ in $$[a,b]$$, then the area is $\int_a^b \left[ f(x)-g(x)\right]\;dx.$

Break $$[a,b]$$ into $$n$$ subintervals $$[x_0,x_1],[x_1,x_2],\ldots,[x_{n-1},x_n]$$ where $$x_i=a+i \Delta x$$ and $$\Delta x=(b-a)/n$$. Choose $$x_i^*$$ from $$[x_{i-1},x_i]$$, $$i=1,2,\ldots,n$$. Let $$R$$ be the region bounded by $$y=f(x)$$, $$y=g(x)$$, and the vertical lines $$x=a$$, $$x=b$$. Partition $$R$$ into $$n$$ regions $$R_1,R_2,\ldots,R_n$$ by intersecting $$R$$ by the vertical lines at $$x_1,x_2,\ldots,x_{n-1}$$. Then the area of $$R_i$$ is approximately $$|f(x_i^*)-g(x_i^*)| \Delta x$$. Therefore the area of $$R$$ is $A(R)\approx \sum_{i=1}^n |f(x_i^*)-g(x_i^*)| \Delta x.$ This approximation of $$A(R)$$ gets better as $$n\to \infty$$. Thus $A(R)=\lim_{n\to \infty} \sum_{i=1}^n |f(x_i^*)-g(x_i^*)| \Delta x =\int_a^b |f(x)-g(x)|\;dx.$

Note that $$|f(x)-g(x)|=y_{\text{top}}-y_{\text{bottom}}$$ on each subinterval of $$[a,b]$$.

Example. Find the area of the region enclosed by the curves $$y=x^2$$ and $$y=4x-x^2$$.

Solution. At the points of intersection of $$y=x^2$$ and $$y=4x-x^2$$, $x^2=4x-x^2 \implies 2x^2-4x=0 \implies 2x(x-2)=0 \implies x=0,2.$ So the points of intersection are $$(0,0)$$ and $$(2,4)$$.

The area of the region enclosed by $$y=x^2$$ and $$y=4x-x^2$$ is \begin{align*} \int_0^2 \left\vert (4x-x^2)-x^2\right\vert\;dx &= \int_0^2 \left(4x-x^2 -x^2\right) \;dx\\ &= \int_0^2 \left(4x-2x^2\right)\;dx\\ &=\left. \left( 2x^2-\frac{2}{3}x^3 \right) \right\vert_0^2\\ &= 2\cdot 2^2-\frac{2}{3}2^3-0\\ &=\frac{8}{3}. \end{align*}

Example. Find the area of the region bounded by the curves $$y=\cos x$$ and $$y=1-\cos x$$ on $$[0,\pi]$$.

Solution. At the points of intersection of $$y=\cos x$$ and $$y=1-\cos x$$ on $$[0,\pi]$$, $\cos x=1-\cos x \implies 2\cos x=1 \implies \cos x=\frac{1}{2} \implies x=\frac{\pi}{3}.$ So the point of intersection is $$\left(\frac{\pi}{3},\frac{1}{2}\right)$$.

Note that $$\cos x \geq 1-\cos x$$ on $$\left[0,\frac{\pi}{3} \right]$$ and $$\cos x \leq 1-\cos x$$ on $$\left[\frac{\pi}{3}, \pi \right]$$. The area of the region enclosed by $$y=\cos x$$ and $$y=1-\cos x$$ on $$[0,\pi]$$ is \begin{align*} &\int_0^{\pi} \left\vert (1-\cos x)-\cos x\right\vert\;dx\\ =& \int_0^{\frac{\pi}{3}} \left\vert (1-\cos x)-\cos x\right\vert\;dx + \int_{\frac{\pi}{3}}^{\pi} \left\vert (1-\cos x)-\cos x\right\vert\;dx\\ =& \int_0^{\frac{\pi}{3}} \left[ -(1-\cos x)+\cos x\right]\;dx + \int_{\frac{\pi}{3}}^{\pi} \left[ (1-\cos x)-\cos x\right]\;dx\\ =& \int_0^{\frac{\pi}{3}} \left[ 2\cos x -1\right]\;dx + \int_{\frac{\pi}{3}}^{\pi} \left[ 1-2\cos x \right]\;dx\\ =&\left. \left( 2\sin x -x \right) \right\vert_0^{\frac{\pi}{3}} +\left. \left( x-2\sin x \right) \right\vert_{\frac{\pi}{3}}^{\pi}\\ =& \left[ \left( 2\sin\left( \frac{\pi}{3} \right) -\frac{\pi}{3} \right) - \left( 2\sin 0 -0 \right) \right] +\left[ \left( \pi -2\sin \pi \right) - \left( \frac{\pi}{3} -2\sin\left( \frac{\pi}{3} \right) \right) \right]\\ =& \left[ 2\frac{\sqrt{3}}{2}-\frac{\pi}{3} \right] + \left[\pi- \frac{\pi}{3}+2\frac{\sqrt{3}}{2} \right] \\ =&\; 2\sqrt{3}+\frac{\pi}{3}.\\ \end{align*}

If $$f$$ and $$g$$ are continuous functions on $$[c,d]$$, then the area of the region bounded by the curves $$x=f(y)$$, $$x=g(y)$$, and the horizontal lines $$y=c$$, $$y=d$$ is $\int_c^d |f(y)-g(y)|\;dy.$ Note that $$|f(y)-g(y)|=x_{\text{right}}-x_{\text{left}}$$ on each subinterval of $$[c,d]$$.

Example. Find the area of the region enclosed by the curves $$y^2=2x+6$$ and $$y=x-1$$.

Solution. At the points of intersection of $$y^2=2x+6$$ and $$y=x-1$$, $2x+6=(x-1)^2 \implies x^2-4x-5=0 \implies x=-1,5.$ So the points of intersection are $$(-1,-2)$$ and $$(5,4)$$.

The area of the region enclosed by $$x=\frac{y^2}{2}-3$$ and $$x=y+1$$ is \begin{align*} \int_{-2}^4 \left( x_{\text{right}}-x_{\text{left}} \right)\;dy &= \int_{-2}^4 \left[ (y+1)-\left( \frac{y^2}{2}-3 \right) \right]\;dy\\ &= \int_{-2}^4 \left[ -\frac{y^2}{2}+y+4 \right]\;dy\\ &=\left. \left( -\frac{y^3}{6}+\frac{y^2}{2}+4y \right) \right\vert_{-2}^4\\ &= \left( -\frac{4^3}{6}+\frac{4^2}{2}+4\cdot 4 \right)- \left( -\frac{(-2)^3}{6}+\frac{(-2)^2}{2}+4(-2) \right)\\ &=18. \end{align*}

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