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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 \[\begin{align*} \int_{-2}^4 \left( x_{\text{right}}-x_{\text{left}} \right)\;dy &= \int_{-2}^4 \left[ (y+1)-\left( \frac{y^2}{2}-3 \right) \right]\;dy\\ &= \int_{-2}^4 \left[ -\frac{y^2}{2}+y+4 \right]\;dy\\ &=\left. \left( -\frac{y^3}{6}+\frac{y^2}{2}+4y \right) \right\vert_{-2}^4\\ &= \left( -\frac{4^3}{6}+\frac{4^2}{2}+4\cdot 4 \right)- \left( -\frac{(-2)^3}{6}+\frac{(-2)^2}{2}+4(-2) \right)\\ &=18. \end{align*}\]

Area Between Curves