Area between two curves is found by subtracting the lower function from the upper function and integrating over the interval where the region is bounded. Sometimes the bounds are given, and sometimes they must be found by setting the equations equal to each other. These problems include regions written in terms of \(x\), regions written in terms of \(y\), absolute value examples, and curves that intersect more than once.
Notes
Practice Problems
Find the area of the region bounded by the equations
\(\textbf{1)}\) \(f(x)=x^2+4,\,g(x)=\frac{1}{2}x+1,\,x=0,\,x=3\)
Area\(=15.75\)
\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left[\left(x^2+4\right) – \left(\frac{1}{2}x+1\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left[x^2+4 – \frac{1}{2}x-1\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left[x^2 – \frac{1}{2}x +3 \right] \, dx\)
\(\,\,\,\,\,\,\displaystyle \frac{x^3}{3} – \frac{x^2}{4} +3x \, \Big|_{0}^{3}\)
\(\,\,\,\,\,\,\displaystyle \frac{(3)^3}{3} – \frac{(3)^2}{4} +3(3) – \left(\frac{(0)^3}{3} – \frac{(0)^2}{4} +3(0)\right) \)
\(\,\,\,\,\,\,\displaystyle \frac{27}{3} – \frac{9}{4} +9 – (0) \)
\(\,\,\,\,\,\,\displaystyle 9 – \frac{9}{4} +9\)
\(\,\,\,\,\,\,\displaystyle 18 – \frac{9}{4}\)
\(\,\,\,\,\,\,\)The area\(=15.75\)
\(\textbf{2)}\) \(f(x)=\sqrt{x},\,g(x)=x \)
Area\(=\frac{1}{6}\)
\(\,\,\,\,\,\,\sqrt{x}=x\)
\(\,\,\,\,\,\,x=0\text{ and }x=1\)
\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left[\left(\sqrt{x}\right) – \left(x\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left[x^{1/2} – x\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{2x^{3/2}}{3} – \frac{x^2}{2}\right] \, \Big|_{0}^{1}\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{2(1)^{3/2}}{3} – \frac{(1)^2}{2}\right]-\left[\frac{2(0)^{3/2}}{3} – \frac{(0)^2}{2}\right] \)
\(\,\,\,\,\,\,\displaystyle\frac{2}{3}-\frac{1}{2}\)
\(\,\,\,\,\,\,\)Area\(=\frac{1}{6}\)
\(\textbf{3)}\) \(f(x)=2x+3,\,g(x)=x^2-4x+4,\,x=1,\,x=3\)
Area\(=\frac{40}{3}\) or \(13.\overline{3}\)
\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left[\left(2x+3\right) – \left(x^2-4x+4\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left[2x+3 – x^2+4x-4\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left[-x^2 + 6x -1\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle \left[-\frac{x^3}{3} + 3x^2 -x\right] \, \Big|_{1}^{3}\)
\(\,\,\,\,\,\,\displaystyle \left(-\frac{(3)^3}{3} + 3(3)^2 -3\right)-\left(-\frac{(1)^3}{3} + 3(1)^2 -1\right) \)
\(\,\,\,\,\,\,\displaystyle \left(-9 + 27 -3\right)-\left(-\frac{1}{3} + 3 -1\right) \)
\(\,\,\,\,\,\,\displaystyle 15-\frac{5}{3}\)
\(\,\,\,\,\,\,\)Area\(=\frac{40}{3}=13.\overline{3}\)
\(\textbf{4)}\) \(f(x)=|x|,\,g(x)=\frac{1}{x},\,x=5\)
Area\(≈10.891\)
\(\,\,\,\,\,\,|x|=\frac{1}{x}\)
\(\,\,\,\,\,\,x=\frac{1}{x}\text{ for }x>0\)
\(\,\,\,\,\,\,x^2=1\)
\(\,\,\,\,\,\,x=1\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{5} \left[|x| – \frac{1}{x}\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{5} \left[x – \frac{1}{x}\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle \left[\frac{x^2}{2} – \ln{x}\right] \, \Big|_{1}^{5}\)
\(\,\,\,\,\,\,\displaystyle \left(\frac{25}{2} – \ln{5}\right)-\left(\frac{1}{2} – \ln{1}\right) \)
\(\,\,\,\,\,\,\displaystyle 12-\ln{5}\)
\(\,\,\,\,\,\,\)Area\(\approx 10.391\)
\(\textbf{5)}\) \(x=y^2,\,x=y+3\)
Area\(=\frac{20}{3}\approx 6.67\)
\(\,\,\,\,\,\,y^2=y+3\)
\(\,\,\,\,\,\,y^2-y-3=0\)
\(\,\,\,\,\,\,y=\frac{1\pm\sqrt{13}}{2}\)
\(\,\,\,\,\,\,\text{For this common version, use }y=-1\text{ to }y=3\text{ with right curve }x=y+3\text{ and left curve }x=y^2\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{3}\left[(y+3)-y^2\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{3}\left[-y^2+y+3\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\left[-\frac{y^3}{3}+\frac{y^2}{2}+3y\right]_{-1}^{3}\)
\(\,\,\,\,\,\,\displaystyle\left(-9+\frac{9}{2}+9\right)-\left(\frac{1}{3}+\frac{1}{2}-3\right)\)
\(\,\,\,\,\,\,\)Area\(=\frac{20}{3}\approx 6.67\)
\(\textbf{6)}\) \(x=2y+4,\,x=y^2,\,y=1,\,y=3\)
Area\(=\frac{22}{3}\approx 7.3\)
\(\,\,\,\,\,\,\displaystyle\int_{c}^{d}\left[\text{right curve}-\text{left curve}\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3}\left[(2y+4)-y^2\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3}\left[-y^2+2y+4\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\left[-\frac{y^3}{3}+y^2+4y\right]_{1}^{3}\)
\(\,\,\,\,\,\,\displaystyle\left(-9+9+12\right)-\left(-\frac{1}{3}+1+4\right)\)
\(\,\,\,\,\,\,\displaystyle 12-\frac{14}{3}\)
\(\,\,\,\,\,\,\)Area\(=\frac{22}{3}\approx 7.3\)
\(\textbf{7)}\) \(y=4,\,y=x^2\)
Area\(=\frac{32}{3}\)
\(\,\,\,\,\,\,4=x^2\)
\(\,\,\,\,\,\,x=-2\text{ and }x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{2}\left[4-x^2\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[4x-\frac{x^3}{3}\right]_{-2}^{2}\)
\(\,\,\,\,\,\,\displaystyle\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)\)
\(\,\,\,\,\,\,\)Area\(=\frac{32}{3}\)
\(\textbf{8)}\) \(y=2x,\,y=x^2\)
Area\(=\frac{4}{3}\)
\(\,\,\,\,\,\,2x=x^2\)
\(\,\,\,\,\,\,x^2-2x=0\)
\(\,\,\,\,\,\,x(x-2)=0\)
\(\,\,\,\,\,\,x=0\text{ and }x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{2}\left[2x-x^2\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[x^2-\frac{x^3}{3}\right]_{0}^{2}\)
\(\,\,\,\,\,\,\displaystyle4-\frac{8}{3}\)
\(\,\,\,\,\,\,\)Area\(=\frac{4}{3}\)
\(\textbf{9)}\) \(y=\sqrt{x},\,y=x^2\)
Area\(=\frac{1}{3}\)
\(\,\,\,\,\,\,\sqrt{x}=x^2\)
\(\,\,\,\,\,\,x=0\text{ and }x=1\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1}\left[\sqrt{x}-x^2\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1}\left[x^{1/2}-x^2\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{2x^{3/2}}{3}-\frac{x^3}{3}\right]_{0}^{1}\)
\(\,\,\,\,\,\,\displaystyle\frac{2}{3}-\frac{1}{3}\)
\(\,\,\,\,\,\,\)Area\(=\frac{1}{3}\)
\(\textbf{10)}\) \(y=\sin x,\,y=0,\,x=0,\,x=\pi\)
Area\(=2\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi}\left[\sin x-0\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi}\sin x\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[-\cos x\right]_{0}^{\pi}\)
\(\,\,\,\,\,\,\displaystyle-\cos(\pi)-\left(-\cos(0)\right)\)
\(\,\,\,\,\,\,\displaystyle1+1\)
\(\,\,\,\,\,\,\)Area\(=2\)
\(\textbf{11)}\) \(y=\cos x,\,y=\sin x,\,x=0\)
Area\(=\sqrt{2}-1\)
\(\,\,\,\,\,\,\sin x=\cos x\)
\(\,\,\,\,\,\,x=\frac{\pi}{4}\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\pi/4}\left[\cos x-\sin x\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\sin x+\cos x\right]_{0}^{\pi/4}\)
\(\,\,\,\,\,\,\displaystyle\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}\right)-\left(0+1\right)\)
\(\,\,\,\,\,\,\)Area\(=\sqrt{2}-1\)
\(\textbf{12)}\) \(y=e^x,\,y=1,\,x=0,\,x=\ln 3\)
Area\(=2-\ln 3\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{\ln 3}\left[e^x-1\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[e^x-x\right]_{0}^{\ln 3}\)
\(\,\,\,\,\,\,\displaystyle\left(e^{\ln 3}-\ln 3\right)-\left(e^0-0\right)\)
\(\,\,\,\,\,\,\displaystyle3-\ln 3-1\)
\(\,\,\,\,\,\,\)Area\(=2-\ln 3\)
\(\textbf{13)}\) \(x=y^2,\,x=4\)
Area\(=\frac{32}{3}\)
\(\,\,\,\,\,\,y^2=4\)
\(\,\,\,\,\,\,y=-2\text{ and }y=2\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{2}\left[4-y^2\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\left[4y-\frac{y^3}{3}\right]_{-2}^{2}\)
\(\,\,\,\,\,\,\displaystyle\left(8-\frac{8}{3}\right)-\left(-8+\frac{8}{3}\right)\)
\(\,\,\,\,\,\,\)Area\(=\frac{32}{3}\)
\(\textbf{14)}\) \(x=y^2,\,x=2y+3\)
Area\(=\frac{32}{3}\)
\(\,\,\,\,\,\,y^2=2y+3\)
\(\,\,\,\,\,\,y^2-2y-3=0\)
\(\,\,\,\,\,\,(y-3)(y+1)=0\)
\(\,\,\,\,\,\,y=-1\text{ and }y=3\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{3}\left[(2y+3)-y^2\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\left[y^2+3y-\frac{y^3}{3}\right]_{-1}^{3}\)
\(\,\,\,\,\,\,\displaystyle(9+9-9)-\left(1-3+\frac{1}{3}\right)\)
\(\,\,\,\,\,\,\)Area\(=\frac{32}{3}\)
\(\textbf{15)}\) \(y=|x|,\,y=2\)
Area\(=4\)
\(\,\,\,\,\,\,|x|=2\)
\(\,\,\,\,\,\,x=-2\text{ and }x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{2}\left[2-|x|\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle2\int_{0}^{2}\left[2-x\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle2\left[2x-\frac{x^2}{2}\right]_{0}^{2}\)
\(\,\,\,\,\,\,\displaystyle2(4-2)\)
\(\,\,\,\,\,\,\)Area\(=4\)
\(\textbf{16)}\) \(y=x^3,\,y=x\)
Area\(=\frac{1}{2}\)
\(\,\,\,\,\,\,x^3=x\)
\(\,\,\,\,\,\,x^3-x=0\)
\(\,\,\,\,\,\,x(x-1)(x+1)=0\)
\(\,\,\,\,\,\,x=-1,\,0,\,1\)
\(\,\,\,\,\,\,\displaystyle\int_{-1}^{0}\left[x^3-x\right]\,dx+\int_{0}^{1}\left[x-x^3\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_{-1}^{0}+\left[\frac{x^2}{2}-\frac{x^4}{4}\right]_{0}^{1}\)
\(\,\,\,\,\,\,\displaystyle\frac{1}{4}+\frac{1}{4}\)
\(\,\,\,\,\,\,\)Area\(=\frac{1}{2}\)
\(\textbf{17)}\) \(x=y^2,\,x=6-y\)
Area\(=\frac{125}{6}\)
\(\,\,\,\,\,\,y^2=6-y\)
\(\,\,\,\,\,\,y^2+y-6=0\)
\(\,\,\,\,\,\,(y+3)(y-2)=0\)
\(\,\,\,\,\,\,y=-3\text{ and }y=2\)
\(\,\,\,\,\,\,\displaystyle\int_{-3}^{2}\left[(6-y)-y^2\right]\,dy\)
\(\,\,\,\,\,\,\displaystyle\left[6y-\frac{y^2}{2}-\frac{y^3}{3}\right]_{-3}^{2}\)
\(\,\,\,\,\,\,\displaystyle\left(12-2-\frac{8}{3}\right)-\left(-18-\frac{9}{2}+9\right)\)
\(\,\,\,\,\,\,\)Area\(=\frac{125}{6}\)
\(\textbf{18)}\) \(y=x^2-4,\,y=0\)
Area\(=\frac{32}{3}\)
\(\,\,\,\,\,\,x^2-4=0\)
\(\,\,\,\,\,\,x=-2\text{ and }x=2\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{2}\left[0-(x^2-4)\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{-2}^{2}\left[4-x^2\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[4x-\frac{x^3}{3}\right]_{-2}^{2}\)
\(\,\,\,\,\,\,\)Area\(=\frac{32}{3}\)
\(\textbf{19)}\) \(y=4x-x^2,\,y=x\)
Area\(=\frac{9}{2}\)
\(\,\,\,\,\,\,4x-x^2=x\)
\(\,\,\,\,\,\,-x^2+3x=0\)
\(\,\,\,\,\,\,x(3-x)=0\)
\(\,\,\,\,\,\,x=0\text{ and }x=3\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3}\left[(4x-x^2)-x\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3}\left[3x-x^2\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_{0}^{3}\)
\(\,\,\,\,\,\,\displaystyle\frac{27}{2}-9\)
\(\,\,\,\,\,\,\)Area\(=\frac{9}{2}\)
\(\textbf{20)}\) \(y=\ln x,\,y=0,\,x=1,\,x=e\)
Area\(=1\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{e}\left[\ln x-0\right]\,dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{e}\ln x\,dx\)
\(\,\,\,\,\,\,\displaystyle x\ln x-x\,\Big|_{1}^{e}\)
\(\,\,\,\,\,\,\displaystyle(e\ln e-e)-\left(1\ln 1-1\right)\)
\(\,\,\,\,\,\,\displaystyle(e-e)-(0-1)\)
\(\,\,\,\,\,\,\)Area\(=1\)
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