Notes

Questions

Find the area of the region bounded by the equations

\(\textbf{1)}\) \(f(x)=x^2+4,\,g(x)=\frac{1}{2}x+1,\,x=0 \, x=3\)

Show Area

Area\(=15.75\)

Show Work

\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left[\left(x^2+4\right) – \left(\frac{1}{2}x+1\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left[x^2+4 – \frac{1}{2}x-1\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{3} \left[x^2 – \frac{1}{2}x +3 \right] \, dx\)
\(\,\,\,\,\,\,\displaystyle \frac{x^3}{3} – \frac{x^2}{4} +3x \, \Big|_{0}^{3}\)
\(\,\,\,\,\,\,\displaystyle \frac{(3)^3}{3} – \frac{(3)^2}{4} +3(3) – \left(\frac{(0)^3}{3} – \frac{(0)^2}{4} +3(0)\right) \)
\(\,\,\,\,\,\,\displaystyle \frac{27}{3} – \frac{9}{4} +9 – (0) \)
\(\,\,\,\,\,\,\displaystyle 9 – \frac{9}{4} +9 – (0) \)
\(\,\,\,\,\,\,\displaystyle 18 – \frac{9}{4}\)
\(\,\,\,\,\,\,\)The area\(=15.75\)

 

\(\textbf{2)}\) \(f(x)=\sqrt{x},\,g(x)=x \)

Show Area

Area\(=\frac{1}{6}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left[\left(\sqrt{x}\right) – \left(x\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{0}^{1} \left[\left(x^{1/2}\right) – \left(x\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\left[\left(\frac{2x^{3/2}}{3}\right) – \left(\frac{x^2}{2}\right)\right] \, \Big|_{0}^{1}\)
\(\,\,\,\,\,\,\displaystyle\left[\left(\frac{2(1)^{3/2}}{3}\right) – \left(\frac{(1)^2}{2}\right)\right]-\left[\left(\frac{2(0)^{3/2}}{3}\right) – \left(\frac{(0)^2}{2}\right)\right] \)
\(\,\,\,\,\,\,\displaystyle\left[\left(\frac{2}{3}\right) – \left(\frac{1}{2}\right)\right]-\left[0\right] \)
\(\,\,\,\,\,\,\displaystyle\frac{4}{6}-\frac{3}{6}\)
\(\,\,\,\,\,\,\)Area\(=\frac{1}{6}\)

 

\(\textbf{3)}\) \(f(x)=2x+3,\,g(x)=x^2-4x+4,\,x=1,\,x=3\)

Show Area

Area\(=\frac{40}{3}\) or \(13.\overline{3}\)

Show Work

\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left[\left(2x+3\right) – \left(x^2-4x+4\right)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left[2x+3 – x^2+4x-4\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{3} \left[-x^2 + 6x -1\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle \left[-\frac{x^3}{3} + \frac{6x^2}{2} -1x\right] \, \Big|_{1}^{3}\)
\(\,\,\,\,\,\,\displaystyle \left(-\frac{(3)^3}{3} + \frac{6(3)^2}{2} -1(3)\right)-\left(-\frac{(1)^3}{3} + \frac{6(1)^2}{2} -1(1)\right) \)
\(\,\,\,\,\,\,\displaystyle \left(-\frac{27}{3} + \frac{54}{2} -3\right)-\left(-\frac{1}{3} + \frac{6}{2} -1\right) \)
\(\,\,\,\,\,\,\displaystyle \left(-9 + 27 -3\right)-\left(-\frac{1}{3} + 3 -1\right) \)
\(\,\,\,\,\,\,\displaystyle \left(15\right)-\left(1\frac{2}{3}\right) \)
\(\,\,\,\,\,\,\displaystyle \left(15\right)-\left(1\frac{2}{3}\right) \)
\(\,\,\,\,\,\,\)Area\(=\frac{40}{3}=13.\overline{3}\)

 

\(\textbf{4)}\) \(f(x)=|x|,\,g(x)=\frac{1}{x},\,x=5\)

Show Area

Area\(≈10.391\)

Show Work

\(\,\,\,\,\,\,\displaystyle\int_{a}^{b} \left[f(x) – g(x)\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{5} \left[|x| – \frac{1}{x}\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle\int_{1}^{5} \left[x – \frac{1}{x}\right] \, dx\)
\(\,\,\,\,\,\,\displaystyle \left[\frac{x^2}{2} – \ln{x}\right] \, \Big|_{1}^{5}\)
\(\,\,\,\,\,\,\displaystyle \left(\frac{(5)^2}{2} – \ln{(5)}\right)-\left(\frac{(1)^2}{2} – \ln{(1)}\right) \)
\(\,\,\,\,\,\,\displaystyle \frac{25}{2} – \ln{(5)}-\frac{1}{2} + \ln{(1)} \)
\(\,\,\,\,\,\,\displaystyle 12 – \ln{(5)}- \ln{(1)} \)
\(\,\,\,\,\,\,\)Area\(≈10.391\)

 

\(\textbf{5)}\) \(x=y^2,\,x=y+3\)

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Area\(≈7.81\)

 

\(\textbf{6)}\) \(x=2y+4,\,x=y^2 ,\, f(x)=1,\,g(x)=3\)

Show Area

Area\(=\frac{22}{3}\approx 7.3\)

 

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Trapezoidal Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{b-a}{2n}\left[f(a)+2f(x_1)+2f(x_2)+…+2fx_{n-1}+f(b)\right]…\)

\(\bullet\text{ Properties of Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}cf(x) \, dx=c\displaystyle \int_{a}^{b}f(x) \,dx…\)

\(\bullet\text{ Indefinite Integrals- Power Rule}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int x^n \, dx = \displaystyle \frac{x^{n+1}}{n+1}+C…\)

\(\bullet\text{ Indefinite Integrals- Trig Functions}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int \cos{x} \, dx=\sin{x}+C…\)

\(\bullet\text{ Definite Integrals}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{5}^{7} x^3 \, dx…\)

\(\bullet\text{ Integration by Substitution}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int (x^2+3)^3(2x) \,dx…\)

\(\bullet\text{ Area of Region Between Two Curves}\)
\(\,\,\,\,\,\,\,\,A=\displaystyle \int_{a}^{b}\left[f(x)-g(x)\right]\,dx…\)

\(\bullet\text{ Arc Length}\)
\(\,\,\,\,\,\,\,\,\displaystyle \int_{a}^{b}\sqrt{1+\left[f'(x)\right]^2} \,dx…\)

\(\bullet\text{ Average Function Value}\)
\(\,\,\,\,\,\,\,\,\displaystyle\frac{1}{b-a} \int_{a}^{b}f(x) \,dx\)

\(\bullet\text{ Volume by Cross Sections}\)
\(\,\,\,\,\,\,\,\,\)

\(\bullet\text{ Disk Method}\)
\(\,\,\,\,\,\,\,\,V=\displaystyle \int_{a}^{b}\left[f(x)\right]^2\,dx…\)

\(\bullet\text{ Cylindrical Shells}\)
\(\,\,\,\,\,\,\,\,V=2 \pi \displaystyle \int_{a}^{b} y f(y) \, dy…\)

 

In Summary

We use integrals to find the area of a region between two curves because it is a useful tool for solving problems in a variety of fields, including physics, engineering, and economics.