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Practice
Application of Integrals
Application of Integrals: Problems with Solutions
By
Prof. Hernando Guzman Jaimes (University of Zulia - Maracaibo, Venezuela)
Problem 1
Write the definite integral that gives the area of the region between $y_{1}, y_{2}$
$y_{1}=x^{2}+2x+1, y_{2}=2x+5$.
What is the area?
$\mathbf{Area}= 10$
$\mathbf{Area}= 12$
$\mathbf{Area}= \frac{32}{3}$
$\mathbf{Area}= \frac{22}{3}$
Solution:
We must find the intersection points of the graphs.
$y_{1}=y_{2}\Longrightarrow x^{2}+2x+1=2x+5\Longrightarrow x^{2}-4=0$ then $x=\pm 2$
The area of this region we can be calculated in this way:
Area $=\overset{2}{\underset{-2}{\int }}\left[ 2x+5-x^{2}-2x-1\right] dx$
Area $=\overset{2}{\underset{-2}{\int }}\left[ 4-x^{2}\right] dx=\left[ 4x-\frac{x^{3}}{3}\right] _{-2}^{2}=\left( 8-\frac{8}{3}+8-\frac{8}{3}\right) =16-\frac{16}{3}= \frac{32}{3}$
Problem 2
Find the integral that allows to determine the area of the region between
$y_{1}=x^{2}-4x+3$ and $y_{2}=-x^{2}+2x+3$
What is the area?
$A=9$
$A=12$
$A=6$
$A=30$
Solution:
We must find the intersection points of the graphs.
$y_{1}=y_{2}\Longrightarrow x^{2}-4x+3=-x^{2}+2x+3\Longrightarrow 2x^{2}-6x=0 $ then $x=0$ and $x=3$
So $A=\overset{3}{\underset{0}{\int }}\left[ -x^{2}+2x+3-x^{2}+4x-3\right] dx=\overset{3}{\underset{0}{\int }}\left[ -2x^{2}+6x\right] dx$
$A=\left[ -\frac{2}{3}x^{3}+3x^{2}\right]_{0}^{3}=-18+27=9$
Problem 3
Consider the graph, calculate the area of the given region.
$y_{1}=3(x^{3}-x), y_{2}=0$
$\mathbf{A}=\frac{7}{2}$
$\mathbf{A}=\frac{1}{2}$
$\mathbf{A}=\frac{5}{2}$
$\mathbf{A}=\frac{3}{2}$
Solution:
We must find the intersection points of the graphs with $x-axis$.
$y_{1}=3(x^{3}-x)=0\Longrightarrow x=-1,0,1$
Then $A=\underset{}{\overset{0}{\underset{-1}{\int }}3(x^{3}-x)dx-}\underset{}{\overset{1}{\underset{0}{\int }}3(x^{3}-x)dx}$
$A=\left[ 3\left( \frac{x^{4}}{4}-\frac{x^{2}}{2}\right) \right]_{-1}^{0}-\left[ 3\left( \frac{x^{4}}{4}-\frac{x^{2}}{2}\right) \right]_{0}^{1}=3\left[ -\left( \frac{1}{4}-\frac{1}{2}\right) -\left( \frac{1}{4}-\frac{1}{2}\right) \right] $
$A=3\left( \frac{1}{4}+\frac{1}{4}\right) =\frac{3}{2}$
Problem 4
Find the area of the figure between the functions
$f(x)=-x^{2}+\frac{9}{2}x+1$ and $g(x)=\frac{1}{2}x+1$
$\mathbf{A}= 10$
$\mathbf{A}= 32$
$\mathbf{A}= 8$
$\mathbf{A}= \frac{32}{3}$
Solution:
We must find the intersection points of the graphs.
$f(x)=g(x)\Longrightarrow -x^{2}+\frac{9}{2}x+1=\frac{1}{2}x+1\Longrightarrow -x^{2}+4x=0$
then $x=0$ and $x=4$.
So $A=\underset{0}{\overset{4}{\int }}\left[ \left( -x^{2}+\frac{9}{2}x+1\right) -\left( \frac{1}{2}x+1\right) \right] dx$
$A=\underset{0}{\overset{4}{\int }}\left[ -x^{2}+4x\right] dx=\left[ -\frac{x^{3}}{3}+2x^{2}\right]_{0}^{4} = -\frac{64}{3}+32= \frac{32}{3}$
Problem 5
Determine the area of the region limited by $y=x, y=2-x, y=0$.
$\mathbf{A}= 2$
$\mathbf{A}= 1$
$\mathbf{A}= 3$
$\mathbf{A}= 5$
Solution:
We must find the intersection points of the straight lines, and the intersection between them and $x-axis.$
So $y_{1}=y_{2}\Longrightarrow x=2-x\Longrightarrow x=1$ then $x=0$ and $x=2$ are the intersections with $x-axis$
Area $=\overset{1}{\underset{0}{\int }}xdx+\underset{1}{\overset{2}{\int }}(2-x)dx=\left[ \frac{x^{2}}{2}\right]_{0}^{1}+\left[ 2x-\frac{x^{2}}{2}\right]_{1}^{2}$
Area $=\frac{1}{2}+4-2-2+\frac{1}{2}= 1$
Problem 6
Given functions $f(x)=\sqrt{x}+3,~g(x)=\frac{1}{2}x+3$.
Determine the area of the region obtained between them.
$\mathbf{A=}\frac{4}{3}$
$\mathbf{A=}4$
$\mathbf{A=}\frac{5}{3}$
$\mathbf{A=}2$
Solution:
We must find the points of intersection.
$f(x)=g(x)\Longrightarrow \sqrt{x}+3=\frac{1}{2}x+3\Longrightarrow \sqrt{x}=\frac{1}{2}x$ then $x=0$ or $x=4$
So area $=\overset{4}{\underset{0}{\int }}\left( f(x)-g(x)\right) dx=\overset{4}{\underset{0}{\int }}\left( \sqrt{x}+3-\frac{1}{2}x-3\right) dx=\overset{4}{\underset{0}{\int }}\left( \sqrt{x}-\frac{1}{2}x\right) dx$
Area $=\left[ \frac{2}{3}x^{\frac{3}{2}}-\frac{1}{4}x^{2}\right]_{0}^{4}=\left[ \frac{16}{3}-4\right] = \frac{4}{3}$
Problem 7
Calculate the area enclosed by the following functions
$f(x)=\sqrt[3]{x-1}$ and $g(x)=x-1$.
$\mathbf{A}=\frac{1}{3}$
$\mathbf{A}= 1$
$\mathbf{A}=\frac{1}{2}$
$\mathbf{A}=\frac{5}{6}$
Solution:
We must find the intersection points of the graphs.
$f(x)=g(x)\Longrightarrow \sqrt[3]{x-1}=x-1\Longrightarrow\\ x=0, x=1, x=2$
The enclosed area is $A=\underset{0}{\overset{1}{\int }}\left( x-1-\sqrt[3]{x-1}\right) dx+\overset{2}{\underset{1}{\int }}\left( \sqrt[3]{x-1}-x+1\right) dx$
$A=\left[ \frac{1}{2}x^{2}-x-\frac{3}{4}(x-1)^{\frac{4}{3}}\right]_{0}^{1}+\left[\frac{3}{4}(x-1)^{\frac{4}{3}}-\frac{1}{2}x^{2}+x\right] _{1}^{2}$
$A=\frac{1}{2}-1+\frac{3}{4}+\frac{3}{4}-2+2+\frac{1}{2}-1= \frac{1}{2}$
Problem 8
Find the area of the region determined by the following graphs
$f(x)=x^{2}-4x+3$ and $g(x)=3+4x-x^{2}$.
$A= 20$
$A= \frac{32}{3}$
$A= \frac{64}{3}$
$A= 15$
Solution:
First we find the points of intersection of the graphs.
$f(x)=g(x)\Longrightarrow x^{2}-4x+3=3+4x-x^{2}\Longrightarrow 2x^{2}-8x=0\Longrightarrow 2x(x-4)=0$
so $x=0\qquad x=4$
Area $=\underset{0}{\overset{4}{\int }}\left[ \left(3+4x-x^{2}\right) -\left( x^{2}-4x+3\right) \right] dxu.$
$A=\underset{0}{\overset{4}{\int }}\left[ 4x-x^{2}+4x-x^{2}\right] dx=\underset{0}{\overset{4}{\int }}\left[ 8x-2x^{2}\right] dx=\left[ 4x^{2}-\frac{2}{3}x^{3}\right] _{0}^{4}=64-\frac{128}{3}= \frac{64}{3}$
$A= \frac{64}{3}$
Problem 9
Determine the area of the region bounded by $y=x^{4}-2x^{2}$ and $y=2x^{2}$
$A=\frac{128}{15}$
$A=\frac{32}{3}$
$A=\frac{32}{5}$
$A=10$
Solution:
We must find the points of intersection of the graphs.
$f(x)=g(x)\Longrightarrow x^{4}-2x^{2}=2x^{2}\Longrightarrow x^{4}-4x^{2}=0\Longrightarrow x^{2}(x^{2}-4)=0$
So $x=-2\qquad x=0\qquad x=2$ are the points of intersection.
Considering the simetry of the graph
$A=2\overset{2}{\underset{0}{\int }}\left[ \left( 2x^{2}\right) -\left(x^{4}-2x^{2}\right) \right] dx=2\left[ \frac{4}{3}x^{3}-\frac{1}{5}x^{5}\right] _{0}^{2}=2\left( \frac{32}{3}-\frac{32}{5}\right) =\frac{128}{15}$
$A= \frac{128}{15}$
Problem 10
Determine the area of the region bounded by $f(x)=x^{4}-4x^{2}$ and $g(x)=x^{2}-4$.
$A=7$
$A=6$
$A=5$
$A=8$
Solution:
We must find the intersection points of the graphs.
$f(x)=g(x)\Longrightarrow x^{4}-4x^{2}=x^{2}-4\Longrightarrow x^{4}-5x^{2}+4=0$
Sustitute $u=x^{2}$ then $u^{2}-5u+4=0\Longrightarrow (u-4)(u-1)=0$
$u=4\qquad u=1$ and since $u=x^{2}\Longrightarrow x=\pm 1\qquad x=\pm 2$
are the points of intersection, if we considerer the simetry of graph
$A=2\left[ \underset{0}{\overset{1}{\int }}\left[ \left( x^{4}-4x^{2}\right) -\left( x^{2}-4\right) \right] dx+\underset{1}{\overset{2}{\int }}\left[ \left( x^{2}-4\right) -\left( x^{4}-4x^{2}\right) \right] dx\right] $
$A=2\left[ \underset{0}{\overset{1}{\int }}\left( x^{4}-5x^{2}+4\right) dx+\underset{1}{\overset{2}{\int }}\left( -x^{4}+5x^{2}-4\right) dx\right]$
$A=2\left( \left[ \frac{1}{5}x^{5}-\frac{5}{3}x^{3}+4x\right] _{0}^{1}+\left[-\frac{1}{5}x^{5}+\frac{5}{3}x^{3}-4x\right] _{1}^{2}\right)$
$A=2\left( \frac{1}{5}-\frac{5}{3}+4-\frac{1}{5}32+\frac{5}{3}8-8+\frac{1}{5}-\frac{5}{3}+4\right)=8$
Problem 11
Find the volume of the solid formed by revolving the region bounded by the curve
$y=4-x^{2}$ about the x−axis.
$V=2\pi$
$V=\frac{12}{15}\pi$
$V=\frac{512}{15}\pi$
$V=5\pi$
Solution:
$x-axis$ intersections are $x=\pm 2$ and the radius of disk is
$R(x)=f(x)$ so $V=\pi \overset{2}{\underset{-2}{\int }}$ $R(x)^{2}dx=\pi \overset{2}{\underset{-2}{\int }}$ $\left( 4-x^{2}\right) ^{2}dx=\pi \overset {2}{\underset{-2}{\int }}$ $\left( 16-8x^{2}+x^{4}\right) dx$
$V=\pi \left[ 16x-\frac{8}{3}x^{3}+\frac{1}{5}x^{5}\right]_{-2}^{2}=\pi \left[ 32-\frac{64}{3}+\frac{32}{5}+32-\frac{64}{3}+\frac{32}{5}\right] =\frac{512}{15}\pi $
$V=\frac{512}{15}\pi $
Problem 12
Find the volume of the solid formed by revolving the region bounded by the curve $y=\sqrt{x}$, $x \in [0, 3]$ about the x−axis.
$\frac{3}{2}\pi$
$\frac{5}{2}\pi$
$\frac{9}{2}\pi$
$\frac{1}{2}\pi$
Solution:
Since $R(x)=\sqrt{x}$, the volume is $V=\pi \underset{0}{\overset{3}{\int }}R^{2}(x)dx$
$V=\pi \underset{0}{\overset{3}{\int }}xdx=\pi \left[ \frac{1}{2}x^{2}\right]_{0}^{3}=\frac{9}{2}\pi $
Problem 13
Find the volume of the solid formed by revolving the region bounded by the curve
$f(x)=\sqrt{\sin x}$, $0< x < \pi $ about the x-axis.
$V=\pi$
$V=2\pi$
$V=3\pi $
$V=4\pi$
Solution:
$R(x)=f(x)=\sqrt{\sin x}$ so the volume is
$V=\pi \underset{0}{\overset{\pi }{\int }}\left( \sqrt{\sin x}\right)^{2}dx$
$V=\pi \underset{0}{\overset{\pi }{\int }}\sin xdx=-\pi \left[ \cos x\right] _{0}^{\pi }=-\pi \left[ \cos \pi -\cos 0\right] =-\pi \left(-1-1\right) =2\pi $
$V=2\pi $
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