PROBLEM 1: Compute the area of the enclosed region bounded by the graphs of the equations $y=x$, $y=2x$, and $x=4$.

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SOLUTION 1: Compute the area of the enclosed region bounded by the graphs of the equations $y=x$, $y=2x$ and $x=4$ . Begin by finding the points of intersection of the two graphs. From $y=x$ and $y=2x$ we get that

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$x = 2x \ \ \ \ \longrightarrow$

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$- x = 0 \ \ \ \ \longrightarrow$

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$${-x} = 0 \ \ \ \ \longrightarrow$$

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$x = 0 \ \ \ \ \longrightarrow$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{0}^{4} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{0}^{4} (2x - x) \ dx }$

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$= \displaystyle { \int_{0}^{4} x \ dx }$

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$= \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{4} }$

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$= \displaystyle { \frac{4^{2}}{2} - \frac{0^{2}}{2} }$

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$= \displaystyle { 8 - 0 }$

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$= \displaystyle { 8 }$

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PROBLEM 2: Compute the area of the enclosed region bounded by the graphs of the equations $y=x^2$ and $y=x+2$ .

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SOLUTION 2: Compute the area of the enclosed region bounded by the graphs of the equations $y=x^2$ and $y=x+2$ . Begin by finding the points of intersection of the two graphs. From $y=x^2$ and $y=x+2$ we get that

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$x^2=x+2 \ \ \ \ \longrightarrow$

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$x^2-x-2=0 \ \ \ \ \longrightarrow$

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$(x-2)(x+1)=0 \ \ \ \ \longrightarrow$

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$x=2$ or $x=-1$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{-1}^{2} (top \ - \ bottom) \ dx }$

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$= \displaystyle{ \int_{-1}^{2} ((x+2)-x^2) \ dx }$

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$= \displaystyle{ \Big( {x^2 \over 2}+2x-{x^3 \over 3} \Big) \Big\vert_{-1}^{2} }$

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$= \displaystyle{ \Big( {2^2 \over 2}+2(2)-{2^3 \over 3} \Big) - \Big( {(-1)^2 \over 2}+2(-1)-{(-1)^3 \over 3} \Big) }$

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$= \displaystyle{ \Big(2+4-{8 \over 3}\Big) - \Big({1 \over 2}-2+{1 \over 3}\Big) }$

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$= \displaystyle{ \Big(6-{8 \over 3}\Big) - \Big({3 \over 6}-{12 \over 6}+{2 \over 6}\Big) }$

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$= \displaystyle{ \Big({36 \over 6}-{16 \over 6}\Big) - \Big({-7 \over 6}\Big) }$

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$= \displaystyle{ {20 \over 6}+{7 \over 6} }$

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$= \displaystyle{ {27 \over 6} }$ .

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PROBLEM 3: Compute the area of the enclosed region bounded by the graphs of the equations $y=e^x, y=e^{-x},$ and $x=\ln 3$ .

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SOLUTION 3: Compute the area of the enclosed region bounded by the graphs of the equations $y=e^x$, $y=e^{-x}$, and $x=\ln 3$ . Begin by finding the points of intersection of the two graphs. From $y=e^{x}$ and $y=e^{-x}$ we get that

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$e^{x} = e^{-x} \ \ \ \ \longrightarrow$

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$e^{2x} = 1 \ \ \ \ \longrightarrow$

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$x = 0 \ \ \ \ \longrightarrow$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{0}^{\ln 3} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{0}^{\ln 3} (e^{x} - e^{-x}) \ dx }$

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$= \displaystyle { \Big(e^{x} - (-e^{-x})\Big) \Big\vert_{0}^{\ln 3} }$

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$= \displaystyle { \Big(e^{x} + e^{-x}\Big) \Big\vert_{0}^{\ln 3} }$

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$= \displaystyle { \Big(e^{\ln 3} + e^{-\ln 3}\Big) - \Big(e^{0} + e^{-0}\Big) }$

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$= \displaystyle { \Big(3 + \frac{1}{3}\Big) - \Big(1 + 1\Big) }$

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$= \displaystyle { \Big(\frac{10}{3}\Big) - \Big(2\Big) }$

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$= \displaystyle { \frac{10}{3} - \frac{6}{3} }$

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$= \displaystyle { \frac{4}{3} }$

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PROBLEM 4: Compute the area of the enclosed region bounded by the graphs of the equations $y=x^2$ and $y=-x^2+4x+6$ .

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SOLUTION 4: Compute the area of the enclosed region bounded by the graphs of the equations $y=x^2$ and $y=-x^{2}+4x+6$ . Begin by finding the points of intersection of the two graphs. From $y=x^2$ and $y=-x^{2}+4x+6$ we get that

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$x^{2} = -x^{2} + 4x + 6 \ \ \ \ \longrightarrow$

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$2x^{2} - 4x - 6 = 0 \ \ \ \ \longrightarrow$

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$x^{2} - 2x - 3 = 0 \ \ \ \ \longrightarrow$

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$(x-3)(x+1) = 0 \ \ \ \ \longrightarrow$

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$x = 3$ or $x=-1$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{-1}^{3} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{-1}^{3} ((-x^{2}+4x+6)-(x^{2})) \ dx }$

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$= \displaystyle { \int_{-1}^{3} (-x^{2}+4x+6) \ dx }$

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$= \displaystyle { \Big(\frac{-2x^{3}}{3} + \frac{4x^{2}}{2} + 6x \Big) \Big\vert_{-1}^{3} }$

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$= \displaystyle { \Big(-\frac{2}{3}x^{3} + 2x^{2} + 6x \Big) \Big\vert_{-1}^{3} }$

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$= \displaystyle { \Big( -\frac{2}{3}(3)^{3} + 2(3)^{2} + 6(3) \Big) - \Big( -\frac{2}{3}(-1)^{3} + 2(-1)^{2} + 6(-1) \Big) }$

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$= \displaystyle { \Big( -18 + 18 + 18 \Big) - \Big( \frac{2}{3} + 2 - 6 \Big) }$

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$= \displaystyle { \Big( 18 \Big) - \Big( \frac{2}{3} - \frac{6}{3} - \frac{18}{3} \Big) }$

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$= \displaystyle { \Big( 18 \Big) - \Big( - \frac{22}{3} \Big) }$

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$= \displaystyle { 18 + \frac{22}{3} }$

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$= \displaystyle { \frac{54}{3} + \frac{22}{3} }$

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$= \displaystyle {\frac{76}{3}}$

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PROBLEM 5: Compute the area of the enclosed region bounded by the graphs of the equations $y=x^3+x^2$ and $y=3x^2+3x$ .

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SOLUTION 2: Compute the area of the enclosed region bounded by the graphs of the equations $y=x^3+x^2$ and $y=3x^2+3x$ . Begin by finding the points of intersection of the two graphs. From $y=x^3+x^2$ and $y=3x^2+3x$ we get that

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$x^{3} + x^{2} = 3x^{2} + 3x \ \ \ \ \longrightarrow$

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$x^{3} - 2x^{2} - 3x = 0 \ \ \ \ \longrightarrow$

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$x(x^{2} - 2x - 3) = 0 \ \ \ \ \longrightarrow$

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$x(x-3)(x+1) = 0 \ \ \ \ \longrightarrow$

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$x=0$, $x = 3$, or $x=-1$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{-1}^{0} (top \ - \ bottom) \ dx + \int_{0}^{3} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{-1}^{0} ((x^{3}+x^{2})-(3x^{2}+3x)) \ dx + \int_{0}^{3} ((3x^{2}+3x) -(x^{3}+x^{2})) \ dx }$

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$= \displaystyle { \int_{-1}^{0} (x^{3}-2x^{2}-3x) \ dx + \int_{0}^{3} (-x^{3}+2x^{2}+3x) \ dx }$

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$= \displaystyle { \Big( \frac{x^{4}}{4} - \frac{2x^{3}}{3} - \frac{3x^{2}}{2} ... ...\frac{x^{4}}{4} + \frac{2x^{3}}{3} + \frac{3x^{2}}{2} \Big) \Big\vert_{0}^{3}}$

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$= \displaystyle { \Big( \frac{0^{4}}{4} - \frac{2(0)^{3}}{3} - \frac{3(0)^{2}}... ...ig) + \Big( -\frac{3^{4}}{4} + \frac{2(3)^{3}}{3} + \frac{3(3)^{2}}{2} \Big) }$

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$${ \ \ \ \ - \Big( -\frac{0^{4}}{4} + \frac{2(0)^{3}}{3} + \frac{3(0)^{2}}{2} \Big) }$$

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$= \displaystyle { \Big( 0 \Big) - \Big( \frac{1}{4} + \frac{2}{3} - \frac{3}{2} \Big) + \Big( -\frac{81}{4} + 18 + \frac{27}{2} \Big) - \Big( 0 \Big) }$

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$= \displaystyle { - \Big( \frac{3}{12} + \frac{8}{12} - \frac{18}{12} \Big) + \Big( -\frac{81}{4} + \frac{72}{4} + \frac{54}{4} \Big) }$

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$= \displaystyle { - \Big( -\frac{7}{12} \Big) + \Big( \frac{45}{4} \Big) }$

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$= \displaystyle { \frac{7}{12} + \frac{45}{4} }$

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$= \displaystyle { \frac{7}{12} + \frac{135}{12} }$

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$= \displaystyle { \frac{142}{12} }$

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$= \displaystyle { \frac{71}{6} }$

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PROBLEM 6: Compute the area of the enclosed region bounded by the graphs of the equations $y= \ln x, y=1,$ and $x=e^2$ .

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SOLUTION 6: Compute the area of the enclosed region bounded by the graphs of the equations $y=\ln x$, $y=1$ and $y=e^{2}$ . Begin by finding the points of intersection of the two graphs. From $y=\ln x$ and $y=1$ we get that

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$\ln x = 1 \ \ \ \ \longrightarrow$

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$e^{\ln x} = e^{1}$

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$x = e$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{e}^{e^2} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{e}^{e^{2}} (\ln x - 1) \ dx }$

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$= \displaystyle { \int_{e}^{e^{2}} \ln x \ dx - \int_{e}^{e^{2}} 1 \ dx }$

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Let $A = \displaystyle { \int \ln x \ dx }$.

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Use integration by parts. Let $u = \ln x$ and $dv = dx$ such that $du = \frac{1}{x} \ dx$ and $v = x$.

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$A = \displaystyle { x \ln x - \int (\frac{1}{x})(x) \ dx }$

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$= \displaystyle { x \ln x - \int \ dx }$

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$= \displaystyle { x \ln x - x + C }$

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$${ A - \int_{e}^{e^{2}} 1 \ dx }$$

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$= \displaystyle { \Big( x \ln x - x - x \Big) \Big\vert_{e}^{e^{2}} }$

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$= \displaystyle { \Big( x \ln x - 2x \Big) \Big\vert_{e}^{e^{2}} }$

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$= \displaystyle { \Big( e^{2} \ln e^{2} - 2e^{2} \Big) - \Big( e \ln e - 2e \Big) }$

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$= \displaystyle { \Big( e^{2}(2) - 2e^{2} \Big) - \Big( e - 2e \Big) }$

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$= \displaystyle { \Big( 0 \Big) - \Big( -e \Big) }$

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$= \displaystyle { e }$

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PROBLEM 7: Compute the area of the enclosed region bounded by the graphs of the equations $y= \cos x, y= \sin x,$ and $x=0$ .

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SOLUTION 7: Compute the area of the enclosed region bounded by the graphs of the equations $y = \cos x$, $y = \sin x$ and $x=0$ . Begin by finding the points of intersection of the two graphs. From $y=x^2$ and $y=x+2$ we get that

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$\cos x = \sin x \ \ \ \ \longrightarrow$

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$\frac{\cos x}{\sin x} = 1 \ \ \ \ \longrightarrow$

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$\cot x = 1 \ \ \ \ \longrightarrow$

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$x = \frac{\pi}{4}$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{0}^{\pi / 4} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{0}^{\pi / 4} ( \cos x - \sin x ) \ dx }$

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$= \displaystyle { \Big( \sin x - (-\cos x) \Big) \Big\vert_{0}^{\pi / 4} }$

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$= \displaystyle { \Big( \sin x + \cos x \Big) \Big\vert_{0}^{\pi / 4} }$

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$= \displaystyle { \Big( \sin \frac{\pi}{4} + \cos \frac{\pi}{4} \Big) - \Big( \sin 0 + \cos 0 \Big) }$

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$= \displaystyle { \Big( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \Big) - \Big( 0 + 1 \Big) }$

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$= \displaystyle { \sqrt{2} - 1 }$

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PROBLEM 8: Compute the area of the enclosed region bounded by the graphs of the equations $y=8/x, y=2x,$ and $y=2$ .

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SOLUTION 8: Compute the area of the enclosed region bounded by the graphs of the equations $y = \frac{8}{x}$, $y=2x$ and $y=2$ . Begin by finding the points of intersection of the two graphs. From $y = \frac{8}{x}$ and $y=2x$ we get that

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$\frac{8}{x} = 2x \ \ \ \ \longrightarrow$

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$8 = 2x^{2} \ \ \ \ \longrightarrow$

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$4 = x^{2} \ \ \ \ \longrightarrow$

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$x=2$ or $x = -2 \ \ \ \ \longrightarrow$

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$y = 4$ or $y = -4$

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Using horizontal cross-sections such that $x = \frac{8}{y}$ and $x = \frac{y}{2}$ we get that

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AREA $= \displaystyle{ \int_{2}^{4} (right \ - \ left ) \ dy }$

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$${ \int_{2}^{4} ( \frac{8}{y} - \frac{y}{2} ) \ dy }$$

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$= \displaystyle { \Big( 8 \ln y - \frac{y^{2}}{4} \Big) \Big\vert_{2}^{4} }$

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$= \displaystyle { \Big( 8 \ln 4 - \frac{4^{2}}{4} \Big) - \Big( 8 \ln 2 - \frac{2^{2}}{4} \Big) }$

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$= \displaystyle { \Big( 8 \ln 4 - 4 \Big) - \Big( 8 \ln 2 - 1 \Big) }$

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$= \displaystyle { \Big( 8 \ln 2^{2} - 4 \Big) - \Big( 8 \ln 2 - 1 \Big) }$

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$= \displaystyle { \Big( 16 \ln 2 - 4 \Big) - \Big( 8 \ln 2 - 1 \Big) }$

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$= \displaystyle { 8 \ln 2 - 3 }$

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PROBLEM 9: Compute the area of the enclosed region bounded by the graphs of the equations $y=\ln x$ and $y= (\ln x)^2$ .

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SOLUTION 9: Compute the area of the enclosed region bounded by the graphs of the equations $y=\ln x$ and $y= (\ln x)^2$ . Begin by finding the points of intersection of the two graphs. From $y=\ln x$ and $y= (\ln x)^2$ we get that

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$\ln x = (\ln x)^{2} \ \ \ \ \longrightarrow$

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$\ln x - (\ln x)^{2} = 0 \ \ \ \ \longrightarrow$

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$\ln x (1 - \ln x) \ \ \ \ \longrightarrow$

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$\ln x = 0$ or $\ln x = 1$ $\ \ \ \ \longrightarrow$

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$x = 1$ or $x = e$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{1}^{e} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{1}^{e} (\ln x - (\ln x)^{2}) \ dx }$

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$= \displaystyle { \int_{1}^{e} \ln x \ dx - \int_{1}^{e} (\ln x)^{2} \ dx }$

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Let $A = \int_{1}^{e} \ln x \ dx$ and $B = \displaystyle { \int (\ln x)^{2} \ dx }$.

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(Recall from PROBLEM 6 that $${ \int \ln x \ dx = x\ln - x + C }$$ )

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$A = \displaystyle { \Big( x \ln x - x \Big) \Big\vert_{1}^{e} }$

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$= \displaystyle { \Big( e \ln e - 1 \ln 1 \Big) - \Big( e - 1 \Big) }$

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$= \displaystyle { \Big( e - 0 \Big) - \Big( e - 1 \Big) }$

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$= \displaystyle { 1 }$

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Use integration by parts. Let $u = (\ln x)^{2}$ and $dv = dx$ such that $du = 2 \ln x (\frac{1}{x}) \ dx$ and $v = x$.

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$B = \displaystyle { \Big( x(\ln x)^{2} \Big) \Big\vert_{1}^{e} - \int_{1}^{e} 2 \ln x \ dx }$

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$= \displaystyle { \Big( x (\ln x)^{2} \Big) \Big\vert_{1}^{e} - 2 \Big( x \ln x - x \Big) \Big\vert_{1}^{e} }$

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$= \displaystyle { \Big( e(\ln e)^{2} - 1(\ln 1)^{2} \Big) - 2 \Big( (e \ln e - e) - (1 \ln 1 - 1) \Big) }$

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$= \displaystyle { \Big( (e - 0) \Big) - 2 \Big( (e - e) - (0 - 1) \Big) }$

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$= \displaystyle { e - 2 }$

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$A - B = 1 - (e - 2)$

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$= 3 - e$

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PROBLEM 10: Compute the area of the enclosed region bounded by the graphs of the equations $y=\tan^2 x$, $y=0$ and $x = 1$ .

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SOLUTION 10: Compute the area of the enclosed region bounded by the graphs of the equations $y=\tan^2 x$, $y=0$, and $x = 1$ . Begin by finding the points of intersection of the two graphs. From $y = \tan^{2} x$ and $y=0$ we get that

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$\tan^{2} x = 0 \ \ \ \ \longrightarrow$

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$\tan x = 0 \ \ \ \ \longrightarrow$

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$x=0$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{0}^{1} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{0}^{1} \tan^{2} x \ dx }$

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$= \displaystyle { \int_{0}^{1} ( \sec^{2} x - 1 ) \ dx }$

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$= \displaystyle { \Big( \tan x - x \Big) \Big\vert_{0}^{1} }$

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$= \displaystyle { \Big( \tan 1 - 1 \Big) - \Big( \tan 0 - 0 \Big) }$

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$= \displaystyle { \tan 1 - 1 }$

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PROBLEM 11: Compute the area of the enclosed region bounded by the graphs of the equations $y=x, y=2x,$ and $y=6-x$ .

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SOLUTION 11: Compute the area of the enclosed region bounded by the graphs of the equations $y=x$, $y=2x$, and $y=6-x$ . Begin by finding the points of intersection of the two graphs.

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From $y=x$ and $y=2x$ we get that

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$x = 2x \ \ \ \ \longrightarrow$

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$- x = 0 \ \ \ \ \longrightarrow$

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$x=0$

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From $y=x$ and $y=6-x$ we get that

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$x = 6-x \ \ \ \ \longrightarrow$

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$2x = 6 \ \ \ \ \longrightarrow$

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$x = 3$

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From $y=2x$ and $y=6-x$ we get that

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$2x = 6-x \ \ \ \ \longrightarrow$

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$3x = 6 \ \ \ \ \longrightarrow$

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$x=2$

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Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{0}^{2} (top \ - \ bottom) \ dx + \int_{2}^{3} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{0}^{2} (2x - x) \ dx + \int_{2}^{3} ((6-x) - x) \ dx }$

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$= \displaystyle { \int_{0}^{2} x \ dx + \int_{2}^{3} (6-2x) \ dx }$

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$= \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{2} + \Big( 6x - \frac{2x^{2}}{2} \Big) \Big\vert_{2}^{3} }$

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$= \displaystyle { \frac{x^{2}}{2} \Big\vert_{0}^{2} + \Big( 6x - x^{2} \Big) \Big\vert_{2}^{3} }$

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$= \displaystyle { \Big( \frac{2^{2}}{2} - \frac{0^{2}}{2} \Big) + \Big( (6 \cdot 3 - 3^{2}) - (6 \cdot 2 - 2^{2}) \Big) }$

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$= \displaystyle { \Big( 2 - 0 \Big) + \Big( 9 - 8 \Big) }$

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$= \displaystyle { 2 + 1 }$

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$= \displaystyle { 3 }$

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PROBLEM 12: Compute the area of the enclosed region bounded by the graphs of the equations $y = \sin \sqrt{x}$, $y=0$, and $x = \pi / 2$ .

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SOLUTION 12: Compute the area of the enclosed region bounded by the graphs of the equations $y = \sin \sqrt{x}$, $y=0$, and $x = \pi / 2$. Begin by finding the points of intersection of the two graphs. From $y = \sin \sqrt{x}$ and $y=0$ we get that

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$\sin \sqrt{x} = 0 \ \ \ \ \longrightarrow$

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$x=0$

Using vertical cross-sections we get that

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AREA $= \displaystyle{ \int_{0}^{\pi / 2} (top \ - \ bottom) \ dx }$

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$= \displaystyle { \int_{0}^{\pi / 2} ( \sin \sqrt{x} ) \ dx }$

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Use integration by parts. Let $u = \sin \sqrt{x}$ and $dv = dx$ such that $du = \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \ dx$ and $v = x$.

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$= \displaystyle { x \sin \sqrt{x} \Big\vert_{0}^{\pi / 2} - \int_{0}^{\pi / 2} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot x \ dx }$

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$= \displaystyle { \Big( (\frac{\pi}{2}) \sin \sqrt{\frac{\pi}{2}} - (0) \sin \... ...) - \int_{0}^{\pi / 2} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot x \ dx }$

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$= \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} - \int_{0}^{\pi / 2} \cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \cdot x \ dx }$

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Use substitution. Let

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$w = \sqrt{x}$

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so that

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$dw = \frac{1}{2 \sqrt{x}} \ dx = \frac{\sqrt{x}}{2x} \ dx$

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$x \ dw = \frac{\sqrt{x}}{2} \ dx \ \ \ \ \longrightarrow$

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$w^{2} \ dw = \frac{\sqrt{x}}{2} \ dx$

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$= \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} - \int_{0}^{\sqrt{\pi / 2}} \cos w \cdot w^{2} \ dw }$

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Use integration by parts. Let $u' = w^{2}$ and $dv' = \cos w \ dw$ such that $du' = 2w \ dw$ and $v' = \sin w$.

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$= \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} - \Big( w^{2} \sin w... ...g\vert_{0}^{\sqrt{\pi / 2}} - \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw \Big) }$

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$= \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} - w^{2} \sin w \Big\vert_{0}^{\sqrt{\pi / 2}} + \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw }$

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$= \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} - \Big( (\sqrt{\frac... ...c{\pi}{2}} - (0)^{2} \sin 0 \Big) + \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw }$

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$= \displaystyle { \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} - \frac{\pi}{2} \sin \sqrt{\frac{\pi}{2}} + \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw}$

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$= \displaystyle { \int_{0}^{\sqrt{\pi / 2}} 2w \sin w \ dw }$

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Use integration by parts. Let $u'' = 2w$ and $dv'' = \sin w \ dw$ such that $du'' = 2 \ dw$ and $v'' = - \cos w$

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$= \displaystyle { -2 w \cos w \Big\vert_{0}^{\sqrt{\pi / 2}} + \int_{0}^{\sqrt{\pi / 2}} 2 \cos w \ dw }$

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$= \displaystyle { -2 \Big( (\sqrt{\frac{\pi}{2}}) \cos \sqrt{\frac{\pi}{2}} - (0) \cos 0 \Big) + 2 \sin w \Big\vert_{0}^{\sqrt{\frac{\pi}{2}}} }$

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$= \displaystyle { -2(\sqrt{\frac{\pi}{2}}) \cos \sqrt{\frac{\pi}{2}} + 2 \Big( \sin \sqrt{\frac{\pi}{2}} - \sin 0 \Big) }$

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$= \displaystyle { -\sqrt{2 \pi} \cos \sqrt{\frac{\pi}{2}} + 2 \sin \sqrt{\frac{\pi}{2}} }$

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PROBLEM 13: Compute the area of the enclosed region bounded by the graphs of the equations $x=y^2$ and $x=4$ .

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SOLUTION 13: Compute the area of the enclosed region bounded by the graphs of the equations $x=y^{2}$ and $x=4$ . Begin by finding the points of intersection of the two graphs. From $x=y^{2}$ and $x=4$ we get that

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$y^{2} = 4 \ \ \ \ \longrightarrow$

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$y = -2$ or $y=2$

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Using horizontal cross-sections we get that

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AREA $= \displaystyle{ \int_{-2}^{2} (right \ - \ left) \ dy }$

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$= \displaystyle { \int_{-2}^{2} (4 - y^{2}) \ dy }$

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$= \displaystyle { \Big( 4y - \frac{y^{3}}{3} \Big) \Big\vert_{-2}^{2} }$

truein

$= \displaystyle { \Big( 4 \cdot 2 - \frac{2^{3}}{3} \Big) - \Big( 4 \cdot (-2) - \frac{(-2)^{3}}{3} \Big) }$

truein

$= \displaystyle { \Big( 8 - \frac{8}{3} \Big) - \Big( -8 + \frac{8}{3} \Big) }$

truein

$= \displaystyle { 16 - \frac{16}{3} }$

truein

$= \displaystyle { \frac{48}{3} - \frac{16}{3} }$

truein

$= \displaystyle { \frac{32}{3} }$

trueintrueintruein

PROBLEM 14: Compute the area of the enclosed region bounded by the graphs of the equations $x=y+3$ and $x=y^2-y$ .

trueintruein

SOLUTION 14: Compute the area of the enclosed region bounded by the graphs of the equations $x=y+3$ and $x=y^{2}-y$ . Begin by finding the points of intersection of the two graphs. From $x=y+3$ and $x=y^{2}-y$ we get that

truein

$y+3 = y^{2}-y \ \ \ \ \longrightarrow$

truein

$0 = y^{2} - 2y - 3 \ \ \ \ \longrightarrow$

truein

$0 = (y-3)(y+1) \ \ \ \ \longrightarrow$

truein

$y=3$ or $y=-1$

truein

Using horizontal cross-sections we get that

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AREA $= \displaystyle{ \int_{-1}^{3} (right \ - \ left) \ dy }$

truein

$= \displaystyle { \int_{-1}^{3} ((y+3)-(y^{2}-y)) \ dy }$

truein

$= \displaystyle { \int_{-1}^{3} (-y^{2}+2y+3) \ dy }$

truein

$= \displaystyle { \Big( -\frac{y^{3}}{3} + \frac{2y^{2}}{2} +3y \Big) \Big\vert_{-1}^{3} }$

truein

$= \displaystyle { \Big( -\frac{y^{3}}{3} + y^{2} +3y \Big) \Big\vert_{-1}^{3} }$

truein

$= \displaystyle { \Big( -\frac{3^{3}}{3} + 3^{2} + 3(3) \Big) - \Big( -\frac{(-1)^{3}}{3} + (-1)^{2} + 3(-1) \Big) }$

truein

$= \displaystyle { \Big( -9 + 9 + 9 \Big) - \Big( \frac{1}{3} + 1 - 3 \Big) }$

truein

$= \displaystyle { \Big( -9 + 9 + 9 \Big) - \Big( \frac{1}{3} + \frac{3}{3} - \frac{9}{3} \Big) }$

truein

$= \displaystyle { \Big( 9 \Big) - \Big( -\frac{5}{3} \Big) }$

truein

$= \displaystyle { \frac{27}{3} + \frac{5}{3} }$

truein

$= \displaystyle { \frac{32}{3} }$

trueintrueintruein

PROBLEM 15: Compute the area of the enclosed region bounded by the graphs of the equations $x=y^3$ and $x=y^2+2y$.

trueintruein

SOLUTION 15: Compute the area of the enclosed region bounded by the graphs of the equations $x=y^3$ and $x=y^{2}+2y$ . Begin by finding the points of intersection of the two graphs. From $x=y^3$ and $x=y^2+2y$ we get that

truein

$y^{3} = y^2 + 2y \ \ \ \ \longrightarrow$

truein

$y^{3} - y^2 - 2y = 0 \ \ \ \ \longrightarrow$

truein

$y ( y^{2} - y - 2 ) = 0 \ \ \ \ \longrightarrow$

truein

$y ( y-2 ) ( y+1 ) = 0 \ \ \ \ \longrightarrow$

truein

$y=0$, $y=2$, or $y=-1$

truein

Using horizontal cross-sections we get that

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AREA $= \displaystyle{ \int_{-1}^{0} (right \ - \ left) \ dy + \int_{0}^{2} (right \ - \ left) \ dy }$

truein

$= \displaystyle { \int_{-1}^{0} (y^{3} - (y^{2}+2y)) \ dy + \int_{0}^{2} (y^{2}+2y - y^{3}) \ dy }$

truein

$= \displaystyle { \int_{-1}^{0} (y^{3} - y^{2} - 2y) \ dy + \int_{0}^{2} (y^{2}+2y - y^{3}) \ dy }$

truein

$= \displaystyle { \Big( \frac{y^{4}}{4} - \frac{y^{3}}{3} - \frac{2y^{2}}{2} \... ...\frac{y^{3}}{3} + \frac{2y^{2}}{2} - \frac{y^{4}}{4} \Big) \Big\vert_{0}^{2} }$

truein

$= \displaystyle { \Big( \frac{y^{4}}{4} - \frac{y^{3}}{3} - y^{2} \Big) \Big\v... ...0} + \Big( \frac{y^{3}}{3} + y^{2} - \frac{y^{4}}{4} \Big) \Big\vert_{0}^{2} }$

truein

$= \displaystyle { \Big( \frac{0^{4}}{4} - \frac{0^{3}}{3} - 0^{2} \Big) - \Big... ...frac{2^{4}}{4} \Big) - \Big( \frac{0^{3}}{3} + 0^{2} - \frac{0^{4}}{4} \Big) }$

truein

$= \displaystyle { \Big( 0 \Big) - \Big( \frac{1}{4} + \frac{1}{3} - 1 \Big) + \Big( \frac{8}{3} + 4 - 4 \Big) - \Big( 0 \Big) }$

truein

$= \displaystyle { -\Big( -\frac{7}{12} \Big) + \Big( \frac{8}{3} \Big) }$

truein

$= \displaystyle { \frac{7}{12} + \frac{8}{3} }$

truein

$= \displaystyle { \frac{7}{12} + \frac{32}{12} }$

truein

$= \displaystyle { \frac{39}{12} }$