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Express the area between the graphs of \(y=x^{2}\) and \(y=\sqrt{x}\) as an integral using a partition of the \(x\)-axis. Do not find the integral.

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Express the area between the graph of \(x=y^{2}\) and the line \(x=1\) as an integral using a partition of the \(y\)-axis. Do not find the integral.

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\(y = x, y =2x, x=1\)

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\(y = x, y =3x, x=3\)

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\(y=x^{2}, y=x\)

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\(y=x^{2}, y=4x\)

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\(y =e^{x}, y =e^{-x}, x=\ln 2\)

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\(y =e^{x}, y =-x+1, x=1\)

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\(y=x^{2}, y=x^{4}\)

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\(y=x, y=x^{3}\)

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\(y=\cos x, y=\dfrac{1}{2}, 0\leq x\leq \dfrac{\pi}{3}\)

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\(y=\sin x, \ y=\dfrac{1}{2}, \dfrac{\pi}{6}\leq x\leq \dfrac{5\pi}{6}\)

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\(x=y^{2}, x=2-y\)

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

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\(x=9-y^{2}, x=5\)

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\(x=16-y^{2}, x=7\)

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\(x=y^{2}+4, y=x-6\)

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\(x=y^{2}+6, y=8-x\)

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\(y=\ln x, x=1, y=2\)

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\(y=\ln x, x=e, y=0\)

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\(y=\sqrt{x}, y=x^{3}\)

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\(y=\sqrt{x}, y=x^{2}\)

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\(y=x^{2}+1, y=x+1\)

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\(y=x^{2}+1, y=4x+1\)

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\(y=\sqrt{9-x}, y=\sqrt{9-3x}, x\)-axis

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\(y=\sqrt{16-2x}, y=\sqrt{16-4x}, x\)-axis

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\(y=\sqrt{2x-6}, y=\sqrt{x-2}, x\)-axis

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\(y=\sqrt{2x-5}, y=\sqrt{4x-17}, x\)-axis

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\(y=4-x^{2}, y=x^{2}\)

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\(y=9-x^{2}, y=x^{2}\)

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\(x=y^{2}-4, x=4-y^{2}\)

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\(x=y^{2}, x=16-y^{2}\)

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\(y=\ln x^{2}\), the \(x\)-axis, and the line \(x=e\)

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\(y=\ln x, y=1-x\), and the line \(y=1\)

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\(y=\cos x, y=1-\dfrac{3}{\pi}x, x=\dfrac{\pi}{3}\)

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\(y=\sin x, y=1, 0\leq x\leq \dfrac{\pi}{2}\)

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\(y=e^{2x}\) and the lines \(x=1\) and \(y=1\)

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\(y=e^{x}, y=e^{3x}, x=2\)

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\(y^{2}=4x, 4x-3y-4=0\)

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\(y^{2}=4x+1, x=y+1\)

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\(y=\sin x, y=\dfrac{2x}{\pi}, x\geq 0\)

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\(y=\cos x, x\geq 0, y=\dfrac{3x}{\pi}\)

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An Archimedean Result Show that the area of the shaded region in the figure is two-thirds of the area of the parallelogram ABCD. (This illustrates a result due to Archimedes concerning sectors of parabolas.)

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Equal Areas Find \(h\) so that the area of the region enclosed by the graphs of \(y=x, y=8x\), and \(y=\dfrac{1}{x^{2}}\) is equal to that of an isosceles triangle of base \(1\) and height \(h\). See the figure.

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Cost of Health Care The cost of health care varies from one country to another. Between 2000 and 2010, the average cost of health insurance for a family of four in the United States was modeled by \begin{equation*} A(x) =8020.6596( 1.0855^{x})\qquad 0\leq x\leq 10 \end{equation*}

where \(x=0\) corresponds to the year 2000 and \(A(x)\) is measured in U.S. dollars. During the same years, the average cost of health care in Canada was given by \begin{equation*} C(x) =4944.6424( 1.0711^{x}) \qquad 0\leq x\leq 10 \end{equation*}

where \(x=0\) corresponds to the year 2000 and \(C(x)\) is measured in U.S. dollars.

1. Find the area between the graphs from \(x=0\) to \(x=10\). Round the answer to the nearest dollar.
2. Interpret the answer to (a).

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Area Find the area in the first quadrant enclosed by the graphs of \(y=\sin (2x)\) and \(y=\cos (2x), 0\leq x\leq \dfrac{\pi}{8}\).

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Area Find the area of the region enclosed by the graphs of \(y=\sin ^{-1}x, y=x, 0\leq x\leq \dfrac{1}{2}\). Which axis did you choose to partition? Explain your choice.

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Area

1. Graph \(y=x^{2}\) and \(y=\sin x\).
2. Find the points of intersection of the graphs.
3. Find the area enclosed by the graphs of \(y=x^{2}\) and \(y=\sin x\) using a partition of the \(x\)-axis.
4. Find the area enclosed by the graphs of \(y=x^{2}\) and \(y=\sin x\) using a partition of the \(y\)-axis.

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1. Graph \(y=\sin ^{-1}x, x+y=1\), and \(y=0\).
2. Find the points of intersection of the graphs.
3. Find the area enclosed by the graphs.

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1. Graph \(y=\cos ^{-1}x, y=x^{3}+1\), and \(y=0\).
2. Find the points of intersection of the graphs.
3. Find the area enclosed by the graphs.

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1. Graph \(2x+y=3, y=\dfrac{1}{1+x^{2}}\), and \(y=1\).
2. Find the points of intersection of the graphs.
3. Find the area enclosed by the graphs.

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1. Graph \(y=\dfrac{5}{1+x^{2}}, x=\sqrt{y+2}\), and \(x=0\).
2. Find the points of intersection of the graphs.
3. Find the area enclosed by the graphs.

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1. Graph \(y=\sin ^{-1}x, y=1-x^{2}\), and \(y=0\).
2. Find the points of intersection of the graphs.
3. Find the area enclosed by the graphs.

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Find the area enclosed by the graph of \(y^{2}=x^{2}-x^{4}\).

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1. Express the area \(A\) of the region in the first quadrant enclosed by the \(y\)-axis and the graphs of \(y=\tan x\) and \(y=k\) for \(k>0\) as a function of \(k\).

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2. What is the value of \(A\) when \(k=1\)?
3. If the line \(y=k\) is moving upward at the rate of \(\dfrac{1}{10}\) unit per second, at what rate is \(A\) changing when \(k=1\)?

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The area \(A\) of the shaded region in the figure is \(A = \dfrac{t}{2}\). Prove this as follows:

1. Let \(A^{\ast}\) be twice the area outlined in the figure, and \((x,y)\) be the coordinates of \(P\). Explain why \begin{equation*} A^{\ast }=x\sqrt{x^{2}-1}-2\int_{1}^{x}\sqrt{u^{2}-1}\,{\it du} \end{equation*}
2. Differentiate \(A^*\) to show that \(\dfrac{dA^{\ast }}{{\it dx}}=\dfrac{1}{\sqrt{x^{2}-1}}\).
3. Show that \(A^{\ast }=\cosh ^{-1}x+C\), and explain why \(C=0\).
4. Why does it follow that \(A^{\ast }=t\)?