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Question 1 The area under the curve \(y=x^4+3\) from \(x=0\) to \(x=1\) is \(\frac{16}{5}\).
Step 1: Since \(y=x^4+3\) is positive on \([0,1]\), the area under the curve is \(\int_0^1(x^4+3)\,dx\). Step 2: Split the integral: \(\int_0^1x^4\,dx+\int_0^1 3\,dx\). Step 3: Find the antiderivative: \(\frac{x^5}{5}+3x\). Step 4: Evaluate from \(0\) to \(1\): \(\left[\frac{x^5}{5}+3x\right]_0^1=\frac{1}{5}+3=\frac{16}{5}\). Step 5: The statement is true. Answer: B. True
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Question 2 The area enclosed between the curves \(y=e^x\) and \(y=e^{-x}\) from \(x=-2\) to \(x=0\), is \(e^2+\frac{1}{e^2}-2\).
Step 1: On \([-2,0]\), \(e^{-x}\) is above \(e^x\). Step 2: The area between the curves is \(\int_{-2}^{0}(e^{-x}-e^x)\,dx\). Step 3: Find the antiderivative: \(\int(e^{-x}-e^x)\,dx=-e^{-x}-e^x\). Step 4: Evaluate from \(-2\) to \(0\): \([-e^{-x}-e^x]_{-2}^{0}=(-1-1)-(-e^2-e^{-2})\). Step 5: Simplify: \(-2+e^2+e^{-2}=e^2+\frac{1}{e^2}-2\). Step 6: The statement is true. Answer: A. True
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Question 3 The region enclosed by the curve \(y=\sin x\) and the x-axis from \(x=0\) to \(x=\pi\) is rotated about the x-axis to create a solid. The volume of this solid is approximately \(4.93\) units\(^3\).
Step 1: Rotating the region under \(y=\sin x\) about the x-axis creates disks with radius \(r=\sin x\). Step 2: Use the disk method: \(V=\pi\int_0^{\pi}r^2\,dx=\pi\int_0^{\pi}\sin^2 x\,dx\). Step 3: Use the identity \(\sin^2 x=\frac{1-\cos(2x)}{2}\). Step 4: Then \(\int_0^{\pi}\sin^2 x\,dx=\frac{\pi}{2}\). Step 5: Therefore, \(V=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}\approx 4.93\). Step 6: The statement is true. Answer: A. True
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Question 4 The region enclosed by the curves \(y=2x\) and \(y=\sqrt{x}+1\) from \(x=0\) to \(x=1\) is rotated around the y-axis to create a solid. The volume of the solid is:
Step 1: The region is rotated around the y-axis, so use the shell method with respect to \(x\). Step 2: The shell radius is \(x\). Step 3: The shell height is upper curve minus lower curve: \((\sqrt{x}+1)-2x\). Step 4: Set up the volume: \(V=\int_0^1 2\pi x(\sqrt{x}+1-2x)\,dx\). Step 5: Expand: \(V=2\pi\int_0^1(x^{\frac{3}{2}}+x-2x^2)\,dx\). Step 6: Integrate: \(V=2\pi\left[\frac{2}{5}x^{\frac{5}{2}}+\frac{x^2}{2}-\frac{2x^3}{3}\right]_0^1\). Step 7: Substitute \(x=1\): \(V=2\pi\left(\frac{2}{5}+\frac{1}{2}-\frac{2}{3}\right)=2\pi\left(\frac{7}{30}\right)=\frac{7\pi}{15}\). Answer: D. \(\frac{7\pi}{15}\)
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