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Question 1 The area under the curve \(y=x^3+1\) from \(x=0\) to \(x=2\) is \(8\).
Step 1: Since \(y=x^3+1\) is positive on \([0,2]\), the area under the curve is \(\int_0^2 (x^3+1)\,dx\). Step 2: Split the integral: \(\int_0^2 x^3\,dx+\int_0^2 1\,dx\). Step 3: Find the antiderivative: \(\frac{x^4}{4}+x\). Step 4: Evaluate from \(0\) to \(2\): \(\left[\frac{x^4}{4}+x\right]_0^2=\frac{2^4}{4}+2-0=4+2=6\). Step 5: Since \(6\neq 8\), the statement is false. Answer: A. False
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Question 2 What is the area between the curves \(y=\sin x\) and \(y=\cos x\) on the interval from \(x=0\) to \(x=\frac{\pi}{4}\)?
Step 1: On \([0,\frac{\pi}{4}]\), \(\cos x\) is above \(\sin x\). Step 2: The area between the curves is \(\int_0^{\frac{\pi}{4}}(\cos x-\sin x)\,dx\). Step 3: Find the antiderivative: \(\int(\cos x-\sin x)\,dx=\sin x+\cos x\). Step 4: Evaluate from \(0\) to \(\frac{\pi}{4}\): \(\left[\sin x+\cos x\right]_0^{\frac{\pi}{4}}\). Step 5: Substitute the bounds: \(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}-(\sin 0+\cos 0)=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}-(0+1)=\sqrt{2}-1\). Answer: B. \(\sqrt{2}-1\)
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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 formula: \(V=\pi\int_0^{\pi}r^2\,dx=\pi\int_0^{\pi}\sin^2 x\,dx\). Step 3: Use the identity \(\sin^2 x=\frac{1-\cos(2x)}{2}\). Step 4: Then \(V=\pi\int_0^{\pi}\frac{1-\cos(2x)}{2}\,dx\). Step 5: Evaluate: \(\int_0^{\pi}\sin^2 x\,dx=\frac{\pi}{2}\). Step 6: Therefore, \(V=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}\approx 4.93\). Answer: B. 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: Since the region is rotated around the y-axis and the functions are written in terms of \(x\), use the shell method. Step 2: The shell radius is \(x\). Step 3: The shell height is upper function minus lower function: \((\sqrt{x}+1)-2x\). Step 4: Set up the volume: \(V=\int_0^1 2\pi x(\sqrt{x}+1-2x)\,dx\). Step 5: Expand the integrand: \(2\pi\int_0^1 (x^{\frac{3}{2}}+x-2x^2)\,dx\). Step 6: Integrate: \(2\pi\left[\frac{2}{5}x^{\frac{5}{2}}+\frac{x^2}{2}-\frac{2x^3}{3}\right]_0^1\). Step 7: Substitute \(x=1\): \(2\pi\left(\frac{2}{5}+\frac{1}{2}-\frac{2}{3}\right)=2\pi\left(\frac{12}{30}+\frac{15}{30}-\frac{20}{30}\right)=2\pi\cdot\frac{7}{30}=\frac{7\pi}{15}\). Answer: C. \(\frac{7\pi}{15}\) units\(^3\)
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