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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: Compute the area using \(\int_0^2(x^3+1)dx\). Step 2: Integrate to obtain \(\frac{x^4}{4}+x\). Step 3: Evaluate: \([\frac{x^4}{4}+x]_0^2=4+2=6\). Step 4: Since \(6\neq8\), the statement is false. Answer: B. False
2.
Question 2 What is the area between \(y=\sin x\) and \(y=\cos x\) on \([0,\frac{\pi}{4}]\)?
Step 1: On the interval, \(\cos x\ge\sin x\). Step 2: Area=\(\int_0^{\pi/4}(\cos x-\sin x)dx\). Step 3: Antiderivative is \(\sin x+\cos x\). Step 4: Evaluate to get \(\sqrt2-1\). Answer: D. \(\sqrt2-1\)
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Question 4 The region enclosed by \(y=x\) and \(y=x^2\) is rotated about \(x=-2\). The volume is:
Step 1: Use shells about \(x=-2\). Step 2: Radius=\(x+2\), height=\(x-x^2\). Step 3: \(V=2\pi\int_0^1(x+2)(x-x^2)dx\). Step 4: Evaluate the integral to obtain \(\frac{5\pi}{6}\). Answer: B. \(\frac{5\pi}{6}\)
4.
Question 3 The region enclosed by \(y=1-x^2\), \(x=0\), and \(y=0\) is rotated about \(x=-1\). The volume is \(\frac{2\pi}{3}\) units\(^3\).
Step 1: Use the shell method about the vertical line \(x=-1\). Step 2: Radius=\(x+1\), height=\(1-x^2\). Step 3: \(V=2\pi\int_0^1(x+1)(1-x^2)dx\). Step 4: Evaluating gives \(\frac{11\pi}{6}\), not \(\frac{2\pi}{3}\). Answer: B. False
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