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Question 1 Use the rectangular approximation with 4 rectangles, using the right endpoint for the height, to approximate the area under the curve of \(f(x)=4x^4\) on the interval \([0,4]\).
Step 1: The interval is \([0,4]\), and there are \(4\) rectangles, so \(\Delta x=\frac{4-0}{4}=1\). Step 2: Since the right endpoint is used, the right endpoints are \(x=1,2,3,4\). Step 3: Evaluate the function at each right endpoint: \(f(1)=4(1)^4=4\), \(f(2)=4(2)^4=64\), \(f(3)=4(3)^4=324\), and \(f(4)=4(4)^4=1024\). Step 4: Add the rectangle areas: \(4(1)+64(1)+324(1)+1024(1)=1416\). Answer: C. 1416
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Question 2 The sum \(\sum_{k=1}^{4} 2k^2\) is equal to \(60\).
Step 1: Substitute \(k=1,2,3,4\) into \(2k^2\). Step 2: Compute each term: \(2(1)^2=2\), \(2(2)^2=8\), \(2(3)^2=18\), and \(2(4)^2=32\). Step 3: Add the terms: \(2+8+18+32=60\). Step 4: The statement is true. Answer: A. True
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Question 3 If \(\int_{-2}^{5} f(x)\,dx=1\) and \(\int_{-2}^{5} g(x)\,dx=-4\), then \(\int_{-2}^{5} [2f(x)-4g(x)]\,dx=16\).
Step 1: Use the linearity property of definite integrals: \(\int_{-2}^{5}[2f(x)-4g(x)]\,dx=2\int_{-2}^{5}f(x)\,dx-4\int_{-2}^{5}g(x)\,dx\). Step 2: Substitute the given values: \(2(1)-4(-4)\). Step 3: Simplify: \(2+16=18\). Step 4: Since \(18\neq 16\), the statement is false. Answer: B. False
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Question 4 The integral \(\int_a^b 3\sin x\,dx\) can be written as the following limit of Riemann sums on the interval from \(a\) to \(b\): \(\lim_{\max \Delta x_k^*\to 0}\sum_{k=1}^{n}3\sin(x_k^*)\Delta x_k^*\).
Step 1: Recall the definition of the definite integral: \(\int_a^b f(x)\,dx=\lim_{\max \Delta x_k^*\to 0}\sum_{k=1}^{n} f(x_k^*)\Delta x_k^*\). Step 2: Here the function is \(f(x)=3\sin x\). Step 3: Substitute \(f(x_k^*)=3\sin(x_k^*)\). Step 4: This gives \(\lim_{\max \Delta x_k^*\to 0}\sum_{k=1}^{n}3\sin(x_k^*)\Delta x_k^*\), so the statement is true. Answer: B. True
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Question 5 \(\frac{d}{dx}\int_{-2}^{x}|4t+1|\,dt=\)
Step 1: Use the Fundamental Theorem of Calculus: \(\frac{d}{dx}\int_a^x f(t)\,dt=f(x)\). Step 2: In this problem, \(f(t)=|4t+1|\). Step 3: Replace \(t\) with \(x\): \(f(x)=|4x+1|\). Step 4: Therefore, \(\frac{d}{dx}\int_{-2}^{x}|4t+1|\,dt=|4x+1|\). Answer: C. \(|4x+1|\)
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Question 6 \(\int_{-2}^{2}(|x|-1)\,dx=2\).
Step 1: Break the absolute value function into pieces: \(|x|=-x\) for \(x<0\), and \(|x|=x\) for \(x\ge 0\). Step 2: Rewrite the integral: \(\int_{-2}^{2}(|x|-1)\,dx=\int_{-2}^{0}(-x-1)\,dx+\int_{0}^{2}(x-1)\,dx\). Step 3: Evaluate the first integral: \(\int_{-2}^{0}(-x-1)\,dx=\left[-\frac{x^2}{2}-x\right]_{-2}^{0}=0\). Step 4: Evaluate the second integral: \(\int_{0}^{2}(x-1)\,dx=\left[\frac{x^2}{2}-x\right]_{0}^{2}=0\). Step 5: The total is \(0+0=0\). Since \(0\neq 2\), the statement is false. Answer: A. False
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Question 7 \(\int_{1}^{4}x\,dx=7.5\).
Step 1: Find the antiderivative of \(x\): \(\int x\,dx=\frac{x^2}{2}\). Step 2: Evaluate from \(1\) to \(4\): \(\left[\frac{x^2}{2}\right]_{1}^{4}\). Step 3: Substitute the bounds: \(\frac{4^2}{2}-\frac{1^2}{2}=\frac{16}{2}-\frac{1}{2}=8-0.5=7.5\). Step 4: The statement is true. Answer: A. True
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