1.
Question 1 Use the rectangular approximation with 4 rectangles, using the left endpoint for the height, to approximate the area under the curve of \(f(x)=2x^2\) 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: Using left endpoints, the \(x\)-values are \(0,1,2,3\). Step 3: Evaluate the function: \(f(0)=2(0)^2=0\), \(f(1)=2(1)^2=2\), \(f(2)=2(2)^2=8\), and \(f(3)=2(3)^2=18\). Step 4: Add the rectangle areas: \(A\approx \Delta x[f(0)+f(1)+f(2)+f(3)]=1(0+2+8+18)=28\). Answer: C. \(28\)
2.
Question 2 The sum \(\sum_{k=1}^{4}(5k+1)\) is equal to \(54\).
Step 1: Substitute each integer value from \(k=1\) to \(k=4\). Step 2: Compute each term: \(5(1)+1=6\), \(5(2)+1=11\), \(5(3)+1=16\), and \(5(4)+1=21\). Step 3: Add the terms: \(6+11+16+21=54\). Step 4: Since the sum is \(54\), the statement is true. Answer: A. True
3.
Question 3 If \(\int_{1}^{5} f(x)\,dx=-3\) and \(\int_{1}^{5} g(x)\,dx=2\), then \(\int_{1}^{5}[5f(x)-g(x)]\,dx=-17\).
Step 1: Use the linearity property of definite integrals. Step 2: Rewrite the integral as \(\int_{1}^{5}[5f(x)-g(x)]\,dx=5\int_{1}^{5}f(x)\,dx-\int_{1}^{5}g(x)\,dx\). Step 3: Substitute the given values: \(5(-3)-2=-15-2=-17\). Step 4: Since the value is \(-17\), the statement is true. Answer: A. True
4.
Question 4 The integral \(\int_a^b 4\ln x\,dx\) can be written as the following limit of Riemann sums on the interval from \(a\) to \(b\): \(\lim_{\max \Delta x_k \to 0}\sum_{k=1}^{n}4\ln(x_k^*)\Delta x_k\).
Step 1: Recall the definition of a 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: In this problem, \(f(x)=4\ln x\). Step 3: Substitute \(f(x_k^*)=4\ln(x_k^*)\) into the definition. Step 4: This gives \(\lim_{\max \Delta x_k\to 0}\sum_{k=1}^{n}4\ln(x_k^*)\Delta x_k\), so the statement is true. Answer: B. True
5.
Question 5 \(\frac{d}{dx}\int_{3}^{x}\frac{5t}{\ln(t+1)}\,dt=\)
Step 1: Use the Fundamental Theorem of Calculus. If \(F(x)=\int_a^x h(t)\,dt\), then \(F'(x)=h(x)\). Step 2: Here, \(h(t)=\frac{5t}{\ln(t+1)}\). Step 3: Replace \(t\) with \(x\) because the upper limit is \(x\). Step 4: Therefore, \(\frac{d}{dx}\int_{3}^{x}\frac{5t}{\ln(t+1)}\,dt=\frac{5x}{\ln(x+1)}\). Answer: A. \(\frac{5x}{\ln(x+1)}\)
6.
Question 6 Find the area between the curve and the x-axis for the function \(f(x)=x^3-x\) on \([0,3]\).
Step 1: Factor the function: \(f(x)=x^3-x=x(x-1)(x+1)\). Step 2: On \([0,1]\), the function is below the x-axis, and on \([1,3]\), the function is above the x-axis. Step 3: Area must be positive, so compute \(A=-\int_0^1(x^3-x)\,dx+\int_1^3(x^3-x)\,dx\). Step 4: Find an antiderivative: \(\int(x^3-x)\,dx=\frac{x^4}{4}-\frac{x^2}{2}\). Step 5: Compute the first part: \(\int_0^1(x^3-x)\,dx=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_0^1=\frac14-\frac12=-\frac14\), so the positive area is \(\frac14\). Step 6: Compute the second part: \(\int_1^3(x^3-x)\,dx=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^3=\left(\frac{81}{4}-\frac{9}{2}\right)-\left(\frac14-\frac12\right)=16\). Step 7: Total area is \(\frac14+16=16.25\). Answer: C. \(16.25\)
7.
Question 7 The area under the curve \(f(x)=x^{-\frac{1}{3}}\) on \([1,3]\) is \(1.74\).
Step 1: Since \(x^{-\frac13}\) is positive on \([1,3]\), the area is \(\int_1^3 x^{-\frac13}\,dx\). Step 2: Use the power rule: \(\int x^{-\frac13}\,dx=\frac{x^{\frac23}}{\frac23}=\frac32x^{\frac23}\). Step 3: Evaluate the definite integral: \(A=\left[\frac32x^{\frac23}\right]_1^3=\frac32(3^{\frac23})-\frac32(1^{\frac23})\). Step 4: Approximate the value: \(\frac32(3^{\frac23})-\frac32\approx 1.62\). Step 5: Since \(1.62\neq 1.74\), the statement is false. Answer: A. False
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