1.
Question 1. Use the rectangular approximation with 4 rectangles, using the midpoint for the height, to approximate the area under the curve of \(f(x)=x^2\) on the interval \([0,4]\).
Step 1: Divide the interval \([0,4]\) into \(4\) equal rectangles. Step 2: The width is \(\Delta x=\frac{4-0}{4}=1\). Step 3: Using midpoints, the midpoint values are \(0.5\), \(1.5\), \(2.5\), and \(3.5\). Step 4: Evaluate the function \(f(x)=x^2\) at each midpoint: \(f(0.5)=0.25\), \(f(1.5)=2.25\), \(f(2.5)=6.25\), and \(f(3.5)=12.25\). Step 5: Multiply each function value by \(\Delta x=1\) and add: \(0.25+2.25+6.25+12.25=21\). Answer: 21
2.
Question 2. The following series, \(e^2+e^4+e^6+e^8+e^{10}\), expressed in sigma notation is \(\sum_{k=1}^{5}e^{2k}\).
Step 1: Look at the sigma expression \(\sum_{k=1}^{5}e^{2k}\). Step 2: Substitute \(k=1\): \(e^{2(1)}=e^2\). Step 3: Substitute \(k=2\): \(e^{2(2)}=e^4\). Step 4: Substitute \(k=3\): \(e^{2(3)}=e^6\). Step 5: Substitute \(k=4\): \(e^{2(4)}=e^8\). Step 6: Substitute \(k=5\): \(e^{2(5)}=e^{10}\). Step 7: The sigma expression expands to \(e^2+e^4+e^6+e^8+e^{10}\). Answer: True
3.
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. Step 2: \(\int_{-2}^{5}[2f(x)-4g(x)]\,dx=2\int_{-2}^{5}f(x)\,dx-4\int_{-2}^{5}g(x)\,dx\). Step 3: Substitute the given values: \(2(1)-4(-4)\). Step 4: Simplify: \(2+16=18\). Step 5: The statement says the value is \(16\), but the correct value is \(18\). Answer: False
4.
Question 4. The integral \(\int_a^b2x^2\,dx\) can be written as the following limit of Riemann sums on the interval from \(a\) to \(b\): \(\lim_{\max \Delta x_k\to0}\sum_{k=1}^{n}2(x_k^*)^2\Delta x_k\), where \(x_k^*\) is any \(x\)-value in the \(k\)-th interval and \(\Delta x_k\) is the width of the \(k\)-th interval.
Step 1: Recall the definition of a definite integral as a limit of Riemann sums. Step 2: For a function \(f(x)\), \(\int_a^b f(x)\,dx=\lim_{\max \Delta x_k\to0}\sum_{k=1}^{n}f(x_k^*)\Delta x_k\). Step 3: Here the function is \(f(x)=2x^2\). Step 4: Substitute \(f(x_k^*)=2(x_k^*)^2\). Step 5: This gives \(\int_a^b2x^2\,dx=\lim_{\max \Delta x_k\to0}\sum_{k=1}^{n}2(x_k^*)^2\Delta x_k\). Step 6: This matches the statement. Answer: True
5.
Question 5. \(\frac{d}{dx}\int_{-2}^{x}|4t+1|\,dt=?\)
Step 1: Use the Fundamental Theorem of Calculus. Step 2: If \(F(x)=\int_a^x f(t)\,dt\), then \(F'(x)=f(x)\). Step 3: Here \(f(t)=|4t+1|\). Step 4: Replace \(t\) with \(x\) because the upper limit is \(x\). Step 5: Therefore, \(\frac{d}{dx}\int_{-2}^{x}|4t+1|\,dt=|4x+1|\). Answer: \(|4x+1|\)
6.
Question 6. Find the unsigned area between the curve and the \(x\)-axis for the function \(f(x)=x^3-x\) on \([0,3]\).
Step 1: Find where \(f(x)=x^3-x\) crosses the \(x\)-axis on \([0,3]\). Step 2: Factor: \(x^3-x=x(x^2-1)=x(x-1)(x+1)\). Step 3: On \([0,3]\), the zeros are \(x=0\) and \(x=1\). Step 4: The function is negative on \((0,1)\) and positive on \((1,3)\). Step 5: Unsigned area is \(-\int_0^1(x^3-x)\,dx+\int_1^3(x^3-x)\,dx\). Step 6: Compute the antiderivative: \(\int(x^3-x)\,dx=\frac{x^4}{4}-\frac{x^2}{2}\). Step 7: On \([0,1]\), \(\int_0^1(x^3-x)\,dx=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_0^1=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}\), so the unsigned area is \(\frac{1}{4}\). Step 8: On \([1,3]\), \(\int_1^3(x^3-x)\,dx=\left[\frac{x^4}{4}-\frac{x^2}{2}\right]_1^3=(\frac{81}{4}-\frac{9}{2})-(\frac{1}{4}-\frac{1}{2})=16\). Step 9: Add the positive areas: \(\frac{1}{4}+16=16.25\). Answer: 16.25
7.
Question 7. \(\int_{-2}^{2}(|x|-1)\,dx=2\).
Step 1: Break the absolute value into a piecewise function. Step 2: For \(x<0\), \(|x|=-x\). For \(x\ge0\), \(|x|=x\). Step 3: Rewrite the integral: \(\int_{-2}^{2}(|x|-1)\,dx=\int_{-2}^{0}(-x-1)\,dx+\int_{0}^{2}(x-1)\,dx\). Step 4: Compute the first integral: \(\int_{-2}^{0}(-x-1)\,dx=\left[-\frac{x^2}{2}-x\right]_{-2}^{0}=0-(-2+2)=0\). Step 5: Compute the second integral: \(\int_{0}^{2}(x-1)\,dx=\left[\frac{x^2}{2}-x\right]_0^2=(2-2)-0=0\). Step 6: Add them: \(0+0=0\). Step 7: The statement says the integral equals \(2\), but the correct value is \(0\). Answer: False
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