1.
Question 1 What is the area between the curve \(y=x^2\) and the x-axis on the interval \([0,2]\)?
Step 1: The area between \(y=x^2\) and the x-axis on \([0,2]\) is \(\int_0^2 x^2\,dx\). Step 2: Use the power rule: \(\int x^2\,dx=\frac{x^3}{3}\). Step 3: Evaluate from \(0\) to \(2\): \(\left[\frac{x^3}{3}\right]_0^2=\frac{2^3}{3}-0=\frac{8}{3}\). Answer: D. \(\frac{8}{3}\)
2.
Question 2 What is the area between the curves \(y=x^2\) and \(y=-x^2+2x\)? Solve algebraically by integrating with respect to \(x\).
Step 1: Find the intersection points: \(x^2=-x^2+2x\). Step 2: Solve: \(2x^2-2x=0\), so \(2x(x-1)=0\), giving \(x=0\) and \(x=1\). Step 3: On \([0,1]\), the upper curve is \(-x^2+2x\) and the lower curve is \(x^2\). Step 4: Set up the area: \(\int_0^1[(-x^2+2x)-x^2]\,dx=\int_0^1(-2x^2+2x)\,dx\). Step 5: Integrate: \(\left[-\frac{2x^3}{3}+x^2\right]_0^1=-\frac{2}{3}+1=\frac{1}{3}\). Answer: D. \(\frac{1}{3}\)
3.
Question 3 What is the approximate volume of the solid created when the region under the curve \(y=\cos^2 x\) on the interval \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) is rotated around the x-axis?
Step 1: Rotating around the x-axis uses the disk method: \(V=\pi\int_a^b [f(x)]^2\,dx\). Step 2: Here \(f(x)=\cos^2 x\), so \(V=\pi\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4 x\,dx\). Step 3: Use symmetry: \(\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^4 x\,dx=2\int_0^{\frac{\pi}{2}}\cos^4 x\,dx\). Step 4: The standard value is \(\int_0^{\frac{\pi}{2}}\cos^4 x\,dx=\frac{3\pi}{16}\). Step 5: Therefore, \(V=\pi\cdot 2\cdot\frac{3\pi}{16}=\frac{3\pi^2}{8}\approx 3.70\). Answer: A. 3.70
4.
Question 4 What is the volume of the solid created when the region under the curve \(y=\sqrt{\sin x}+x\) on the interval \((0,\pi)\) is rotated around the line \(x=-4\)?
Step 1: The axis of rotation \(x=-4\) is vertical, so use the shell method with respect to \(x\). Step 2: The shell radius is \(x+4\). Step 3: The shell height is \(\sqrt{\sin x}+x\). Step 4: Set up the volume: \(V=\int_0^{\pi}2\pi(x+4)(\sqrt{\sin x}+x)\,dx\). Step 5: Evaluate this integral using a calculator or numerical integration. Step 6: \(V\approx 272.84\). Answer: A. 272.84
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Question 5 A spring is stretched a distance of 0.5 m from its natural resting position when a 220 N force is applied. How much work is done to stretch it that far?
Step 1: Use Hooke's Law \(F=kx\). Step 2: Since \(F=220\) N when \(x=0.5\) m, \(k=\frac{220}{0.5}=440\). Step 3: Work required to stretch a spring is \(W=\frac{1}{2}kx^2\). Step 4: Substitute: \(W=\frac{1}{2}(440)(0.5)^2=55\) J. Answer: D. 55 J
6.
Question 6 If a variable force \(F(x)=\frac{60}{x^5}\) N were applied to an object to move it over the distance \(x=7\) m to \(x=14\) m, then the work done over that distance is given by \(W=\int_7^{14}\frac{60}{x^5}\,dx\).
Step 1: Work done by a variable force over an interval \([a,b]\) is \(W=\int_a^b F(x)\,dx\). Step 2: Here \(F(x)=\frac{60}{x^5}\), \(a=7\), and \(b=14\). Step 3: Substitute into the work formula: \(W=\int_7^{14}\frac{60}{x^5}\,dx\). Step 4: This matches the given statement. Answer: A. True
7.
Question 7 Find the average value of \(f(x)=\sin x\) over the interval \([0,\pi]\).
Step 1: Use the average value formula: \(f_{avg}=\frac{1}{b-a}\int_a^b f(x)\,dx\). Step 2: Substitute \([0,\pi]\): \(f_{avg}=\frac{1}{\pi-0}\int_0^{\pi}\sin x\,dx\). Step 3: Integrate: \(\int\sin x\,dx=-\cos x\). Step 4: Evaluate: \(\int_0^{\pi}\sin x\,dx=[-\cos x]_0^{\pi}=1-(-1)=2\). Step 5: Divide by the interval length: \(f_{avg}=\frac{2}{\pi}\). Answer: A. \(\frac{2}{\pi}\)
8.
Question 8 For an object whose velocity is given by \(v(t)=-2t^2+8\), what is its distance on the interval \(t=0\) to \(t=3\)?
Step 1: Distance is \(\int_0^3 |v(t)|\,dt\). Step 2: Find where velocity changes sign: \(-2t^2+8=0\), so \(t=2\) on the interval. Step 3: On \([0,2]\), \(v(t)\ge 0\), and on \([2,3]\), \(v(t)\le 0\). Step 4: Set up the distance: \(\int_0^2(-2t^2+8)\,dt-\int_2^3(-2t^2+8)\,dt\). Step 5: An antiderivative is \(-\frac{2t^3}{3}+8t\). Step 6: Evaluate: \(\int_0^2(-2t^2+8)\,dt=\frac{32}{3}\), and \(\int_2^3(-2t^2+8)\,dt=-\frac{14}{3}\). Step 7: Distance is \(\frac{32}{3}-\left(-\frac{14}{3}\right)=\frac{46}{3}\approx 15.33\). Answer: A. 15.33
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Question 9 If \(v(t)=-t+2\) and \(d(0)=-3\), then \(d(t)=-\frac{t^2}{2}+2t-3\).
Step 1: Position/displacement function \(d(t)\) is an antiderivative of velocity \(v(t)\). Step 2: Integrate \(v(t)=-t+2\): \(d(t)=\int(-t+2)\,dt=-\frac{t^2}{2}+2t+C\). Step 3: Use \(d(0)=-3\): \(-\frac{0^2}{2}+2(0)+C=-3\), so \(C=-3\). Step 4: Therefore, \(d(t)=-\frac{t^2}{2}+2t-3\). Answer: A. True
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