EBUS CAL12 Chapter 6 Applications of Integrals Test-2

Step 1: The curve \(y=x^2\) is above the x-axis on \([0,2]\), so the area 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 \(2\): \(\left[\frac{x^3}{3}\right]_0^2=\frac{8}{3}-0=\frac{8}{3}\). Answer: B. \(\frac{8}{3}\)

Step 1: Find the intersection points by solving \(x^2=-x^2+2x\). Step 2: This gives \(2x^2-2x=0\), so \(2x(x-1)=0\), and \(x=0,1\). Step 3: On \([0,1]\), the upper function is \(-x^2+2x\), and the lower function 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: B. \(\frac{1}{3}\)

Step 1: Rotating around the x-axis uses the disk method: \(V=\pi\int_a^b [f(x)]^2\,dx\). Step 2: Here \(f(x)=\sin x\), so \(V=\pi\int_0^{\pi}\sin^2 x\,dx\). Step 3: Use \(\int_0^{\pi}\sin^2 x\,dx=\frac{\pi}{2}\). Step 4: Therefore, \(V=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}\approx 4.93\). Answer: D. 4.93

Step 1: The axis of rotation \(x=-1\) is vertical, so use the shell method with respect to \(x\). Step 2: The shell radius is \(x+1\). Step 3: The shell height is \(\sin x+x\). Step 4: Set up the volume: \(V=2\pi\int_0^{\pi}(x+1)(\sin x+x)\,dx\). Step 5: Expand the integrand: \((x+1)(\sin x+x)=x\sin x+\sin x+x^2+x\). Step 6: Evaluate: \(\int_0^{\pi}x\sin x\,dx=\pi\), \(\int_0^{\pi}\sin x\,dx=2\), \(\int_0^{\pi}x^2\,dx=\frac{\pi^3}{3}\), and \(\int_0^{\pi}x\,dx=\frac{\pi^2}{2}\). Step 7: Thus \(V=2\pi\left(\pi+2+\frac{\pi^3}{3}+\frac{\pi^2}{2}\right)\approx 128.25\). Answer: B. 128.25

Step 1: Use Hooke's Law \(F=kx\). Step 2: Since \(F=300\) N when \(x=0.6\) m, \(k=\frac{300}{0.6}=500\). Step 3: The work required to stretch a spring is \(W=\frac{1}{2}kx^2\). Step 4: Substitute: \(W=\frac{1}{2}(500)(0.6)^2=90\) J. Step 5: The computed work is \(90\) J, which corresponds to option C. Answer: C. 90 J

Step 1: Work done by a variable force over an interval \([a,b]\) is written as \(W=\int_a^b F(x)\,dx\). Step 2: Here \(F(x)=\frac{15}{x^3}\), \(a=0\), and \(b=6\). Step 3: Substituting into the work formula gives \(W=\int_0^6\frac{15}{x^3}\,dx\). Step 4: This matches the statement. Answer: A. True

Step 1: Use the average value formula: \(f_{avg}=\frac{1}{b-a}\int_a^b f(x)\,dx\). Step 2: Substitute \([0,\pi]\): \(f_{avg}=\frac{1}{\pi}\int_0^{\pi}\sin x\,dx\). Step 3: Evaluate the integral: \(\int_0^{\pi}\sin x\,dx=[-\cos x]_0^{\pi}=2\). Step 4: Therefore, \(f_{avg}=\frac{2}{\pi}\). Answer: C. \(\frac{2}{\pi}\)

Step 1: Distance is \(\int_0^3 |v(t)|\,dt\). Step 2: Find where velocity changes sign: \(-2t^2+8=0\), so \(t=2\). Step 3: On \([0,2]\), \(v(t)\ge 0\), and on \([2,3]\), \(v(t)\le 0\). Step 4: Set up 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: C. 15.33

Step 1: Since \(v(t)=d'(t)\), find \(d(t)\) by integrating \(v(t)=t^2-5\). Step 2: \(d(t)=\int(t^2-5)\,dt=\frac{t^3}{3}-5t+C\). Step 3: Use \(d(0)=-2\): \(C=-2\). Step 4: So \(d(t)=\frac{t^3}{3}-5t-2\), not \(2t-2\). Step 5: The statement is false. Answer: B. False

Step 1: The region is bounded by \(x=0\), \(y=0\), and \(y=x^2-3\). Step 2: Solve for \(x\) in terms of \(y\): \(x=\sqrt{y+3}\). Step 3: For cross sections perpendicular to the y-axis, the diameter of each semicircle is \(d(y)=\sqrt{y+3}\). Step 4: The area of a semicircle with diameter \(d\) is \(A=\frac{\pi d^2}{8}\). Step 5: Therefore, \(A(y)=\frac{\pi}{8}(y+3)\). Step 6: The y-values go from \(-3\) to \(0\). Step 7: The volume is \(V=\int_{-3}^{0}\frac{\pi}{8}(y+3)\,dy\). Step 8: Evaluate: \(V=\frac{\pi}{8}\left[\frac{y^2}{2}+3y\right]_{-3}^{0}=\frac{9\pi}{16}\approx 1.77\). Answer: 1.77