Mathematics 265 Introduction to Calculus I

Solutions to Sample Final Exam

Sample Final Examination 1

Time: 3.5 hours
Passing grade: 55%
Total points: 124

Solutions and Marking Scheme

Take note of how the points are assigned in each question.

1. Compute the derivatives of each of the functions given below. You may not need to simplify your answers.   (20 points)
   1. ${\displaystyle y=\sin x\cdot \cos (\sin x^2)}$
   2. ${\displaystyle y={\left({\displaystyle \frac{1+x^3}{1-x^2}}\right)}^{{\displaystyle \frac{1}{3}}}}$
   3. ${\displaystyle y=\sqrt{1+\sqrt{1+x}}}$
   4. ${\displaystyle y={\displaystyle \frac{\sqrt{x^2-1}}{x^2-2x-8}}}$
   5. ${\displaystyle y={\sec }^2\left({\displaystyle \frac{x+1}{x-2}}\right)}$

   Solution

   1. By the Product and Chain rules:

      ${\displaystyle {\displaystyle \frac{d}{dx}}\sin (x)\cdot \cos (\sin x^2)}$

      ${\displaystyle =\cos x\cos (\sin x^2)+\sin x(-\sin (\sin x^2)\cos (x^2)(2x))}$    (4 pts)

      ${\displaystyle y^{\prime }=\cos x\cos (\sin x^2)-2x\sin x\cos (x^2)\sin (\sin x^2)}$

   2. By the General Power and Quotient rules:

      ${\displaystyle {\displaystyle \frac{d}{dx}}{\left({\displaystyle \frac{1+x^3}{1-x^2}}\right)}^{{\displaystyle \frac{1}{3}}}}$

      ${\displaystyle ={\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{1+x^3}{1-x^2}}\right)}^{-2/3}{\displaystyle \frac{3x^3(1-x^2)+2x(1+x^3)}{{(1-x^2)}^2}}}$    (4 pts)

      ${\displaystyle y^{\prime }={\displaystyle \frac{1}{3}}{\left({\displaystyle \frac{1-x^2}{1+x^3}}\right)}^{2/3}{\displaystyle \frac{x(3x-x^3+2)}{{(1-x^2)}^2}}}$

   3. By the General Power Rule:

      ${\displaystyle \begin{array}{ll}{\displaystyle \frac{d}{dx}}\sqrt{1+\sqrt{1+x}}\hfill & \hfill \\ ={\displaystyle \frac{1}{2}}{\left(1+\sqrt{1+x}\right)}^{-1/2}{\displaystyle \frac{d}{dx}}\left(1+{(1+x)}^{1/2}\right)\hfill & \text{(2}\text{pts)}\hfill \\ y^{\prime }={\displaystyle \frac{1}{2\sqrt{1+\sqrt{1+x}}}}\left({\displaystyle \frac{1}{2\sqrt{1+x}}}\right)={\displaystyle \frac{1}{4\sqrt{1+\sqrt{1+x}}\sqrt{1+x}}}\hfill & \text{(2}\text{pts)}\hfill \end{array}}$

   4. By the Quotient and General Power rules:

      ${\displaystyle \begin{array}{ll}{\displaystyle \frac{d}{dx}}{\displaystyle \frac{\sqrt{x^2-1}}{x^2-2x-8}}\hfill & \hfill \\ ={\displaystyle \frac{\left({\displaystyle \frac{2x}{2}}\left({(x^2-1)}^{-1/2}\right)\right)(x^2-2x-8)-2\sqrt{x^2-1}(x-1)}{{(x^2-2x-8)}^2}}\hfill & \text{(4}\text{pts)}\hfill \end{array}}$

   5. By the Chain and Quotient rules:

      ${\displaystyle \begin{array}{ll}{\displaystyle \frac{d}{dx}}{\sec }^2\left({\displaystyle \frac{x+1}{x-2}}\right)\hfill & \hfill \\ =2{\sec }^2\left({\displaystyle \frac{x+1}{x-2}}\right)\tan \left({\displaystyle \frac{x+1}{x-2}}\right)\left({\displaystyle \frac{(x-2)-(x+1)}{{(x-2)}^2}}\right)\hfill & \begin{array}{l}\\ \text{(4}\text{pts)}\end{array}\hfill \\ y^{\prime }=2{\sec }^2\left({\displaystyle \frac{x+1}{x-2}}\right)\tan \left({\displaystyle \frac{x+1}{x-2}}\right)\left({\displaystyle \frac{(x-2)-(x+1)}{{(x-2)}^2}}\right)\hfill & \hfill \end{array}}$

2. Use the method of differentials to estimate each of the values below to four decimal places (show the process).   (8 points)
   1. ${\displaystyle \cos 62\text{°}}$
   2. ${\displaystyle \sqrt{16.4}}$

   Solution

   1. ${\displaystyle \begin{array}{ll}\cos 62\text{°}=\cos (60\text{°}+2\text{°})=\cos \left({\displaystyle \frac{\pi }{3}}+{\displaystyle \frac{\pi }{90}}\right)\hfill & \text{(1}\text{pt)}\hfill \\ \text{Let}f(x)=\cos x\text{,}a=60\text{°}={\displaystyle \frac{\pi }{3}}\text{,}\Delta a=2\text{°}={\displaystyle \frac{2\pi }{180}}\hfill & \text{(1}\text{pt)}\hfill \\ \cos 62\text{°}=f(a)+f^{\prime }(a)\Delta x=\cos \left({\displaystyle \frac{\pi }{3}}\right)-\sin \left({\displaystyle \frac{\pi }{3}}\right)\left({\displaystyle \frac{\pi }{90}}\right)\hfill & \text{(1}\text{pt)}\hfill \\ \approx {\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{3}}{2}}{\displaystyle \frac{\pi }{90}}={\displaystyle \frac{1}{2}}-{\displaystyle \frac{\sqrt{3}\pi }{180}}\hfill & \text{(1}\text{pt)}\hfill \end{array}}$
   2. ${\displaystyle \begin{array}{ll}\sqrt{16.4}=\sqrt{16+0.4}\hfill & \text{(1}\text{pt)}\hfill \\ \text{Let}f(x)=\sqrt{x}\text{,}a=16\text{,}\Delta a=0.4\hfill & \text{(1}\text{pt)}\hfill \\ \sqrt{16.4}\approx f(a)+f^{\prime }(a)\Delta x=\sqrt{16}+{\displaystyle \frac{1}{2\sqrt{16}}}\left({\displaystyle \frac{4}{10}}\right)\hfill & \text{(2}\text{pts)}\hfill \\ \phantom{\sqrt{16.4}}\approx 4+{\displaystyle \frac{1}{5(4)}}=4+{\displaystyle \frac{1}{20}}={\displaystyle \frac{81}{20}}\hfill & \hfill \end{array}}$
3. Use the definition of the derivative as a limit to compute the derivative of ${\displaystyle \cot x}$.   (6 points)

   Solution

   By the definition of the derivative:

   ${\displaystyle \begin{array}{ll}{\displaystyle \frac{d}{dx}}\cot x\hfill & =\underset{h\to 0}{\lim }{\displaystyle \frac{\cot (x+h)-\cot x}{h}}\hfill \\ \hfill & =\underset{h\to 0}{\lim }{\displaystyle \frac{1}{h}}\left({\displaystyle \frac{\cos (x+h)}{\sin (x+h)}}-{\displaystyle \frac{\cos (x)}{\sin (x)}}\right)\hfill \end{array}}$

   and we have that

   ${\displaystyle \begin{array}{ll}{\displaystyle \frac{\cos (x+h)}{\sin (x+h)}}\hfill & -{\displaystyle \frac{\cos (x)}{\sin (x)}}\hfill \\ \hfill & ={\displaystyle \frac{\cos (x+h)\sin (x)-\cos (x)\sin (x+h)}{\sin (x+h)\sin (x)}}\hfill \\ \hfill & ={\displaystyle \frac{\left(\cos (x)\cos (h)-\sin (x)\sin (h)\right)\sin (x)}{\sin (x+h)\sin (x)}}\hfill \\ \hfill & -{\displaystyle \frac{\left(\sin (x)\cos (h)+\cos (x)\sin (h)\right)\cos x}{\sin (x+h)\sin (x)}}\hfill \\ \hfill & ={\displaystyle \frac{\cos x\cos h\sin x-{\sin }^2\text{}x\sin h-\cos x\cos h\sin x-{\cos }^2\text{}x\sin h}{\sin (x+h)\sin (x)}}\hfill \\ \hfill & ={\displaystyle \frac{-({\cos }^2\text{}x+{\sin }^2\text{}x)\sin h}{\sin (x+h)\sin (x)}}\hfill \\ \hfill & =-{\displaystyle \frac{\sin h}{\sin (x+h)\sin (x)}}\left(3\text{ pts}\right)\hfill \end{array}}$

   Hence

   ${\displaystyle \begin{array}{lll}\underset{h\to 0}{\lim }{\displaystyle \frac{\cot (x+h)-\cot x}{h}}\hfill & =\underset{h\to 0}{\lim }-{\displaystyle \frac{\sin h}{h\sin (x+h)\sin (x)}}\hfill & \hfill \\ \hfill & =\underset{h\to 0}{\lim }-\left({\displaystyle \frac{\sin h}{h}}\right){\displaystyle \frac{1}{\sin (x+h)\sin x}}\hfill & \hfill \\ \hfill & =-{\displaystyle \frac{1}{{\sin }^2\text{}x}}\hfill & \hfill \\ \hfill & =-{\csc }^2\text{}x\hfill & \text{(3}\text{pts)}\hfill \end{array}}$

4. Sketch the graph of each of the functions below. State all critical points, cusps, vertical asymptotes and points of inflection.   (20 points)
   1. ${\displaystyle f(x)={\displaystyle \frac{x^2}{x^2-1}}\text{.}}$
   2. ${\displaystyle f(x)=x+\sin x\text{.}}$

   Solution

   1. Step A

      We find the domain and range of the function ${\displaystyle f(x)={\displaystyle \frac{x^2}{x^2-1}}}$:

      ${\displaystyle \begin{array}{ll}x\text{is}\text{in}\text{domain}\hfill & \iff x^2-1\ne 0\hfill \\ \hfill & \iff (x-1)(x+1)\ne 0\hfill \\ \hfill & \iff x-1\ne 0\text{and}x+1\ne 0\hfill \\ \hfill & \iff x\ne 1\text{and}x\ne -1\hfill \end{array}}$

      The domain is all numbers except ${\displaystyle x\ne 1}$ and ${\displaystyle x\ne -1}$   (1 pt)

      Step B

      The x and y intercepts are (0, 0)   (1 pt)

      Step C

      For the symmetry we have ${\displaystyle f(-x)={\displaystyle \frac{{(-x)}^2}{{(-x)}^2-1}}={\displaystyle \frac{x^2}{x^2-1}}}$, hence the function is even.   (1 pt)

      Step D

      For the vertical asymptotes we evaluate

      ${\displaystyle \underset{x\to 1^{+}}{\lim }{\displaystyle \frac{x^2}{x^2-1}}=\underset{x\to 1^{+}}{\lim }-{\displaystyle \frac{x^2}{x+1}}\left({\displaystyle \frac{1}{x-1}}\right)=\infty }$

      Thus the line ${\displaystyle x=1}$ is a vertical asymptote and by symmetry also the line ${\displaystyle x=-1}$   (1 pt)

      For the horizontal asymptotes we have

      ${\displaystyle \underset{x\to \pm \infty }{\lim }{\displaystyle \frac{x^2}{x^2-1}}=1}$

      hence ${\displaystyle y=1}$ is a horizontal asymptote.   (1 pt)

      Step E

      We find ${\displaystyle f^{\prime }}$

      $f^{\prime }(x)=-{\displaystyle \frac{2x}{{(x^2-1)}^2}}$;

      hence, ${\displaystyle f(x)=0}$ if ${\displaystyle x=0}$, and the critical points are ${\displaystyle 0\text{,}-1,1}$.
      We have $f^{\prime }(x)>0$ for $x<-1$ and $-1<x<0$; and $f^{\prime }(x)>0$ for $0<x<1$ and $x>1$.
      On the real line, we have   (1 pt)

      

      Step F

      Next, we find ${\displaystyle f^{″}(x)}$:

      ${\displaystyle \begin{array}{ll}{f}^{″}(x)\hfill & =-2\left({\displaystyle \frac{{(x^2-1)}^2-x(2(x^2-1)2x)}{{(x^2-1)}^4}}\right)\hfill \\ \hfill & =-2\left({\displaystyle \frac{(x^2-1)-(x^2-1-4x^2)}{{(x^2-1)}^4}}\right)\hfill \\ \hfill & ={\displaystyle \frac{2(x^2-1)(3x^2+1)}{{(x^2-1)}^4}}\hfill \\ \hfill & ={\displaystyle \frac{2(3x^2+1)}{{(x^2-1)}^3}}\hfill \\ \hfill & ={\displaystyle \frac{2(3x^2+1)}{{(x-1)}^3{(x+1)}^3}}\hfill \end{array}}$

      Critical points of ${\displaystyle f^{″}(x)}$ are ${\displaystyle 1}$ and ${\displaystyle -1}$.
      When ${\displaystyle x}$ is on the rightmost interval ${\displaystyle (1\text{,}\infty )}$, all the factors of ${\displaystyle f^{″}(x)}$ are positive. So the sign of ${\displaystyle f^{″}(x)}$ is positive. For ${\displaystyle -1<x<1}$ the factor ${\displaystyle x^2-1<0}$, hence ${\displaystyle f^{″}}$ is negative. By symmetry ${\displaystyle f^{″}>0}$ on the interval ${\displaystyle (-\infty \text{,}1)}$.
      Thus, we sketch the sign of ${\displaystyle f^{″}(x)}$ on the real line, below.   (1 pt)

      

      Thus, we have the diagram of the real line, showing where the function ${\displaystyle f}$ is concave downward or convex upward:   (1 pt)

      

      Step G

      We use all of the information derived from the steps above to sketch the graph of ${\displaystyle y={\displaystyle \frac{x^2}{x^2-1}}}$ shown below.   (2 pts)

      

   2. Step A

      The domain of the function ${\displaystyle f(x)=x+\sin x}$ is all real numbers.   (1 pt)

      Step B

      The y intercept is (0, 0), we skip the x-intercept since we would have to solve the equation ${\displaystyle \sin x=x}$ and obtain an approximation only.   (1 pt)

      Step C

      Symmetries ${\displaystyle f(-x)=(-x)+\sin (-x)=-x+-\sin (x)=-f(x)}$, thus the function is odd, so the graph of ${\displaystyle y=x+\sin x}$ is symmetric with respect to the origin (0, 0).   (1 pt)

      Step D

      We find ${\displaystyle f^{\prime }(x)=1+\cos x}$ and ${\displaystyle f^{\prime }(x)=1+\cos x\ge 0}$ if ${\displaystyle \cos x=-1}$, the critical points are ${\displaystyle x=(2k+1)\pi }$ for any integer k
      Since ${\displaystyle f^{\prime }(x)=1+\cos x>0}$ for all x except at the critical points the graph is increasing on ${\displaystyle (-\infty \text{,}\infty )}$.   (1 pt)

      Step E

      We find ${\displaystyle f^{″}(x)=-\sin x}$ and ${\displaystyle f^{″}(x)=-\sin x=0}$ for ${\displaystyle x={\displaystyle \frac{(2k+1)\pi }{2}}}$
      Hence ${\displaystyle y=x+\sin x}$ is concave upward on the intervals
      ${\displaystyle (\pi +2n\pi \text{, }2\pi +2n\pi )}$
      and concave downward on the intervals ${\displaystyle (2n\pi \text{, }\pi +2n\pi )}$     (2 pts)
      The function has inflection points at ${\displaystyle x={\displaystyle \frac{(2k+1)\pi }{2}}}$.

      Step F

      The graph is    (4 pts)

      

5. Find all maxima and minima of the functions listed below on the indicated intervals.   (11 points)
   1. ${\displaystyle f(x)=2x^{{\displaystyle \frac{5}{3}}}-5x^{{\displaystyle \frac{4}{3}}}}$ on ${\displaystyle [-1\text{, }20]}$
   2. ${\displaystyle f(x)=x+\cos x}$ on ${\displaystyle [-\pi \text{, }2\pi ]}$.

   Solution

   1. ${\displaystyle \begin{array}{ll}{f}^{\prime }(x)\hfill & =2\cdot {\displaystyle \frac{5}{3}}{x}^{{\displaystyle \frac{5}{3}}-1}-5\cdot {\displaystyle \frac{4}{3}}{x}^{{\displaystyle \frac{4}{3}}-1}\hfill \\ \hfill & ={\displaystyle \frac{10}{3}}{x}^{{\displaystyle \frac{2}{3}}}-{\displaystyle \frac{20}{3}}{x}^{{\displaystyle \frac{1}{3}}}\hfill \\ \hfill & ={\displaystyle \frac{10}{3}}\left(x^{{\displaystyle \frac{2}{3}}}-2x^{{\displaystyle \frac{1}{3}}}\right)\hfill \\ \hfill & ={\displaystyle \frac{10}{3}}{x}^{{\displaystyle \frac{1}{3}}}\left(x^{{\displaystyle \frac{1}{3}}}-2\right)\hfill \end{array}}$

      ${\displaystyle \begin{array}{lll}{f}^{\prime }(x)=0\hfill & \iff {\displaystyle \frac{10}{3}}{x}^{{\displaystyle \frac{1}{3}}}\left(x^{{\displaystyle \frac{1}{3}}}-2\right)=0\hfill & \hfill \\ \hfill & \iff x^{{\displaystyle \frac{1}{3}}}\left(x^{{\displaystyle \frac{1}{3}}}-2\right)=0\hfill & \hfill \\ \hfill & \iff x^{{\displaystyle \frac{1}{3}}}=0\text{or}{x}^{{\displaystyle \frac{1}{3}}}=2\hfill & \hfill \\ \hfill & \iff x=0\text{or}x=8\hfill & \text{(3}\text{pts)}\hfill \end{array}}$

      To find maximum or minimum values, we check ${\displaystyle x=-1}$, ${\displaystyle x=0}$, ${\displaystyle x=8}$, and ${\displaystyle x=20}$:

      ${\displaystyle f(-1)=2{\left(-1\right)}^{{\displaystyle \frac{5}{3}}}-5{\left(-1\right)}^{{\displaystyle \frac{4}{3}}}=-2{\left(1\right)}^{{\displaystyle \frac{5}{3}}}-5{(-1)}^{{\displaystyle \frac{4}{3}}}=-2-5=-7}$

      ${\displaystyle f(0)=2{\left(0\right)}^{{\displaystyle \frac{5}{3}}}-5{\left(0\right)}^{{\displaystyle \frac{4}{3}}}=0}$

      ${\displaystyle f(8)=2{\left(8\right)}^{{\displaystyle \frac{5}{3}}}-5{\left(8\right)}^{{\displaystyle \frac{4}{3}}}=\left(2\right){\left(2\right)}^5-5{\left(2\right)}^5=2^4(2(2)-5)=2^4(4-5)=-16}$

      ${\displaystyle f(20)=2{\left(20\right)}^{{\displaystyle \frac{5}{3}}}-5{\left(20\right)}^{{\displaystyle \frac{4}{3}}}=23.2807582317404}$    (1 pt)

      Thus, the minimum value is ${\displaystyle f(8)=-16}$ and the maximum value is ${\displaystyle f(20)=23.2807582317404}$    (2 pts)

   2. ${\displaystyle f(x)=x+\cos x}$ on ${\displaystyle [-\pi \text{, }2\pi ]}$.

      ${\displaystyle f^{\prime }(x)=1-\sin x\ge 0}$ and ${\displaystyle \sin x\le 1}$; this is true for any real x    (2 pts)

      The function is continuous and increasing on the real line.    (1 pt)

      Hence, the maximum value is ${\displaystyle f(2\pi )=2\pi +\cos (2\pi )=2\pi +1}$ and the minimum value is ${\displaystyle f(\pi )=\pi +\cos (\pi )=\pi -1}$.    (2 pts)

6. Find the dimensions of the rectangle of area ${\displaystyle 220{\text{cm}}^2}$ that has the smallest perimeter. What is the perimeter?   (6 points)

   Solution

   The formula for perimeter is ${\displaystyle P=2x+2y}$; the formula for area is ${\displaystyle A=xy}$.

   ${\displaystyle \begin{array}{ll}\text{We}\text{know}\text{that}xy=220\text{,}\text{and}\text{so}y={\displaystyle \frac{220}{x}}\hfill & \text{(2}\text{pts)}\hfill \\ P(x)=2x+{\displaystyle \frac{440}{x}}\text{and}{P}^{\prime }(x)=2-{\displaystyle \frac{440}{x^2}}=0,\text{so}x=2\sqrt{5}\hfill & \text{(2}\text{pts)}\hfill \\ P^{″}(x)={\displaystyle \frac{880}{x^3}}>0\text{for}\text{any}\text{positive}x\text{.}\hfill & \hfill \\ \text{Hence,}\text{the}\text{perimeter}\text{is}\text{a}\text{minimum}\text{when}\text{the}\text{dimensions}\text{are}\hfill & \hfill \\ x=2\sqrt{55}\text{and}y=\sqrt{220}=2\sqrt{55}\text{.}\text{The}\text{perimeter}\text{is}8\sqrt{55}\text{.}\hfill & \text{(2}\text{pts)}\hfill \end{array}}$

7. Use Newton’s method to approximate the solution of the equation ${\displaystyle x^4+2x-5=0}$ in the interval [1, 2].   (5 points)

   Solution

   ${\displaystyle \begin{array}{ll}{x}_1=1& \text{(1}\text{pt)}\\ x_2=1-{\displaystyle \frac{-2}{6}}=1+{\displaystyle \frac{1}{3}}={\displaystyle \frac{4}{3}}& \text{(2}\text{pts)}\\ x_3={\displaystyle \frac{4}{3}}-{\displaystyle \frac{67}{3(310)}}={\displaystyle \frac{391}{310}}\approx 1.2612& \text{(2}\text{pts)}\end{array}}$

8. Compute the value of each of the integrals listed below.   (20 points)
   1. ${\displaystyle \int (x^2-x)\sqrt{3x}dx}$
   2. $\displaystyle \int {\sin(2x)\cos x\,dx}$
   3. $\displaystyle \int_a^b {\left( x + \cos (2x) \right) dx}$
   4. ${\displaystyle \int \left(x^2-4\right){}^{2}dx}$
   5. ${\displaystyle {\int }_2^{4}x\sqrt{x-1}dx}$

   Solution

   1. ${\displaystyle \begin{array}{lll}\int (x^2-x)\sqrt{3x}dx\hfill & =\sqrt{3}\int x^{5/2}-x^{3/2}dx\hfill & \text{(2}\text{pts)}\hfill \\ \hfill & =\sqrt{3}\left({\displaystyle \frac{2x^{7/2}}{7}}\right)-\sqrt{3}\left({\displaystyle \frac{2x^{5/2}}{5}}\right)+C\hfill & \text{(2}\text{pts)}\hfill \end{array}}$

   2. $\displaystyle \int {\sin (2x)\cos x \, dx} = 2\int \sin x \cos^2 x$

      Since $\displaystyle{\frac{d}{dx} \cos x = -\sin x}$, (2 pts)

      $\displaystyle\int{\sin (2x)\cos x\;dx} = \displaystyle{-\frac{2\cos^3 x}{3} + C}$     (2 pts)

   3. ${\displaystyle {\int }_a^b(x+\cos (2x))dx}$

      ${\displaystyle \begin{array}{lll}{\int }_a^b(x+\cos (2x))dx\hfill & ={\displaystyle \frac{x^2}{2}}+{\displaystyle \frac{1}{2}}\sin (2x)|{}_a^b\hfill & \text{(2}\text{pts)}\hfill \\ \hfill & ={\displaystyle \frac{x^2+\sin (2x)}{2}}|{}_a^b\hfill & \hfill \\ \hfill & ={\displaystyle \frac{b^2+\sin (2b)}{2}}-{\displaystyle \frac{a^2+\sin (2a)}{2}}\hfill & \text{(2}\text{pts)}\hfill \end{array}}$

   4. ${\displaystyle \begin{array}{cc}{\int \left(x^2-4\right)}^{2}dx=\int x^4-8x^2+16dx& \text{(2}\text{pts)}\hfill \\ ={\displaystyle \frac{x^5}{5}}-{\displaystyle \frac{8x^3}{3}}+16x+C& \text{(2}\text{pts)}\hfill \end{array}}$
   5. Let ${\displaystyle u=x-1}$, thus $du=dx$ and $x=u+1\text{,}$ for $x=2\text{,}$ $u=1$ for $x=4\text{,}$ $u=3\text{(1}\text{pts)}\hfill$

      ${\displaystyle \begin{array}{cc}{\int }_2^{4}x\sqrt{x-1}dx={\int }_1^3\left(u+1\right)\sqrt{u}du={\int }_1^3{u}^{3/2}+u^{1/2}du& \text{(1}\text{pts)}\hfill \\ ={\displaystyle \frac{2u^{5/2}}{5}}+{\displaystyle \frac{2u^{3/2}}{3}}|{}_1^3& \text{(1}\text{pts)}\hfill \\ ={\displaystyle \frac{84\sqrt{3}-16}{15}}={\displaystyle \frac{28\sqrt{3}}{5}}-{\displaystyle \frac{16}{15}}& \text{(1}\text{pts)}\hfill \end{array}}$

9. Use the properties of the integral to find an interval where the value of the integral (5 points)
   
   ${\displaystyle {\int }_0^{\pi /3}\sec xdx}$
   
   is located.

   Solution

   The secant function is continuous and increasing on the interval ${\displaystyle \left[0,\pi /3\right]}$, hence

   ${\displaystyle 1=\sec \left(0\right)\le \sec x\le \sec \left({\displaystyle \frac{\pi }{3}}\right)=2\text{(2}\text{pts)}\hfill }$

   hence by the properties of the integral (8 page 303)

   ${\displaystyle \left(\pi /3\right)\le {\int }_0^{\pi /3}\sec xdx\le {\displaystyle \frac{2\pi }{3}}\text{(2}\text{pts)}\hfill }$

   So the value of the integral is in the interval ${\displaystyle \left[{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}\right]\text{(1}\text{pts)}\hfill }$

10. Find the area between the curves ${\displaystyle y=x}$ and ${\displaystyle y=2-x^2}$.   (8 points)

    Solution

    Step 1

    We sketch the graph of ${\displaystyle y=x}$ and ${\displaystyle y=2-x^2}$.    (2 pts)

    Step 2

    We find the ${\displaystyle x}$ values where two curves meet.    (2 pts)

    ${\displaystyle \begin{array}{ll}x=y\text{and}y=2-x^2\hfill & \Rightarrow x=2-x^2\hfill \\ \hfill & \iff x^2+x-2=0\hfill \\ \hfill & \iff (x+2)(x-1)=0\hfill \\ \hfill & \iff x+2=0\text{or}x-1=0\hfill \\ \hfill & \iff x=-2\text{or}x=1\hfill \end{array}}$

    Step 3

    We find the are of the region.    (4 pts)

    

    Let A be the area of the region; then,

    ${\displaystyle \begin{array}{ll}A\hfill & ={\int }_{-2}^1((2-x^2)-x)dx\hfill \\ \hfill & ={\int }_{-2}^1(-x^2-x+2)dx\hfill \\ \hfill & ={\left[-{\displaystyle \frac{1}{3}}{x}^3-{\displaystyle \frac{1}{2}}{x}^2+2x\right]}_{-2}^1\hfill \\ \hfill & ={\left[-{\displaystyle \frac{1}{3}}{x}^3\right]}_{-2}^1+{\left[-{\displaystyle \frac{1}{2}}{x}^2\right]}_{-2}^1+{[2x]}_{-2}^1\hfill \\ \hfill & =-{\displaystyle \frac{1}{3}}{[x^3]}_{-2}^1+-{\displaystyle \frac{1}{2}}{[x^2]}_{-2}^1+2{[x]}_{-2}^1\hfill \\ \hfill & =-{\displaystyle \frac{1}{3}}[1^3-{(-2)}^3]+-{\displaystyle \frac{1}{2}}[1^2-{(-2)}^2]+2[1-(-2)]\hfill \\ \hfill & =-{\displaystyle \frac{1}{3}}\cdot 9+-{\displaystyle \frac{1}{2}}\cdot (-3)+2\cdot 3\hfill \\ \hfill & =-3+{\displaystyle \frac{3}{2}}+6\hfill \\ \hfill & ={\displaystyle \frac{9}{2}}\hfill \end{array}}$

11. Use the Fundamental Theorem of Calculus to evaluate   (4 points)

    ${\displaystyle {\displaystyle \frac{d}{dx}}\underset{2x}{\overset{x}{\int }}\sin (t^2)dt}$.

    Solution

    ${\displaystyle \begin{array}{lll}{\displaystyle \frac{d}{dx}}\underset{2x}{\overset{x}{\int }}\sin (t^2)dt\hfill & ={\displaystyle \frac{d}{dx}}\underset{0}{\overset{x}{\int }}\sin (t^2)dt+{\displaystyle \frac{d}{dx}}\underset{2x}{\overset{0}{\int }}\sin (t^2)dt\hfill & \text{(1 pt)}\hfill \\ \hfill & ={\displaystyle \frac{d}{dx}}\underset{0}{\overset{x}{\int }}\sin (t^2)dt-{\displaystyle \frac{d}{dx}}\underset{0}{\overset{2x}{\int }}\sin (t^2)dt\hfill & \text{(1 pt)}\hfill \\ \hfill & =\sin (x^2)-2\sin ({(2x)}^2)\hfill & \hfill \\ \hfill & =\sin (x^2)-2\sin (4x^2)\hfill & \text{(2 pts)}\hfill \end{array}}$

12. Water flows from the bottom of a storage tank at a rate of ${\displaystyle r(t)=180-6t}$ liters per minute, where ${\displaystyle 0\le t\le 50\text{.}}$ Find the amount of water that flows from the tank during the first 15 minutes.   (3 points)

    Solution

    ${\displaystyle {\int }_0^{15}180-6tdt=180t-3t^2{0}^{15}=2025\text{(3}\text{pts)}\hfill }$

    2025 liters of water flows during the first 15 minutes.

13. A chain lying on the ground is 10 m long and its mass is 80 kg. How much work is required to raise one end of the chain to a height of 6 m?   (5 points)

    Solution

    After lifting, the chain is L-shaped, with 4 m of the chain lying along the ground. The chain slides without friction along the ground while its end is lifted. The weight density of the chain is constant throughout its length and therefore equals (8 kg/m)(9.8 m/s²) = 78.4 N/m. ${\displaystyle \text{(3}\text{pts)}\hfill }$

    The part of the chain x m from the lifted end is raised ${\displaystyle 6-x\text{m}\text{if }0\le x\le 6\text{m,}}$ and it is lifted 0 m if ${\displaystyle x>6\text{m}.}$ Thus the work needed is

    ${\displaystyle \begin{array}{cc}W={\int }_0^6(6-x)78.4dx=78.4\left[6x-{\displaystyle \frac{x^2}{2}}\right]|\underset{0}{\overset{6}{}}& \text{(1}\text{pts)}\hfill \\ =(78.4)(18)=1411.2J\text{(2}\text{pts)}\end{array}}$

14. The temperature of a metal rod, 6 m long, is 3 x (in degrees centigrade) at a distance x meters from one end of the rod. What is the average temperature of the rod?   (3 points)

    Solution

    ${\displaystyle T_{ave}={\displaystyle \frac{1}{6}}{\int }_0^{6}3xdx={\displaystyle \frac{1}{6}}\left({\displaystyle \frac{3x^2}{2}}\right)|{}_0^6=9\text{(3}\text{pts)}\hfill }$