Mathematics 265 Introduction to Calculus I

Solutions to Sample Final Exam

Sample Final Examination 2

Time: 3 hours
Passing grade: 55%
Total points: 81

Solutions and Marking Scheme

These are our solutions. Your solutions may be different, but equally correct. If you have any doubts, consult your tutor. Full credit is given if

• the question is answered: you find what we are asking for, you describe what we are asking for, you gave what we are asking for.
• the answer is correct.
• you explain and/or justify each of your steps.

1. (10 points)

1. Evaluate the following limits, no credit will be given for unjustified answers. If a limit does not exist explain why.

   1. ${\displaystyle \underset{x\to \infty }{\lim }\left({\displaystyle \frac{3x^2-4}{4+2x+2x^4}}\right)\sin (x)}$

   2. ${\displaystyle \underset{x\to 5\pi ∕2}{\lim }\tan x}$

   Solution

   1. ${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{3x^2-4}{4+2x+2x^4}}=\underset{x\to \infty }{\lim }{\displaystyle \frac{3x^2}{2x^4}}=\underset{x\to \infty }{\lim }{\displaystyle \frac{3}{2x^2}}=0}$ (2 pts)

      and for ${\displaystyle x\to \infty }$ we have ${\displaystyle {\displaystyle \frac{3x^2-4}{4+2x+2x^4}}>0}$ (1 pt)

      so

      ${\displaystyle -{\displaystyle \frac{3x^2-4}{4+2x+2x^4}}\le \left({\displaystyle \frac{3x^2-4}{4+2x+2x^4}}\right)\sin (x)\le {\displaystyle \frac{3x^2-4}{4+2x+2x^4}}}$ (1 pt)

      by the squeeze theorem

      ${\displaystyle \underset{x\to \infty }{\lim }\left({\displaystyle \frac{3x^2-4}{4+2x+2x^4}}\right)\sin (x)=0}$ (1 pt)

   2. We have

      ${\displaystyle \tan x=\tan (2\pi +x)}$ so for ${\displaystyle x\to {{\displaystyle \frac{5\pi }{2}}}^{-}}$ we have ${\displaystyle t=x-2\pi \to {{\displaystyle \frac{\pi }{2}}}^{-}}$ (2 pts)

      hence

      ${\displaystyle \underset{x\to 5\pi ∕2^{-}}{\lim }\tan x=\underset{t\to \pi ∕2^{-}}{\lim }\tan t=\infty }$ (1 pt)

      and for ${\displaystyle x\to {{\displaystyle \frac{5\pi }{2}}}^{+}}$ we have ${\displaystyle t=x-2\pi \to {{\displaystyle \frac{\pi }{2}}}^{+}}$ (1 pt)

      hence

      ${\displaystyle \underset{x\to 5\pi ∕2^{+}}{\lim }\tan x=\underset{t\to \pi ∕2^{+}}{\lim }\tan t=-\infty }$ (1 pt)

      So the limit does not exist.

2. (8 points)

2. Give the indicated derivatives listed below. State which rule(s) you are applying. You may not need to simplify your answer.

   1. ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{\tan (2x)}{\sqrt{x}}}}$

   2. ${\displaystyle {{\displaystyle \frac{d}{dx}}{\cos }^3(x^2)|}_{x=\sqrt{\pi }∕2}}$

   Solution

   1. By the chain rule ${\displaystyle {\displaystyle \frac{d}{dx}}\tan (2x)=2{\sec }^2(2x)}$ and power rule ${\displaystyle {\displaystyle \frac{d}{dx}}\sqrt{x}={\displaystyle \frac{x^{-1/2}}{2}}}$ (2 pts)

      By the quotient rule

      ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{\tan (2x)}{\sqrt{x}}}={\displaystyle \frac{2\sqrt{x}{\sec }^2(2x)-{\displaystyle \frac{x^{-1/2}\tan (2x)}{2}}}{x}}}$
      (2 pts)

      A simplification gives

      ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{\tan \left(2x\right)}{\sqrt{x}}}={\displaystyle \frac{2{\sec }^2\left(2x\right)}{\sqrt{x}}}-{\displaystyle \frac{\tan \left(2x\right)}{2x\sqrt{x}}}}$

   2. By the chain rule

      ${\displaystyle {\displaystyle \frac{d}{dx}}{\cos }^3(x^2)=-3{\cos }^2(x^2)\sin (x^2)(2x)}$ (2 pts)

      For ${\displaystyle x=\sqrt{\pi }/2}$ we have ${\displaystyle {\cos }^2(\pi /4)=1/2}$ and ${\displaystyle \sin (\pi /4)=1/\sqrt{2}}$ (1 pt)

      so

      ${\displaystyle {{\displaystyle \frac{d}{dx}}{\cos }^3(x^2)|}_{x=\sqrt{\pi }/2}=-{\displaystyle \frac{3\sqrt{\pi }}{2\sqrt{2}}}}$ (1 pt)

3. (4 points)

3. 1. Use the Extreme Value Theorem to find the absolute extreme values of the function

      ${\displaystyle f(x)={\displaystyle \frac{1}{x^2+1}}}$   on the interval ${\displaystyle [-1,1]}$.

   2. Use the properties of the definite integral to estimate the value of the integral

      ${\displaystyle {\int }_{-1}^1{\displaystyle \frac{1}{x^2+1}}dx}$.

   Solution

   1. ${\displaystyle f^{\prime }(x)=-{\displaystyle \frac{2x}{{(x^2+1)}^2}}}$ one critical point for ${\displaystyle x=0}$. So ${\displaystyle f(-1)={\displaystyle \frac{1}{2}}=f(1)}$ and ${\displaystyle f\left(0\right)=1}$. So the extreme values are ${\displaystyle 1/2}$ and ${\displaystyle 1}$. (2 pts)

   2. The function is continuous on ${\displaystyle [-1,1]}$.

      ${\displaystyle 1=2\left({\displaystyle \frac{1}{2}}\right)\le {\int }_{-1}^1{\displaystyle \frac{1}{x^2+1}}dx\le 2}$ (2 pts)

4. (8 points)

4. Give the interpretation in terms of the graph of the function ${\displaystyle f\left(x\right)}$ if

   1. ${\displaystyle f(0)=-3}$

   2. ${\displaystyle \underset{x\to \infty }{\lim }f(x)=-2}$

   3. ${\displaystyle \underset{x\to 3}{\lim }f(x)=\infty }$

   4. $f^{\prime }(x)<0$ on the interval $\left[3,\infty \right)$

   5. ${\displaystyle f^{\prime }(x)<0}$ on the interval ${\displaystyle (-\infty ,-2)}$

   6. ${\displaystyle f^{″}(x)>0}$ on the interval ${\displaystyle \left[4,\infty \right)}$

   Give a sketch of the graph of ${\displaystyle f\left(x\right)}$.

   Solution

   1. The graph passes through the point ${\displaystyle (0,-3)}$, or ${\displaystyle (0,-3)}$ is the ${\displaystyle y}$ intercept of ${\displaystyle f}$.
   2. ${\displaystyle y=-2}$ is a horizontal asymptote in the positive direction.
   3. ${\displaystyle x=3}$ is a vertical asymptote.
   4. The function is decreasing on the interval ${\displaystyle \left[3,\infty \right)}$.
   5. The function is decreasing on the interval ${\displaystyle \left(-\infty ,-2\right)}$.
   6. The function is concave up on the interval ${\displaystyle \left[4,\infty \right)}$. (6 pts)

   There are infinitely many possible graphs, this is one of them.

   (2 pts)

5. (10 points)

5. What are the dimensions of the cheapest rectangular box that can be constructed if the material for the box is $1.20 per square cm and the cost for the lid is $1.50 per square cm. The length of the base is twice as long as it is wide and the volume must be 120 cm²? What is the minimum cost?

   Solution

   Let ${\displaystyle w}$ be the width of the box, ${\displaystyle x}$ be the length of the box, and ${\displaystyle h}$ be the height of the box. The volume ${\displaystyle V=xwh}$ so the constrain equation is ${\displaystyle 120=xwh}$.

   Since ${\displaystyle y=2x}$, we have ${\displaystyle 120=2x^{2}h}$. Solving for ${\displaystyle h=60/x^2}$. (1 pt)

   The cost is

   ${\displaystyle C=1.20\left(2x^2+2\left(xh\right)+2\left(2xh\right)\right)+1.50\left(2x^2\right)=5.40x^2+7.20\left(xh\right)}$. (2 pts)

   So we have to minimize the function

   ${\displaystyle C\left(x\right)=5\text{.}40x^2+{\displaystyle \frac{432}{x}}}$

   Using the second derivative test

   $C^{\prime }\left(x\right)=10.8x-{\displaystyle \frac{432}{x^2}}=0$   solving   $x=\sqrt[3]{40}$ (2 pts)

   $C^{″}\left(x\right)=10.8+{\displaystyle \frac{864}{x^3}}>0$   for any positive $x$ (1 pt)

   So the minimum is at ${\displaystyle x=\sqrt[3]{40}}$. The dimensions of the box are

   $\sqrt[3]{40}\times 2\sqrt[3]{40}\times {\displaystyle \frac{15}{\sqrt[3]{25}}}\text{.}$ (1 pt)

   The cost is

   $C\left(\sqrt[3]{40}\right)=5.4\sqrt[3]{{40}^2}+{\displaystyle \frac{432}{\sqrt[3]{40}}}\approx 189.47$ (1 pt)

   Note: Observe that we approximate at the end.

6. (10 points)

6. Integrate each one of the integrals listed below. Indicate the technique you are using.

   1. ${\displaystyle \int {\displaystyle \frac{\cos (\sqrt{2x})}{\sqrt{x}}}dx}$

   2. ${\displaystyle {\int }_0^{\pi /3}\tan x{\sec }^{2}xdx}$

   3. ${\displaystyle \int {\displaystyle \frac{x^3+\sqrt{5x}-4}{x^2}}dx}$

   4. ${\displaystyle \int {\sec }^{3}x\tan xdx}$

      Hint: ${\displaystyle {\sec }^{3}x\tan x={\sec }^{2}x\sec x\tan x}$.

   Solution

   1. By substitution ${\displaystyle u=\sqrt{x}}$, ${\displaystyle du={\displaystyle \frac{dx}{2\sqrt{x}}}={\displaystyle \frac{dx}{2u}}}$, hence ${\displaystyle dx=2udu}$

      ${\displaystyle \int 2}\cos (\sqrt{2}u)du={\displaystyle \frac{2\sin (\sqrt{2}u)}{\sqrt{2}}}=\sqrt{2}\sin \sqrt{2x}+C$ (3 pts)

   2. ${\displaystyle {\displaystyle \frac{d}{dx}}\tan x={\sec }^{2}x}$, hence

      ${\displaystyle {{\int }_0^{\pi /3}\tan x{\sec }^{2}xdx={\displaystyle \frac{{\tan }^{2}x}{2}}|}_0^{\pi /3}={\displaystyle \frac{{\tan }^2\left(\pi /3\right)}{2}}={\displaystyle \frac{3}{2}}}$ (3 pts)

   3. multiplying out

      $$\begin{gathered} {\displaystyle \int {\displaystyle \frac{x^3+\sqrt{5x}-4}{x^2}}dx=\int x+\sqrt{5}{x}^{-3∕2}-4x^{-2}dx \\ \phantom{\int {\displaystyle \frac{x^3+\sqrt{5x}-4}{x^2}}dx}={\displaystyle \frac{x^2}{2}}-{\displaystyle \frac{2\sqrt{5}}{\sqrt{x}}}-{\displaystyle \frac{4}{x}}+C} \end{gathered}$$ (3 pts)

   4. ${\displaystyle {\displaystyle \frac{d}{dx}}\sec x=\sec x\tan x}$, hence

      ${\displaystyle \int {\sec }^{3}x\tan xdx=\int {\sec }^{2}x\sec x\tan xdx={\displaystyle \frac{{\sec }^{3}x}{3}}+C}$ (3 pts)

7. (6 points)

7. 1. Find the absolute extreme values of the function ${\displaystyle f(x)={\left(x^2+2x\right)}^{2/3}}$ on the interval ${\displaystyle [-2,3]}$.

   2. Use the properties of definite integrals to bound de value of the integral

      ${\displaystyle {{\int }_{-2}^3\left(x^2+2x\right)}^{2/3}dx}$

   Solution

   1. ${\displaystyle f\left(-2\right)=0}$ and ${\displaystyle f(3)={\left({\left(3\right)}^2+6\right)}^{2/3}={15}^{2/3}}$. (1 pt)

      Also ${\displaystyle f^{\prime }(x)={\displaystyle \frac{4(x+1)}{3{\left(x^2+2x\right)}^{1/3}}}}$ the critical points are ${\displaystyle x=-1,0,-2}$. (2 pts)

      Thus

      ${\displaystyle f(-1)={\left(1-2\right)}^{2/3}=1}$   and   ${\displaystyle f(0)=f(-1)=0}$. (1 pt)

      The absolute maximum value is ${\displaystyle {15}^{2/3}}$ and the absolute minimum value is ${\displaystyle 0}$.

   2. We have by part (a) that ${\displaystyle 0\le {\left(x^2+x\right)}^{2/3}\le {15}^{2/3}}$ and the length of the interval ${\displaystyle [-2,3]}$ is ${\displaystyle 5}$ so

      $0\le {\displaystyle {\int }_{-2}^3{\left(x^2+x\right)}^{2/3}dx\le 5\left({15}^{2/3}\right)=5\sqrt[3]{225}}$ (2 pts)

8. (6 points)

8. Find the area between the curves ${\displaystyle \cos x}$ and the horizontal line ${\displaystyle y={\displaystyle \frac{1}{\sqrt{2}}}}$ in the interval $\left[-{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{\pi }{2}}\right]$.

   Solution

   The points of intersection are ${\displaystyle \left(\pm {\displaystyle \frac{\pi }{4}},{\displaystyle \frac{1}{\sqrt{2}}}\right)}$. (1 pt)

   By symmetry the area is

   ${\displaystyle 2{\displaystyle {\int }_0^{\pi /4}\cos x-{\displaystyle \frac{1}{\sqrt{2}}}dx={2\left(\sin x-{\displaystyle \frac{x}{\sqrt{2}}}\right)|}_0^{\pi /4}=\left({\displaystyle \frac{2}{\sqrt{2}}}-{\displaystyle \frac{\pi }{2\sqrt{2}}}\right)}}$ (4 pts)

   The area is ${\displaystyle \left({\displaystyle \frac{2}{\sqrt{2}}}-{\displaystyle \frac{\pi }{2\sqrt{2}}}\right)=\left(\sqrt{2}-{\displaystyle \frac{\pi \sqrt{2}}{4}}\right)=\sqrt{2}\left(1-{\displaystyle \frac{\pi }{4}}\right)}$.

9. (4 points)
9. Assume that 20 ft-lb of work is required to stretch a spring 1 ft beyond its natural length.
   1. What is the spring constant?
   2. How much work is required to stretch the spring 2 ft beyond its natural length?

   Solution

   1. The spring constant is ${\displaystyle k}$ and

      ${\displaystyle 20={\displaystyle {\int }_0^{1}kxdx={{\displaystyle \frac{k{x}^2}{2}}|}_0^1={\displaystyle \frac{k}{2}}}}$

      thus ${\displaystyle k=40}$. (2 pts)

   2. The work is given by the integral

      ${\displaystyle {\displaystyle {\int }_0^{2}40xdx={20x^2|}_0^2=80}}$

      Work is 80 ft-lb. (2 pts)

10. (6 points)
10. A particle moves along a line so that its velocity at time ${\displaystyle t}$ is ${\displaystyle v\left(t\right)=t^2-3t+2}$ (meter/sec).
    1. Find the displacement of the particle during the time period ${\displaystyle \left[0,3\right]}$.
    2. Find the distance traveled during this same time period.

    Solution

    1. ${\displaystyle s(3)-s(0)={\displaystyle {\int }_0^{3}v(t)dt={\displaystyle {\int }_0^3{t}^2-3t+2dt={\displaystyle \frac{t^3}{3}}-{\displaystyle \frac{3t^2}{2}}+{2t|}_0^3=1.5}}}$ (2 pts)

    2. Since ${\displaystyle v(t)=0}$ for ${\displaystyle x_1=1}$ and ${\displaystyle x_2=2}$ we have ${\displaystyle v(t)\le 0}$ on ${\displaystyle \left[1,2\right]}$ and ${\displaystyle v\left(t\right)\ge 0}$ on ${\displaystyle \left[0,1\right]}$ and ${\displaystyle [2,3]}$, so the distance traveled is

       ${\displaystyle {\displaystyle {\int }_1^2-t^2+3t-2dt+{\displaystyle {\int }_0^1{t}^2-3t+2dt+{\displaystyle {\int }_2^3{t}^2-3t+2dt}}}}$

       ${\displaystyle ={\displaystyle \frac{t^3}{3}}+{\displaystyle \frac{3t^2}{2}}-{2t|}_1^2+{\displaystyle \frac{t^3}{3}}-{\displaystyle \frac{3t^2}{2}}+{2t|}_0^1+{\displaystyle \frac{t^3}{3}}-{\displaystyle \frac{3t^2}{2}}+{2t|}_2^3}$

       ${\displaystyle ={\displaystyle \frac{1}{6}}-{\displaystyle \frac{5}{6}}+{\displaystyle \frac{5}{6}}}$

       ${\displaystyle ={\displaystyle \frac{11}{6}}}$ (4 pts)

11. (5 points)

11. A sprinter in a 100 m race explodes out of the starting block with an acceleration of 4 m/s², which she sustains for 2 seconds. Her acceleration then drops to zero for the rest of the race. What is her time for the race?

    Solution

    The initial acceleration is ${\displaystyle a=4}$ and the initial velocity and distance is zero, thus ${\displaystyle v\left(t\right)=4t}$ and ${\displaystyle s\left(t\right)=2t^2}$. (2 pts)

    Two seconds later the velocity is ${\displaystyle v\left(2\right)=8}$ and the distance is ${\displaystyle s\left(2\right)=8}$ and the acceleration is zero. Hence for ${\displaystyle t>2}$,

    ${\displaystyle v(t)=8}$   ${\displaystyle s(t)=8+8t}$ (2 pts)

    and she has 92 m more to go. Thus ${\displaystyle s(t)=8+8t=100}$ solving for ${\displaystyle t}$ we have ${\displaystyle t=11.5}$. So her time for the race is ${\displaystyle 13.5}$ seconds. (1 pt)

12. (4 points)

12. Find a positive number ${\displaystyle k}$ such that the average value of the function ${\displaystyle f(x)={\displaystyle \frac{5}{x^2}}}$ over the interval between ${\displaystyle 1}$ and ${\displaystyle k}$ is 32.

    Solution

    The average is

    $32={\displaystyle \frac{1}{k-1}}\underset{1}{\overset{k}{{\displaystyle \int }}}5x^{-2}dx={{\displaystyle \frac{1}{k-1}}(-5x^{-1})|}_1^k=-{\displaystyle \frac{5}{k(k-1)}}+{\displaystyle \frac{5}{k-1}}$ (2 pts)

    solving $32k^2-37k+5=0$ and $k={\displaystyle \frac{37\pm 27}{64}=1}$ or ${\displaystyle \frac{5}{32}}$ both values are positive. (2 pts)

    ${\displaystyle \frac{5}{32}}$ is a valid solution, but ${\displaystyle 1}$ is not a valid solution as it would give an interval ${\displaystyle \left[1,1\right]}$.