Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 7

Applications of the Definite Integral

Finishing This Unit

1. Review the objectives of this unit and make sure you are able to meet all of them.
2. If there is a concept, definition, example or exercise that is not yet clear to you, go back and reread it, then contact your tutor for help.
3. Do the exercises in “Learning from Mistakes” section for this unit.
4. Find the area of the region bounded by the curves given below.
   1. ${\displaystyle y=x^2-x-6\text{,}}$ ${\displaystyle y=0}$
   2. ${\displaystyle y=1-x^2\text{,}}$ ${\displaystyle y=1-\sqrt{x}}$
   3. ${\displaystyle x+y=0\text{,}}$ ${\displaystyle x=y^2+3y}$
   4. ${\displaystyle y=sin\left(\pi x∕2\right)\text{,}}$ ${\displaystyle y=x^2-2x}$
5. A force of 30 N is required to maintain a spring stretched from its natural length of 12 cm to a length of 15 cm. How much work is done in stretching the spring from 12 cm to 20 cm?
6. Find the average value of the function ${\displaystyle f\left(t\right)=tsin\left(t^2\right)}$ on the interval ${\displaystyle \left[0,1\right]\text{.}}$
7. If ${\displaystyle f}$ is a continuous function, what is the limit as ${\displaystyle h\to 0}$ of the average value of ${\displaystyle f}$ on the interval ${\displaystyle \left[x,x+h\right]\text{?}}$

Learning from Mistakes

There are mistakes in each of the following solutions, identify them and give the right answer.

1. A spacecraft uses a sail and the “solar wind” to produce a constant acceleration of ${\displaystyle 0.032}$ m/s${\displaystyle {}^2\text{.}}$ Assuming that the spacecraft has a velocity of ${\displaystyle 600}$ km/h when the sail is first raised, how far will the spacecraft travel in 1 hour, and what will its velocity be at the end of the hour?

   Erroneous Solution

   We define ${\displaystyle t_0=0}$ to be the time the sail is raised, and ${\displaystyle t_1=1}$ to be one hour later. The acceleration ${\displaystyle a=0.032\text{.}}$ At time ${\displaystyle t\text{,}}$ the velocity is ${\displaystyle v\left(t\right)}$ and the displacement is ${\displaystyle s\left(t\right)\text{.}}$ We then have ${\displaystyle v\left(0\right)=600}$ and ${\displaystyle s\left(0\right)=0\text{.}}$

   ${\displaystyle v\left(1\right)-v\left(0\right)={\int }_0^{1}adt={\int }_0^{1}0.032dt=0.032.}$

   Thus,

   ${\displaystyle v\left(1\right)=0.32+v\left(0\right)=0.032+600=600.032\text{ and }v\left(t\right)=60+0.032t.}$

   ${\displaystyle s(1)-<!--\; -\; -->s(0)={\displaystyle {\int <!--\; \int \; -->}_0^{1}600+0.032tdt=600t+.016t^2{|}_0^1=600.016\text{.}}}$

   Answer: The velocity is 600.032 km/h, and the distance is 600.016 km.

2. Find the area under the curve ${\displaystyle y=x^2+x-2}$ on the interval ${\displaystyle \left[0,3\right]\text{.}}$

   Erroneous Solution

   ${\displaystyle {\int <!--\; \int \; -->}_0^3{x}^2+x-<!--\; -\; -->2dx=\frac{x^3}{3}+\frac{x^2}{2}-<!--\; -\; -->2x{|}_0^3=\frac{15}{2}\text{.}}$

3. Find the area between the curves ${\displaystyle y=\frac{1}{x^2}\text{,}}$ ${\displaystyle y=x}$ and ${\displaystyle y=\frac{x}{8}\text{.}}$

   Erroneous Solution

   The point of intersection of ${\displaystyle y=\frac{1}{x^2}}$ and ${\displaystyle y=x}$ is the solution of ${\displaystyle x=\frac{1}{x^2}}$; that is ${\displaystyle x=1}$ and ${\displaystyle \left(1,1\right)\text{.}}$ The point of intersection of ${\displaystyle y=frac1x^2}$ and ${\displaystyle y=\frac{x}{8}}$ is the solution of ${\displaystyle \left(x^2\right)\frac{x}{8}=1}$; that is, ${\displaystyle x=2}$ and ${\displaystyle \left(2,\frac{1}{4}\right)\text{.}}$

   $\begin{array}{rl}{\displaystyle {\int }_0^1{\displaystyle \frac{1}{x^2}}-xdx}+{\displaystyle {\int }_1^2-{\displaystyle \frac{1}{x^2}}+{\displaystyle \frac{x}{8}}dx}& ={-{\displaystyle \frac{1}{x}}-{\displaystyle \frac{x^2}{2}}{|}_0^1+{\displaystyle \frac{1}{x}}+{\displaystyle \frac{x^2}{16}}|}_1^2\\ & =-{\displaystyle \frac{3}{2}}+{\displaystyle \frac{12}{16}}-{\displaystyle \frac{17}{16}}\\ & =-{\displaystyle \frac{35}{16}}\text{.}\end{array}$

4. A spring exerts a force of 100 N when it is stretched 0.2 m beyond its natural length. How much work is required to stretch the spring 0.8 m beyond its natural length?

   Erroneous Solution

   ${\displaystyle {\int }_{.2}^{.8}100xdx=50x^2{|}_{.2}^{.8}=50\left(.64-.04\right)=30.}$

   Answer: 30 J.