Mathematics 265 Introduction to Calculus I

Study Guide :: Unit 6

Integration

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 the “Learning from Mistakes” section for this unit.
4. You may want to do Exercises 53-58 from the “Review” on pages 260-262 of the textbook.
5. You may want to do Exercises 4, 5 and 8, the odd-numbered exercises from 9 to 23, and Exercises 33, 35, 39 and 41 from the “Review” (pages 308-311 of the textbook).

Learning from Mistakes

There are mistakes in each of the following solutions. Identify the errors, and give the correct answer.

1. Integrate ${\displaystyle \int {tan}^2\left(3x\right){sec}^2\left(3x\right)dx\text{.}}$

   Erroneous Solution

   ${\displaystyle f\left(x\right)=tan\left(3x\right)\text{,}}$ ${\displaystyle r=3}$ and ${\displaystyle f^{\prime }\left(x\right)={sec}^2\left(3x\right)\text{,}}$ then

   ${\displaystyle \int {tan}^2\left(3x\right){sec}^2\left(3x\right)dx={\displaystyle \frac{{tan}^3\left(3x\right)}{3}}\text{.}}$

2. Integrate ${\displaystyle {\int }_1^5{x}^3\sqrt{x^2+1}dx\text{.}}$

   Erroneous Solution

   Let ${\displaystyle u=x^2+1\text{,}}$ ${\displaystyle du=2udx}$ and ${\displaystyle x^2=u-1\text{.}}$ Hence,

   \begin{align*} \dfrac{1}{2}\int_1^5 (u - 1)u^{1/2}\,du & = \dfrac{1}{2} \int_1^5 u^{3/2} - u^{1/2}\,du \\[3pt] & = \dfrac{2u^{5/2}}{5} - \dfrac{2u^{3/2}}{3}\Bigg|_1^5 \\[3pt] & = \dfrac{2(5)^{5/2}}{3} - \dfrac{2(5)^{3/2}}{3} - \left(\dfrac{2}{5} - \dfrac{2}{3}\right) \\[3pt] & = \dfrac{4\sqrt{5^3}}{3} + \dfrac{4}{15} \end{align*}

3. Integrate ${\displaystyle {\int }_0^1{\left(x^2-\sqrt{x}\right)}^{2}dx\text{.}}$

   Erroneous Solution

   \begin{align*} \int_0^1 (x^2 - \sqrt{x})^2\,dx & = \int_0^1 x^4 - 2x^{5/2} + x \,dx \\[3pt] & = \dfrac{x^5}{5} - \dfrac{4x^{7/2}}{7} + \dfrac{x^2}{2} \Bigg|_0^1 \\[3pt] & = 0 - \! \left(\dfrac{1}{5} - \dfrac{4}{7} + \dfrac{1}{2}\right) \\[3pt] & = -\dfrac{9}{70} \end{align*}

4. Find a single function $f(x)$ that satisfies the following two conditions:
   1. ${\displaystyle f^{\prime }\left(x\right)=\cos x+\sqrt{3x}\text{,}}$ and
   2. ${\displaystyle f\left(1\right)=1\text{.}}$

   Erroneous Solution

   To find ${\displaystyle f\text{,}}$ we integrate its derivative

   ${\displaystyle \int \cos x+\sqrt{3}\sqrt{x}dx=\sin x+{\displaystyle \frac{2\sqrt{3}{x}^{3∕2}}{3}}\text{.}}$

   Hence,

   ${\displaystyle f\left(x\right)=\sin x+{\displaystyle \frac{2\sqrt{3}{x}^{3∕2}}{3}}\text{.}}$

5. Let ${\displaystyle f\left(x\right)=4x^3-6x^2+3}$ be a continuous function. Find the value $c$ in the interval ${\displaystyle \left[-1,0\right]}$ which satisfies the Mean Value Theorem.

   Erroneous Solution

   ${\displaystyle f^{\prime }\left(x\right)=12x^2-12x}$ and ${\displaystyle {\displaystyle \frac{f\left(0\right)-f\left(-1\right)}{0-\left(-1\right)}}=10\text{.}}$ Hence, ${\displaystyle 12x^2-12x=10\text{,}}$ and solving for $x$ (using the quadratic formula)

   ${\displaystyle x={\displaystyle \frac{3\pm \sqrt{39}}{6}}\text{.}}$

   There are two values satisfying the Mean Value Theorem.

6. Find the area below the curve of the function ${\displaystyle f\left(x\right)=x^3+4x}$ on the interval ${\displaystyle \left[-1,3\right]\text{.}}$

   Erroneous Solution

   ${\displaystyle {\int }_{-1}^3{x}^3+4xdx={\displaystyle \frac{x^4}{4}}+{2x|}_{-1}^4={\displaystyle \frac{4^4}{4}}+2\left(4\right)-\left({\displaystyle \frac{1}{4}}-2\right)={\displaystyle \frac{291}{4}}\text{.}}$

   The area is ${\displaystyle {\displaystyle \frac{291}{4}}\text{.}}$

7. Find the derivative of the function ${\displaystyle g\left(x\right)={\int }_{3x}^{4}t\tan tdt\text{.}}$

   Erroneous Solution

   ${\displaystyle g^{\prime }\left(x\right)=\left(3x\right)tan\left(3x\right)\text{.}}$