Practice Problems 13

Lecture 13: Integration

Problem numbers refer to the textbook (Chiang & Wainwright, 4th edition). These are not graded, but you should attempt them all before the quiz. Try each question yourself first, then expand the Solution to check.

Exercise 14.2

1. Find the following:

1. ${\displaystyle \int 16x^{-3}dx(x\ne 0)}$

Solution

$$\int 16x^{-3}dx=16\cdot \frac{x^{-2}}{-2}+c=-8x^{-2}+c$$

3. ${\displaystyle \int (x^5-3x)dx}$

Solution

$$\int (x^5-3x)dx=\int x^{5}dx-3\int xdx=\frac{x^6}{6}-\frac{3x^2}{2}+c$$

4. ${\displaystyle \int 2e^{-2x}dx}$

Solution

$$\int 2e^{-2x}dx=2\int e^{-2x}dx=2\cdot \frac{e^{-2x}}{-2}+c=-e^{-2x}+c$$

Exercise 14.3

1. Evaluate the following:

1. ${\displaystyle {\int }_1^3\frac{1}{2}{x}^{2}dx}$

Solution

$${\int }_1^3\frac{1}{2}{x}^{2}dx={\left[\frac{x^3}{6}\right]}_1^3=\frac{27-1}{6}=\frac{13}{3}$$

5. ${\displaystyle {\int }_{-1}^1(a{x}^2+bx+c)dx}$

Solution

$$\begin{array}{rl}{\int }_{-1}^1(a{x}^2+bx+c)dx& ={[\frac{a{x}^3}{3}+\frac{b{x}^2}{2}+cx]}_{-1}^1\\ & =(\frac{a}{3}+\frac{b}{2}+c)-(-\frac{a}{3}+\frac{b}{2}-c)\\ & =\frac{2a}{3}+2c\end{array}$$

2. Evaluate the following:

1. ${\displaystyle {\int }_1^2{e}^{-2x}dx}$

Solution

$$\begin{array}{rl}{\int }_1^2{e}^{-2x}dx& ={\left[\frac{e^{-2x}}{-2}\right]}_1^2\\ & =-\frac{1}{2}(e^{-4}-e^{-2})=\frac{1}{2}(e^{-2}-e^{-4})\end{array}$$

4. ${\displaystyle {\int }_e^6(\frac{1}{x}+\frac{1}{1+x})dx}$

Solution

$$\begin{array}{rl}{\int }_e^6(\frac{1}{x}+\frac{1}{1+x})dx& =[\ln x+\ln (1+x){]}_e^6\\ & =\ln 6+\ln 7-\ln e-\ln (1+e)\\ & =\ln 42-1-\ln (1+e)\end{array}$$

5. Verify that a constant $c$ can be equivalently expressed as a definite integral:

1. $c={\displaystyle {\int }_0^b\frac{c}{b}dx}$

Solution

$${\int }_0^b\frac{c}{b}dx={\left[\frac{cx}{b}\right]}_0^b=\frac{cb}{b}-0=c$$

2. $c={\displaystyle {\int }_0^{c}1dt}$

Solution

$${\int }_0^{c}1dt=[t{]}_0^c=c-0=c$$