Chapter 6.3, Problem 2E

(a)

To determine

To find: ${\displaystyle \underset{1}{\overset{9}{\int }}-2f(x)dx}$

Answer to Problem 2E

  ${\displaystyle \underset{1}{\overset{9}{\int }}-2f(x)dx}=2$

Explanation of Solution

Given information:Suppose that $f$ and $h$ are continuous functions and that,

  ${\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1$, ${\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5$, ${\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{9}{\int }}-2f(x)dx}\\ =-2{\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}[\because {\displaystyle \underset{a}{\overset{b}{\int }}kf(x)dx}=k{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}]\\ =-2\times (-1)[\because {\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1]\\ =2\end{array}$

Thus, ${\displaystyle \underset{1}{\overset{9}{\int }}-2f(x)dx}=2$

(b)

To determine

  ${\displaystyle \underset{7}{\overset{9}{\int }}[f(x)+h(x)]dx}$

Answer to Problem 2E

  ${\displaystyle \underset{7}{\overset{9}{\int }}[f(x)+h(x)]dx}=9$

Explanation of Solution

Given information:Suppose that $f$ and $h$ are continuous functions and that,

  ${\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1$, ${\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5$, ${\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{7}{\overset{9}{\int }}[f(x)+h(x)]dx}\\ ={\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx+{\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}[\because {\displaystyle \underset{a}{\overset{b}{\int }}(f(x)\pm g(x))dx}}={\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}\pm {\displaystyle \underset{a}{\overset{b}{\int }}g(x)dx}]\\ =5+4[\because {\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5,{\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4]\\ =9\end{array}$

Thus, ${\displaystyle \underset{7}{\overset{9}{\int }}[f(x)+h(x)]dx}=9$

(c)

To determine

To find: ${\displaystyle \underset{7}{\overset{9}{\int }}[2f(x)-3h(x)]dx}$

Answer to Problem 2E

  ${\displaystyle \underset{7}{\overset{9}{\int }}[2f(x)-3h(x)]dx}=-2$

Explanation of Solution

Given information:Suppose that $f$ and $h$ are continuous functions and that,

  ${\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1$, ${\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5$, ${\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{7}{\overset{9}{\int }}[2f(x)-3h(x)]dx}\\ ={\displaystyle \underset{7}{\overset{9}{\int }}2f(x)dx}-{\displaystyle \underset{7}{\overset{9}{\int }}3h(x)dx}[\because {\displaystyle \underset{a}{\overset{b}{\int }}(f(x)\pm g(x))dx}={\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}\pm {\displaystyle \underset{a}{\overset{b}{\int }}g(x)dx}]\\ =2{\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}-3{\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}[\because {\displaystyle \underset{a}{\overset{b}{\int }}kf(x)dx}=k{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}]\\ =2\times 5-3\times 4[\because {\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5,{\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4]\\ =10-12\\ =-2\end{array}$

Thus, ${\displaystyle \underset{7}{\overset{9}{\int }}[2f(x)-3h(x)]dx}=-2$

(d)

To determine

To find: ${\displaystyle \underset{9}{\overset{1}{\int }}f(x)dx}$

Answer to Problem 2E

  ${\displaystyle \underset{9}{\overset{1}{\int }}f(x)dx}=1$

Explanation of Solution

Given information:Suppose that $f$ and $h$ are continuous functions and that,

  ${\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1$, ${\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5$, ${\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{9}{\overset{1}{\int }}f(x)dx}\\ =-{\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}[\because {\displaystyle \underset{b}{\overset{a}{\int }}f(x)dx}=-{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}]\\ =-(-1)[\because {\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1]\\ =1\end{array}$

Thus, ${\displaystyle \underset{9}{\overset{1}{\int }}f(x)dx}=1$

(e)

To determine

To find: ${\displaystyle \underset{1}{\overset{7}{\int }}f(x)dx}$

Answer to Problem 2E

  ${\displaystyle \underset{1}{\overset{7}{\int }}f(x)dx}=-6$

Explanation of Solution

Given information:Suppose that $f$ and $h$ are continuous functions and that,

  ${\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1$, ${\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5$, ${\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{1}{\overset{7}{\int }}f(x)dx}\\ ={\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}+{\displaystyle \underset{9}{\overset{7}{\int }}f(x)dx}[\because {\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}+{\displaystyle \underset{b}{\overset{c}{\int }}f(x)dx}={\displaystyle \underset{a}{\overset{c}{\int }}f(x)dx}]\\ ={\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}-{\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}[\because {\displaystyle \underset{b}{\overset{a}{\int }}f(x)}dx=-{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}]\\ =-1-5[\because {\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1,{\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5]\\ =-6\end{array}$

Thus, ${\displaystyle \underset{1}{\overset{7}{\int }}f(x)dx}=-6$

(f)

To determine

To find: ${\displaystyle \underset{9}{\overset{7}{\int }}[h(x)-f(x)]dx}$

Answer to Problem 2E

  ${\displaystyle \underset{9}{\overset{7}{\int }}[h(x)-f(x)]dx}=1$

Explanation of Solution

Given information:Suppose that $f$ and $h$ are continuous functions and that,

  ${\displaystyle \underset{1}{\overset{9}{\int }}f(x)dx}=-1$, ${\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5$, ${\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4$

Calculation:

  $\begin{array}{l}{\displaystyle \underset{9}{\overset{7}{\int }}[h(x)-f(x)]dx}\\ ={\displaystyle \underset{9}{\overset{7}{\int }}h(x)dx}-{\displaystyle \underset{9}{\overset{7}{\int }}f(x)dx}[\because {\displaystyle \underset{a}{\overset{b}{\int }}(f(x)\pm g(x))}dx={\displaystyle \underset{a}{\overset{b}{\int }}f(x)}dx\pm {\displaystyle \underset{a}{\overset{b}{\int }}g(x)dx}]\\ =-{\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}+{\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}[\because {\displaystyle \underset{b}{\overset{a}{\int }}f(x)dx}=-{\displaystyle \underset{a}{\overset{b}{\int }}f(x)dx}]\\ =-4+5[\because {\displaystyle \underset{7}{\overset{9}{\int }}h(x)dx}=4,{\displaystyle \underset{7}{\overset{9}{\int }}f(x)dx}=5]\\ =1\end{array}$

Thus, ${\displaystyle \underset{9}{\overset{7}{\int }}[h(x)-f(x)]dx}=1$