Chapter 7.2, Problem 77E

(a)

To determine

To show ${\displaystyle \int \sqrt{x+1}dx}=\frac{2}{3}{\left(x+1\right)}^{3/2}+C$

Explanation of Solution

Given information: $u={\tan }^{-1}x$

Formula used: ${\displaystyle \int \sqrt{x+1}dx}$

Calculation :

The given integral becomes as:

  ${\displaystyle \int \sqrt{x+1}dx}=\left[\frac{{\left(x+1\right)}^{3/2}}{3/2}\right]+C=$

This is the required proof

(b)

To determine

To evaluate that $y_1$ and $y_2$ are the anti-derivative of $\sqrt{x+1}$

Explanation of Solution

Given information: $y_1={\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{t+1}dt}\text{and}{y}_2={\displaystyle \underset{3}{\overset{x}{\int }}\sqrt{t+1}dt}$

Calculation :

  $y_1={\displaystyle \underset{0}{\overset{x}{\int }}\sqrt{t+1}dt}={\left[\frac{2}{3}{\left(t+1\right)}^{3/2}\right]}_0^x=\frac{2}{3}{\left(x+1\right)}^{3/2}-\frac{2}{3}$

  $y_2={\displaystyle \underset{3}{\overset{x}{\int }}\sqrt{t+1}dt}={\left[\frac{2}{3}{\left(t+1\right)}^{3/2}\right]}_3^x=\frac{2}{3}{\left(x+1\right)}^{3/2}-\frac{2}{3}(8)$

The derivative of $y_1$ and $y_2$ is $\sqrt{x+1}$

Hence, $y_1$ and $y_2$ are the anti-derivative of $\sqrt{x+1}$

(c)

To determine

To find the value of $C$

Explanation of Solution

Given information: $y_1=y_2+C$

Calculation :

  $y_1=y_2+C$

  $\frac{2}{3}{\left(x+1\right)}^{3/2}-\frac{2}{3}=\frac{2}{3}{\left(x+1\right)}^{3/2}-\frac{2}{3}(8)+C$

  $\Rightarrow C=\frac{16}{3}-\frac{2}{3}=\frac{14}{3}$

The value of $C$ is $\frac{14}{3}$

(d)

To determine

To find the value of $C={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{x+1}dx}$

Explanation of Solution

Given information: ${\displaystyle \int \sqrt{x+1}dx}=\frac{2}{3}{\left(x+1\right)}^{3/2}+C$

Calculation :

  $C={\displaystyle \underset{0}{\overset{3}{\int }}\sqrt{x+1}dx}={\left[\frac{2}{3}{\left(x+1\right)}^{3/2}\right]}_0^3=\left[\frac{2}{3}(8)-\frac{2}{3}\right]=\frac{14}{3}$

Hence, the value of $C$ is $y_1-y_2$