Chapter 6.2, Problem 77E

$\text{(a)}$

To determine

To prove: The statement ${\displaystyle \int \sqrt{x+1}}dx=\frac{2}{3}{(x+1)}^{3/2}+C$

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \sqrt{x+1}}dx$

Calculation:

  $\begin{array}{c}{\displaystyle \int \sqrt{x+1}}dx={\displaystyle \int \sqrt{u}}du\\ ={\displaystyle \int u^{\frac{1}{2}}}du\\ =\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C\\ =\frac{2}{3}{u}^{\frac{3}{2}}+C\\ =\frac{2}{3}{(}^x+C\end{array}$

Therefore, the required ${\displaystyle \int \sqrt{x+1}}dx=\frac{2}{3}{(x+1)}^{3/2}+C$

  $($ proved $)$ .

$\text{(b)}$

To determine

To prove: The given statement.

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \sqrt{x+1}}dx$

And

  $\begin{array}{c}{y}_1={\displaystyle {\int }_0^x\sqrt{t+1}}dt\\ y_2={\displaystyle {\int }_3^x\sqrt{t+1}}dt\end{array}$

Are anti derivatives of $\sqrt{x+1}$ .

Calculation:

By FTC part $1$ which states $\frac{d}{dx}{\displaystyle {\int }_c^{x}f}(t)dt=f(x)$ , the derivatives of $y_1$ and $y_2$ are as follow:

  $\begin{array}{c}\frac{d{y}_1}{dx}=\frac{d}{dx}{\displaystyle {\int }_0^x\sqrt{t+1}}dt\\ =\sqrt{x+1}\\ \frac{d{y}_2}{dx}=\frac{d}{dx}{\displaystyle {\int }_3^x\sqrt{t+1}}dt\\ =\sqrt{x+1}\end{array}$

Hence, they both have the same derivative of $\sqrt{x+1}$ , they are both antiderivatives of $\sqrt{x+1}$ .

$(c)$

To determine

To find: The value of $C$ .

Answer to Problem 77E

The answer is $C=4\frac{2}{3}$ .

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \sqrt{x+1}}dx$

Calculation:

Use the NINT feature on your graphing calculator to input

  $\begin{array}{c}{Y}_1={\displaystyle {\int }_0^x\sqrt{x+1}}dx\\ Y_2={\displaystyle {\int }_3^x\sqrt{x+1}}dx\end{array}$

Then input,

  $Y_3=Y_1-Y_2$

Then go to the table feature on your graphing calculator to find the values of these three equations from $x=0$

to

  $x=5$ . The table of values for $Y_1,Y_2\text{, }Y$ from $x=0$

to

  $x=5$ is shown below:

$\text{x}$	$Y_1$	$Y_2$	$Y_3=Y_1-Y_2$
0	0	-4.667	4.667
1	1.219	-3.448	4.667
2	2.797	-1.869	4.667
3	4.667	0	4.667
4	6.787	2,120	4.667
5	9.131	4.464	4.667

For every value of $x$ ,

  $\begin{array}{c}{Y}_1-Y_2\approx 4.667\\ =4\frac{2}{3}\end{array}$

So,

  $C=4\frac{2}{3}$

Therefore, the required value of $C=4\frac{2}{3}$ .

$(d)$

To determine

To explain: The statement

Answer to Problem 77E

The answer is

Explanation of Solution

Given information:

The integral is ${\displaystyle \int \sqrt{x+1}}dx$

Calculation:

From part $(c)$ : $C=y_1-y_2$

From part $\text{b)}$

:

  

Put these in:

  

To use the order of integration rule,

  

Then gives,

  

To use the activity rule,

  

Then gives,