Chapter 14, Problem 82RE

(A)

To determine

To find:

The antiderivative of each function

Answer to Problem 82RE

  $\frac{136}{3}$

Explanation of Solution

Given:

The function

  $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}$

  $F\text{'}\left(x\right)=f\left(x\right)$

Concept used:

Antiderivative :- A function $F\left(x\right)$ is antiderivative of on an interval I if $F\text{'}\left(x\right)=f\left(x\right)$ for all $x$ in interval I

The entire family of antiderivative of a function by adding a constant to a known antiderivative

So, if $F\left(x\right)$ is the antiderivative of $f\left(x\right)$ , then the family of the antiderivatives would be $F\left(x\right)+c$

Calculation:

The function ${\displaystyle {\int }_1^4\sqrt{x}dx}\mathrm{.............}\left(1\right)$

if $F\text{'}\left(x\right)=f\left(x\right)$

Integrating the equation (1) by given the limit

  $\begin{array}{l}={\displaystyle {\int }_1^4\sqrt{x}dx}\\ ={\left[\frac{2x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3}\right]}_1^4\\ =\left[\frac{2}{3}{\left(x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)}_1^4\right]\\ =\left[\frac{2}{3}\left(1-{\left(2^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right)\right]\\ =-\frac{14}{3}\end{array}$

(B)

To determine

To draw:

The graph of this area

Explanation of Solution

Given:

The function

  $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}$

  $F\text{'}\left(x\right)=f\left(x\right)$

Concept used:

Antiderivative :- A function $F\left(x\right)$ is antiderivative of on an interval I if $F\text{'}\left(x\right)=f\left(x\right)$ for all $x$ in interval I

The entire family of antiderivative of a function by adding a constant to a known antiderivative

So, if $F\left(x\right)$ is the antiderivative of $f\left(x\right)$ , then the family of the antiderivatives would be $F\left(x\right)+c$

Calculation:

Draw the table

  $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$0$	$1$	$2$	$3$
$y-axis$	$4.66$	$4.66$	$4.66$	$4.66$

  

(C)

To determine

To draw:

The graph of this area

Explanation of Solution

Given:

The function

  $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}$

  $F\text{'}\left(x\right)=f\left(x\right)$

Concept used:

Antiderivative :- A function $F\left(x\right)$ is antiderivative of on an interval I if $F\text{'}\left(x\right)=f\left(x\right)$ for all $x$ in interval I

The entire family of antiderivative of a function by adding a constant to a known antiderivative

So, if $F\left(x\right)$ is the antiderivative of $f\left(x\right)$ , then the family of the antiderivatives would be $F\left(x\right)+c$

Calculation:

The function $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}\mathrm{.............}\left(1\right)$

if $F\text{'}\left(x\right)=f\left(x\right)$

Draw the table

  $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}$

Draw the table

  $f\left(x\right)={\displaystyle {\int }_1^4\sqrt{x}dx}$

Test one point in each of the region formed by the graph

If the point satisfies the function then shade the entire region to denote that every point in the region satisfies the function

$x-axis$	$0$	$1$	$2$	$3$
$y-axis$	$4.66$	$4.66$	$4.66$	$4.66$