Chapter 5.2, Problem 49E

Each of the regions A, B, and C bounded by the graph of f and the x-axis has area 3. Find the value of

${\displaystyle {\int }_{-4}^2[f(x)+2x+5]dx}$

To determine

To calculate: The value of the integral function ${\displaystyle \underset{-4}{\overset{2}{\int }}\left[f\left(x\right)+2x+5\right]}dx$.

Answer to Problem 49E

The value of the integral function ${\displaystyle \underset{-4}{\overset{2}{\int }}\left[f\left(x\right)+2x+5\right]}dx$ is 15.

Explanation of Solution

Given information:

The integral function is ${\displaystyle \underset{-4}{\overset{2}{\int }}\left[f\left(x\right)+2x+5\right]}dx$ (1)

The region lies between $x=-4$ and $x=2$.

Calculation:

The integral property 2 is shown below:

${\displaystyle \underset{a}{\overset{b}{\int }}\left[f\left(x\right)+g\left(x\right)\right]}dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)}dx+{\displaystyle \underset{a}{\overset{b}{\int }}g\left(x\right)}dx$ (2)

The property is valid only, if the functions $f\left(x\right)$ and $g\left(x\right)$ are continuous.

Apply the integral property 2 in Equation (1).

${\displaystyle \underset{-4}{\overset{2}{\int }}\left[f\left(x\right)+2x+5\right]}dx={\displaystyle \underset{-4}{\overset{2}{\int }}f\left(x\right)}dx+2{\displaystyle \underset{-4}{\overset{2}{\int }}x}dx+{\displaystyle \underset{-4}{\overset{2}{\int }}5dx}$ (3)

Rearrange Equation (3) as shown below.

${\displaystyle \underset{-4}{\overset{2}{\int }}\left[f\left(x\right)+2x+5\right]}dx=I_1+2I_2+I_3$ (4)

Consider the regions A, B, and C bounded by the graph of f as in Figure 1.

Refer to Figure 1.

The value of the region which lies in the integral ${\displaystyle \underset{-4}{\overset{-2}{\int }}f\left(x\right)}dx$ is $-3$.

The value of the region which lies in the integral ${\displaystyle \underset{-2}{\overset{0}{\int }}f\left(x\right)}dx$ is 3.

The value of the region which lies in the integral ${\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)}dx$ is $-3$.

Calculate the integral $I_1$ by rearranging the integral function ${\displaystyle \underset{-4}{\overset{2}{\int }}f\left(x\right)}dx$ as shown below:

$\begin{array}{c}{I}_1={\displaystyle \underset{-4}{\overset{-2}{\int }}f\left(x\right)}dx+{\displaystyle \underset{-2}{\overset{0}{\int }}f\left(x\right)}dx+{\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)}dx\\ =-3+3-3\\ =-3\end{array}$

Calculate $I_2$ using the relation:

$I_2={\displaystyle \underset{-4}{\overset{2}{\int }}x}dx$ (5)

Consider the function $g\left(x\right)=x$.

The function $g\left(x\right)$ represents a straight line with slope 1.

Plot the function $g\left(x\right)$ within the limits $\left[-4,2\right]$ as shown in Figure 2.

Refer to Figure 2.

Take the area bounded by the function $g\left(x\right)$ within the limits $\left[-4,2\right]$ as $A_1$ and $A_2$

Modify Equation (5) using the Figure 2.

$\begin{array}{c}{I}_2={\displaystyle \underset{-4}{\overset{2}{\int }}x}dx\\ =A_1+A_2\\ =\left(\frac{1}{2}\times 2\times 2\right)-\left(\frac{1}{2}\times 4\times 4\right)\\ =-6\end{array}$

Calculate $I_3$ using the relation:

$\begin{array}{c}{I}_2={\displaystyle \underset{-4}{\overset{2}{\int }}5}dx\\ =5{\left[x\right]}_{-4}^2\\ =5\left(2-\left(-4\right)\right)\\ =30\end{array}$

Calculate the integral function using Equation (4).

Substitute $-$3 for $I_1$, $-$6 for $I_2$ and 30 for $I_3$ in Equation (4).

$\begin{array}{c}{\displaystyle \underset{-4}{\overset{2}{\int }}\left[f\left(x\right)+2x+5\right]}dx=I_1+2I_2+I_3\\ =-3+2\left(-6\right)+30\\ =-3-12+30\\ =15\end{array}$

Therefore, the value of the integral function is 15.