Chapter 5.5, Problem 59E

To determine

To evaluate: The integral function ${\displaystyle {\int }_{-2}^2\left(x+3\right)\sqrt{4-x^2}dx}$.

Answer to Problem 59E

Theevaluation of the integral is $\underset{\_}{6\pi }$.

Explanation of Solution

Given:

The integral function is ${\displaystyle {\int }_{-2}^2\left(x+3\right)\sqrt{4-x^2}dx}$.

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

Show the theorem 7 (Integrals of symmetric Functions):

Apply the condition as follows if f is continuous on [–a, a].

1. a) If f is even $\left[f\left(-x\right)=f\left(x\right)\right]$, then ${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx}=2{\displaystyle {\int }_0^{a}f\left(x\right)dx}$.
2. b) If f is odd $\left[f\left(-x\right)=-f\left(x\right)\right]$, then ${\displaystyle {\int }_{-a}^{a}f\left(x\right)dx}=0$.

Calculation:

The integral function is,

$\begin{array}{c}{\displaystyle {\int }_{-2}^2\left(x+3\right)\sqrt{4-x^2}dx}={\displaystyle {\int }_{-2}^{2}x\sqrt{4-x^2}+3\sqrt{4-x^2}dx}\\ ={\displaystyle {\int }_{-2}^{2}x\sqrt{4-x^2}dx}+{\displaystyle {\int }_{-2}^{2}3\sqrt{4-x^2}dx}\end{array}$ (1)

Consider $I_1$ and $I_2$ as follows:

$I_1={\displaystyle {\int }_{-2}^{2}x\sqrt{4-x^2}dx}$ (2)

$I_2={\displaystyle {\int }_{-2}^{2}3\sqrt{4-x^2}dx}$ (3)

Substitute $I_1$ for ${\displaystyle {\int }_{-2}^{2}x\sqrt{4-x^2}dx}$ and $I_2$ for ${\displaystyle {\int }_{-2}^{2}3\sqrt{4-x^2}dx}$ in Equation (1).

${\displaystyle {\int }_{-2}^2\left(x+3\right)\sqrt{4-x^2}dx}=I_1+I_2$ (4)

The function $f\left(x\right)$ is,

$f\left(x\right)=x\sqrt{4-x^2}$ (5)

Substitute $\left(-x\right)$ for x in Equation (5).

$\begin{array}{c}f\left(-x\right)=\left(-x\right)\sqrt{4-{\left(-x\right)}^2}\\ =-x\sqrt{4-x^2}\\ =-f\left(x\right)\end{array}$ (6)

Apply the theorem in Equation (7).

The function satisfies the condition as $\left[f\left(-x\right)=-f\left(x\right)\right]$.

Thus, the function is an odd function and the value of $I_1$ is 0.

Interpret the integral function $\left(I_2\right)$

The integral function is equal to three times of the area of the semicircle with radius 2.

Calculate the integral function $\left(I_2\right)$ using Equation (3):

$\begin{array}{c}{I}_2=3\times \left(\text{Area}\text{of}\text{semicircle}\right)\\ =3\times \left(\frac{1}{2}\times \pi \times 2^{{}^2}\right)\\ =6\pi \end{array}$

Substitute 0 for $I_1$ and $6\pi$ for $I_2$ in Equation (4).

$\begin{array}{c}{\displaystyle {\int }_{-2}^2\left(x+3\right)\sqrt{4-x^2}dx}=0+6\pi \\ =6\pi \end{array}$

Hence, the evaluation of the integral is $\underset{\_}{6\pi }$.