Chapter 5.2, Problem 34E

To determine

To Find: The integral of ${\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx$

Answer to Problem 34E

The integral of ${\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx=2\pi$

Explanation of Solution

Given information:

The integral is:

  ${\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx$

Calculation:

  ${\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx$

Interpret this integral as the area under the curve $y=\sqrt{4-x^2}$ from $\text{-2to2}$

But since

  $\begin{array}{l}{y}^2=4-x^2\\ x^2+y^2=4\end{array}$

Which show the graph of $f(x)$ is the half circle with radius $\text{2}$ in figure

  

Therefore

  $\begin{array}{l}{\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx=\frac{1}{2}\left(\pi {\left(2\right)}^2\right)\\ {\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx=2\pi \end{array}$

Hence,

The integral of ${\displaystyle \underset{-2}{\overset{2}{\int }}\sqrt{4-x^2}}dx=2\pi$