Question

Use the Fundamental Theorem of Calculus to find the exact

values of each of the definite integrals in Exercises 19–64. Use

a graph to check your answer. (Hint: The integrands that involve

absolute values will have to be considered piecewise.)

${\int }_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{\sqrt{9-x^2}}dx$

Short answer

${\int }_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{\sqrt{9-x^2}}dx=\frac{\mathrm{\pi }}{6}$.

Step-by-step solution

Step 1. Given information.

A definite integral is given as${\int }_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{\sqrt{9-x^2}}dx$.

Step 2. Using the Fundamental Theory of Calculus.

We get

$$\begin{gathered} {\int }_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{\sqrt{9-x^2}}dx \\ ={\int }_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{3\sqrt{1-{\left({\displaystyle \frac{x}{3}}\right)}^2}}dx \\ =\frac{1}{3}{\int }_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{1}{\sqrt{1-{\left({\displaystyle \frac{x}{3}}\right)}^2}}dx \\ =\frac{1}{3}\left(3\right){\left[{\sin }^{-1}\frac{x}{3}\right]}_0^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ ={\sin }^{-1}\left(\frac{{\displaystyle \frac{3}{2}}}{3}\right)-{\sin }^{-1}0 \\ ={\sin }^{-1}\left(\frac{1}{2}\right)-0 \\ =\frac{\mathrm{\pi }}{6}-0 \\ =\frac{\mathrm{\pi }}{6} \end{gathered}$$

So the exact value of the given definite integral is$\frac{\mathrm{\pi }}{6}$.

Step 3. The graph to verify the answer is

The solution is the area under graph which is

$$\begin{gathered} a\approx 0.5235987 \\ \approx \frac{\mathrm{\pi }}{6} \end{gathered}$$