Q. 29 - Calculus

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.) 

${\int }_0^{\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{1}{\sqrt{1-4x^2}}dx$

Step-by-Step Solution

Answer

The required value is $\frac{\mathrm{\pi }}{8}$.

Step 1. Given Information

We are given the definite integral ${\int }_0^{\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{1}{\sqrt{1-4x^2}}dx$ and we need to use the Fundamental Theorem of Calculus to find the exact value of the integral.

Step 2. Finding the integral

The required value is:

$$\begin{gathered} {\int }_0^{\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{1}{\sqrt{1-4x^2}}dx \\ ={\int }_0^{\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\frac{1}{\sqrt{1-{\left(2x\right)}^2}}dx \\ ={\left[{\sin }^{-1}\left(2x\right)\right]}_0^{\raisebox{1ex}{$\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.} \\ =\frac{1}{2}{\sin }^{-1}\left(\raisebox{1ex}{$2\sqrt{2}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\right)-{\sin }^{-1}0 \\ =\frac{1}{2}{\sin }^{-1}\left(\frac{\sqrt{2}}{2}\right) \\ =\frac{1}{2}\left(\frac{\mathrm{\pi }}{4}\right) \\ =\frac{\mathrm{\pi }}{8} \end{gathered}$$

Step 3. Rechecking solution

The required graph is: 

After plotting the graph we can see that the area under the graph is exactly same as the area obtained from the definite integral. The area under the graph is $\frac{\mathrm{\pi }}{8}$ square units.