Q. 30 - Calculus

Question

Use the Fundamental Theorem of Calculus to find the exact values of the given definite integrals. Use a graph to check your answer. 

 ${\int }_0^{1/2} \frac{1}{1+4x^2}dx$

Step-by-Step Solution

Answer

Ans:   The exact value of  ${\int }_0^{1/2} \frac{1}{1+4x^2}dx =\frac{\mathrm{\pi }}{8}$

Step 1. Given information.

given expression,

       ${\int }_0^{1/2} \frac{1}{1+4x^2}dx$

Step 2. The objective is to determine the exact value of the definite integral.

The exact value is calculated as shown below,

     $\begin{array}{r}{\int }_0^{1/2} \frac{1}{1+4x^2}dx\\ ={\int }_0^{1/2} \frac{1}{1+(2x{)}^2}dx\\ =\frac{1}{2}{\left[{\tan }^{-1}\left(2x\right)\right]}_0^{\frac{1}{2}}\\ =\frac{1}{2}\left[{\tan }^{-1}\left(2\frac{1}{2}\right)-{\tan }^{-1}\left(0\right)\right]\\ =\frac{1}{2}\left[\frac{\pi }{4}-0\right]\\ =\frac{\pi }{8}\end{array}$

Therefore, the exact value is $\frac{\mathrm{\pi }}{8}$.

Step 3. Check using a graph.

The required graph is: