Question

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

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

Short answer

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

Step-by-step solution

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: