First slide

Evaluation of definite integrals

Question

If $I = \int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2 n}}} for n \geq 1$ the value of $I$ is

Moderate

A

less than 1
B

more than 1/2
C

more than 1
D

less than 1/2

Solution

For $n \geq 1, 0 \leq x \leq 1 / 2, x^{2 n} \leq x^{2}$ .
$\begin{array}{l} \Rightarrow 0 \leq 1 - x^{2} \leq 1 - x^{2 n} \Rightarrow \frac{1}{\sqrt{1 - x^{2 n}}} \leq \frac{1}{\sqrt{1 - x^{2}}} \\ \begin{aligned} \Rightarrow \int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2 n}}} \leq \int_{0}^{1 / 2} \frac{d x}{\sqrt{1 - x^{2}}} & {= \sin^{- 1} x]}_{0}^{1 / 2} \\ = \frac{π}{6} < 1 \end{aligned} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App