First slide

Introduction to definite integration

Question

$\int_{0}^{k} \frac{d x}{2 + 8 x^{2}} = \frac{π}{16} then k =$

Moderate

A

0
B

$\frac{1}{2}$
C

$\frac{π}{2}$
D

2

Solution

$\begin{array}{l} G.I = \frac{1}{2} \int_{0}^{K} \frac{d x}{1 + 4 x^{2}} = \frac{1}{2} \int_{0}^{k} \frac{d x}{1 + (2 x)^{2}} \\ = \frac{1}{2} \cdot \frac{1}{2} \cdot [\tan^{- 1} (2 x]_{0}^{k} \\ \Rightarrow \frac{1}{4} [\tan^{- 1} (2 k)] = \frac{π}{16} \end{array}$
$\begin{array}{l} \tan^{- 1} (2 k) = \frac{π}{4} \\ 2 k = 1 \\ k = \frac{1}{2} \end{array}$

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