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# K ∫ 0 1 2 + 8 X 2 D X = π 16 , Find the Value of K. - CBSE (Commerce) Class 12 - Mathematics

ConceptProperties of Definite Integrals

#### Question

$\int\limits_0^k \frac{1}{2 + 8 x^2} dx = \frac{\pi}{16},$ find the value of k.

#### Solution

$\text{We have},$
$\int_0^k \frac{1}{2 + 8 x^2} d x = \frac{\pi}{16}$
$\Rightarrow \frac{1}{8} \int_0^k \frac{1}{\frac{1}{4} + x^2} d x = \frac{\pi}{16}$
$\Rightarrow \frac{1}{4} \left[ \tan^{- 1} 2x \right]_0^k = \frac{\pi}{16}$
$\Rightarrow \tan^{- 1} 2k = \frac{\pi}{4}$
$\Rightarrow 2k = \tan\frac{\pi}{4}$
$\Rightarrow 2k = 1$
$\Rightarrow k = \frac{1}{2}$

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Solution K ∫ 0 1 2 + 8 X 2 D X = π 16 , Find the Value of K. Concept: Properties of Definite Integrals.
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