Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 3

Year : 2013

Let $7^{-4x^2}= e^{-t}$

Taking log on both sides

$-4x^2 \log 7 = -t$

$x^2=\dfrac t{4\log 7}$

$x=\dfrac {\sqrt t}{\sqrt{4\log 7}}$

Differ w. r. t. x

$dx=\dfrac 1{2\sqrt t\sqrt{4\log 7}}dt=\dfrac 1{4\sqrt t\sqrt{\log 7}}$

Substituting all the value we get

$\int\limits_0^{\infty}\dfrac t{4\log 7}e^{-t}\dfrac 1{4\sqrt t\sqrt{\log 7}}dt$

$=\dfrac 1{16\log 7 \sqrt{\log 7}}\int\limits^{\infty}_0e^{-t}t^{1/2}$ dt

By formula of beta gamma function integration is replaced by $\sqrt{\dfrac 12+1}$

$I=\dfrac 1{16\log 7\sqrt{\log 7}}\sqrt{\dfrac 32}$

$=\dfrac 1{16\log 7\sqrt{\log 7}}\times\dfrac 12\sqrt{\dfrac 12}$

$=\dfrac 1{32\log 7\sqrt{\log 7}}\sqrt{\pi}$