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

Marks : 3

Year : 2015

Let $4^x = e^t t = x \log 4$

$\therefore dt=\log 4 dx$

when $x=0\space \space \space t=0$

$x=\infty\space \space \space t=\infty$

$I=\int\limits_0^{\infty}\Bigg(\dfrac t{\log 4}\Bigg)^4\times e^{-t}\times\dfrac {dt}{\log 4}$

$=\dfrac 1{(\log 4)^5}\int\limits_0^{\infty}e^{-t}t^4 dt$

$=\dfrac 1{(\log 4)^5}\sqrt5$