Sum

Evaluate `int_0^oo5^(-4x^2)dx`

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#### Solution

Let I` = int_0^oo5^(-4x^2)dx`

Put `5^(-4x^2)=e^(-t)`

taking log on both sides,

`4x^2log5=t`

`x^2=t/(4log5)=t => x=sqrtt/(2sqrtlog5)`

diff. w.r.t x,

`dx=t^(-1/2)/(4log5)dt` `lim->[0,oo]`

`therefore I=int_0^ooe^(-t)/(4sqrtlog5)t^(-1/2)`

`therefore I=1/(4sqrtlog5)int_0^ooe^(-t).t^(-1/2)dt`

`therefore I=1/(4sqrtlog5)` ................`{int_0^ooe^(-1).t^(-1/2)dt=sqrtpi}`

Concept: Exact Differential Equations

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