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Question: If \(\int_{1}^{2}{e^{x^{2}}dx}\) = a, then the value of \(\int_{e}^{e^{4}}{\sqrt{\mathcal{l}n(x)}dx}...

If ∫12ex2dx\int_{1}^{2}{e^{x^{2}}dx} = a, then the value of ∫ee4ln(x)dx\int_{e}^{e^{4}}{\sqrt{\mathcal{l}n(x)}dx} is-

A

e4 – e

B

e4 – a

C

2e4 – a

D

2e4 – e – a

Answer

2e4 – e – a

Explanation

Solution

Let I = ∫ee4ln(x)\int_{e}^{e^{4}}\sqrt{\mathcal{l}n(x)}dx.

Put ln x = t2 \ dx = 2t et2e^{t^{2}} dt

= ∫122t2et2\int_{1}^{2}{2t^{2}e^{t^{2}}}dt =(tet2)12(te^{t^{2}})_{1}^{2} –∫12et2dt\int_{1}^{2}{e^{t^{2}}dt}

= 2e4 – e – a