If the value of the integral \[\int_{1}^{2}e^{x^2}dx\] is \(a\) then the value of \[\int_{e}^{e^4}\sqrt{lnx}-dx\] is :- \[(1)e^4-e-a\] \[(2)2e^4-e-a\] \[(3)2(e^4-e)-a\] (4)None of these

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## Comments

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TopNewestNote that both functions are inverse of each other, a relation between them is

\(\displaystyle \int_a^{b} f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dy = b.f(b) - a.f(a)\)

Here \(f(x) = e^{x^{2}}\)

\(\displaystyle \int_e^{e^{4}} \sqrt{ln x} dx = 2.e^{4} - e - a\)

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Oh.... Anyways nice solution! @Krishna Sharma

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use part by part integration: \(\int v\, \mathrm{d}u\) = uv - \(\int u\, \mathrm{d}v\)

so \(\int ln(x) ^ (1/2)\, \mathrm{d}x\) = x(ln(x)) ^ (1/2)-\(\int x\, \mathrm{d}ln(x) ^ (1/2)\)

now if ln(x)^(1/2)=y then x=e^(y^2)

so \(\int ln(x) ^ (1/2)\, \mathrm{d}x\) = x(ln(x)) ^ (1/2)-\(\int (e^(y^2))\, \mathrm{d}y\)

a=\(\int (e^(y^2))\, \mathrm{d}y\) (because when x is from e to e^4, y is from 1 to 2)

our last answer will be: e^4\(\sqrt{ln(e)^4}\)-e\(\sqrt{ln(e)}\)-a = 2e^4-e-a

p.s:sorry i couldn't find all the math signs , ^ represents power ;)

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none of these

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How? the answer is 2nd option

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