Can anyone tell me how to approach & solve this integral

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

Note: This is a JEE Main Question. Pls post the solution in DETAIL.....

#Calculus #Integration

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
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`\sqrt{2}`	$\sqrt{2}$
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Comments

Note 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 $f(x) = e^{x^{2}}$

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

Krishna Sharma - 6 years, 6 months ago

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

Parag Zode - 6 years, 6 months ago

none of these

polaki durga - 6 years, 6 months ago

How? the answer is 2nd option

Parag Zode - 6 years, 6 months ago

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 ;)

bahman rouhani - 6 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$