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# 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.....

Note by Parag Zode
3 years ago

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

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

- 3 years ago

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

- 3 years 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 ;)

- 3 years ago

none of these

- 3 years ago

How? the answer is 2nd option

- 3 years ago