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Area bounded by the curve $y=e^{x^2}$, $x$ axis and the lines $x=1$, $x=2$ is given to be equal to $a$ (unit$^2$). Area bounded by the curve $y=\sqrt{\ln x}$, $y$ axis, and the lines $y=e$ and $y=e^4$ is equal to what? (in terms of $a$)

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    $\begingroup$ Are you sure that the square root is on the entire $ln$? Because if the square root would be only on the $x$, then we would have inverses. Just wondering... $\endgroup$
    – imranfat
    May 7 '18 at 15:02
  • $\begingroup$ It is inverse of the function! $\endgroup$ May 7 '18 at 18:57
  • $\begingroup$ Then it should be $y=ln\sqrt{x}$ $\endgroup$
    – imranfat
    May 7 '18 at 19:46
  • $\begingroup$ Taking ln on both sides gives us ln(y)=x² and then taking square root gives us √ln(y)=x. Please correct me if I am wrong. It is e^x² not (e^x)² $\endgroup$ May 8 '18 at 1:57
  • $\begingroup$ Ok, that makes sense. I think making a sketch can do miracles in terms of where $x=1$ and $x=2$ end up after inverting. I think the area gets inverted as well. $\endgroup$
    – imranfat
    May 8 '18 at 2:37
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The functions corresponding to two curves are inverse functions. To see this, let $f(x)=e^{x^2}$ and $g(x)=\sqrt{\ln x}$ $$f(g(x))=e^{\ln x}=x \quad \Rightarrow\quad g(x)=f^{-1}(x)$$ Hence $$\int_{e}^{e^4}g^{-1}(y)dy=\int_{f^{-1}(e)}^{f^{-1}(e^4)}f(x)dx=\int_{1}^{2}f(x)dx=a$$

Viola! The two areas are actually equal. :)

(Note that we don't need absolute value to compute area because functions are nonnegative on relevant intervals)

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  • $\begingroup$ Your answer is wrong. According to the textbook the answer is supposed to be 2e⁴-e-a $\endgroup$ May 11 '18 at 6:03

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