$$ \int_{0}^1 e^{x^2} dx+\int_{1}^e \sqrt {lnx} dx=e$$ Prove the above mentioned without using the imaginary function error "erfi".Solve it by using basic definite integrals properties.
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Hint. They are inverses each other: that is, $f:[0,1]\to [1,e]$, $f(x)=e^{x^2}$ and $g:[1,e]\to [0,1]$, $g(x)=\sqrt{\ln x}$ satisfies $f(g(x))=x$ and $g(f(x))=x$. Plot $y=f(x)$, $x=f(y)$ and consider the geometric meaning of the integral.
Hanul Jeon
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For the first integral use the change of variable $x=\sqrt \ln t$, i.e. $t = e^{x^2}$. The second can be integrated by parts and the two remaining integrals will cancel each other.
user26977
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