compare the two following integral without calculation :
1)$\displaystyle{\int_0^1x{e^{x^2}}dx}$
2)$\displaystyle{\int_0^1 \sqrt{x}{e^{x}}dx}$
I would be interest for any comments or any replies
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$\begingroup$ Compare $xe^{x^2}$ and $xe^x$ on the interval $[0,1]$. $\endgroup$– Alex SchiffJun 25, 2014 at 21:19
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2$\begingroup$ Compare $x$ and $x^2$ over $[0,1]$. $\endgroup$– Pedro ♦Jun 25, 2014 at 21:19
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$\begingroup$ sorry , see i edited the question $\endgroup$– salimmath15Jun 25, 2014 at 21:35
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2 Answers
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We know that $x^2 \leq x$ for $x \in [0,1]$, so $e^{x^2} \leq e^x$, next $x \leq \sqrt{x}$ (because $\sqrt{x}(\sqrt{x}-1) \leq 0$ for $x \in [0,1]$ so $xe^{x^2} \leq \sqrt{x}e^{x}$ and finally $\int_{0}^{1} xe^{x^2} dx \leq \int_{0}^{1} \sqrt{x}e^{x} dx$
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Note that on the interval $[0,1]$, $x^2\leq x$. Since $e^x$ is monotonically increasing, it follows that on the interval $[0,1]$, $e^{x^2}\leq e^x$. You can then compare the two integrals.
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$\begingroup$ @rafik thank you for your edits! $\endgroup$– user155385Jun 25, 2014 at 21:46