0
$\begingroup$

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

$\endgroup$
3
  • $\begingroup$ Compare $xe^{x^2}$ and $xe^x$ on the interval $[0,1]$. $\endgroup$ Commented Jun 25, 2014 at 21:19
  • 2
    $\begingroup$ Compare $x$ and $x^2$ over $[0,1]$. $\endgroup$
    – Pedro
    Commented Jun 25, 2014 at 21:19
  • $\begingroup$ sorry , see i edited the question $\endgroup$ Commented Jun 25, 2014 at 21:35

2 Answers 2

1
$\begingroup$

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 $\int_{0}^{1} xe^{x^2} dx \leq \int_{0}^{1} \sqrt{x}e^{x} dx$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ @rafik thank you for your edits! $\endgroup$
    – user155385
    Commented Jun 25, 2014 at 21:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .