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# $\int \limits_0^1 e^{x^2}. (x- \alpha) dx$. then , $\alpha =?$

$\begin {array} {1 1} (a)\;1 < \alpha < 2 \\ (b)\;\alpha < 0 \\ (c)\;0< 2 < 1 \\ (d)\; 2 < 0 \end {array}$
Can you answer this question?

There will be that point in $(0,1)$ where $e^{x^2} (x- \alpha)=0$
$x=\alpha$
But $0 < x <1$ the $\alpha$ will be => $0 < \alpha <1$
Hence c is the correct answer.
answered Dec 16, 2013 by
edited Oct 13, 2014