Range of a for $\int_{0}^{1} e^{x^2}(2x-a)dx$ = 0 If
$\int_{0}^{1} e^{x^2}(2x-a)dx$ = 0 , where a is any real number, then
(A) a > 2
(B) a is negative
(C) a = 1/e^2
(D) a lies between 1/3 and 2 
How can i integrate $e^{x^2}$?
 A: You don't need to integrate !
if $f(x)=e^{x^2}(2x-a)$, the question is "does it exist a $a$ such that $f$ change of sign when $x\in[0,1]$, and the response is for $0\leq a\leq 2$ (because the sign of $f$ depend of the sign of $2x-a$ and $2x-a<0$ if $x<\frac{a}{2}$ and $f(x)\geq 0$ if $x\geq \frac{a}{2}$)
(A) if $a>2$, then $f$ is always negative. Then the integral can't be nul. 
(B) If $a<0$, the $f$ is always positive, therefore the integral can't be nul.
(C) If $a=\frac{1}{e^2}$, then $f(x)\geq 2x-\frac{1}{e^2}$ if $x\in[0,1]$ and thus 
$$\int_0^1f(x)dx\geq \int_0^1 (2x-e^{-2})dx>0$$
(The last integral is easy to calculate.)
Therefore, if there is a solution, it must be $(D)$.
But to prove that it's really (D), you can set $g(a)=\int_0^1 e^{x^2}(2x-a)dx$, which is continuous. You have that $g(2)\leq 0$ (because we proves that if $a>2$ we have that $f$ is negative) and $g(1/3)\geq 0$ because $f(x)\geq (x-1/3)$ for $a=1/3$. Therefore, by the mean value theorem, there exist a $c\in]1/3,2[$ such that $g(c)=0$ what conclude the proof.
