Question

if $\int\limits^1_0 {e^{ x^{2} } (2x-a)\, dx =0$, where a is any real number, then

Answer 1

$\int\limits^1_0 {e^{ x^{2}}(2x-a)} \, dx= \int\limits^1_0 {e^{ x^{2} }(2x)} \, dx - \int\limits^1_0 {ae^{ x^{2} }} \, dx$
take u=x²
             du=2xdx
$\int\limits {e^{u } }} \ du=e^u$
$\int\limits^1_0 {e^{ x^{2}} (2x)} \, dx=(e^{ x^{2} })^1_0=e-1$
$z=-a\int\limits {e^{ x^{2}}} \, dx$
change variable it will remain same 
$y= -a\int\limits {e^{ y^{2} } \, dy$
multiply both
$z^2=a^2 \int\limits \int\limits {e^{ x^{2}+y^2 }} \, dxdy$
as u know $x^{2} +y^2=r^2 \\ dxdy=rd(theta)$
so
$z^2=a^2 \int\limits \int\limits {e^{ x^{2}+y^2 }} \, dxdy=a^2 \int\limits \int\limits {e^{ r^2 }} \, rd(theta)=a^2 \int\limits d(theta) \int\limits e^{r^2} rdr$
$z^2=\pi a^2e^{r^2} \\ z=a \sqrt{ \pi e^{r^2}}$
so now change variable and sub. limits u get
$z=a \sqrt{ \pi (e-1)}$
so the finally
$e-1-a \sqrt{ \pi (e-1)}$

Answer 2

take u=x²
             du=2xdx



change variable it will remain same 

multiply both

as u know 
so


so now change variable and sub. limits u get

so the finally