Answers

2014-05-05T10:21:39+05:30
 \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)}


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  • Brainly User
2014-05-06T17:18:00+05:30

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
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