# differentiate e^x².hence determine a,b,c for which ∫limit 1to 2 x³e^x² dx=a/b（e^c）

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`d(e^u) = e^udu`

So for derivative of `e^(x^2)` , `e^(x^2) = 2xe^(x^2)`

`int_1^2x^3e^(x^2)dx`

Let set `u = x^2, du = 2xdx`

and `dv = xe^(x^2)dx , v = (1/2)e^(x^2)`

Appy Integration by parts formula:

`intudv = uv -intvdu`

`int_1^2x^3e^(x^2)dx = (1/2)x^2e^(x^2) - intxe^(x^2)dx`

`= (1/2)x^2e^(x^2) - (1/2)e^(x^2) `

`= e^(x^2)((1/2)x^2 - (1/2))`` `

Plug-in the upper limit and lower limit.

`[e^(2^2)((1/2)(2^2) - (1/2))]- [e^(1^2)((1/2)(1^2) - (1/2)] = (3/2)e^4`

So, **a = 3, b = 2, and c = 4**.

` `

let

`t=e^(x^2)`

differentiate t w.r.t. x

`dt/dx=e^(x^2)(d/dx)(x^2)=2xe^(x^2)` (i)``

`y=x^2`

`dy=2xdx , `

`dy/2=xdx`

`x=1 ,y=1 `

`x=2 ,y=2^2=4`

`x^3e^(x^2)=x^2 x e^(x^2)`

`int_1^2x^3e^(x^2)dx=int_1^2x^2e^(x^2)xdx=int_1^4ye^y(dy/2)`

`=(1/2)int_1^4ye^ydy`

,integrating by parts with respect to y ,by considering y as first function and `e^y` second function.

`=(1/2){ye^y-e^y}_1^4`

`=(1/2){(4e^4-e^4)-(1e^1-e^1)}`

`=(1/2)(3e^4)`

`=(3/2)e^4`

`a=3,b=2,` and `c=4`

Ans.