Question

Solve this integral asap​

Answer 1

Question

$\sf \large\displaystyle\int \: {x}^{2} {e}^{x} dx$

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

By using ILATE concept :

x² is an algebraic term so it's a first function and e^x is a second function.

By applying Product form of integration we will solve Question :-

$\sf\boxed{\large \displaystyle\int \: u vdx = u\int \: vdx - \displaystyle \int\bigg( \dfrac{d}{dx} u\displaystyle \int \: vdx\bigg)dx}$

$\large\displaystyle \int {x}^{2} {e}^{x} = {x}^{2}\displaystyle \int {e}^{x} dx - \displaystyle \int(2x. {e}^{x} dx)$

$\sf\large= {x}^{2} {e}^{x} - 2 \displaystyle\int \: x. {e}^{x} dx$

$\sf\large= {x}^{2} {e}^{x} - 2\bigg(x {e}^{x} - \displaystyle \int1. {e}^{x} dx\bigg)$

$\sf\large= {x}^{2} {e}^{x} - 2\bigg(x {e}^{x} - {e}^{x}\bigg )$

$\sf\large= {x}^{2} {e}^{x} - 2x {e}^{x} + 2 {e}^{x}$

Answer 2

By using ILATE concept :

x² is an algebraic term so it's a first function and e^x is a second function.

By applying Product form of integration we will solve Question :-

\sf\boxed{\large \displaystyle\int \: u vdx = u\int \: vdx - \displaystyle \int\bigg( \dfrac{d}{dx} u\displaystyle \int \: vdx\bigg)dx}

∫uvdx=u∫vdx−∫(

dx

d

u∫vdx)dx

\large\displaystyle \int {x}^{2} {e}^{x} = {x}^{2}\displaystyle \int {e}^{x} dx - \displaystyle \int(2x. {e}^{x} dx)∫x

2

e

x

=x

2

∫e

x

dx−∫(2x.e

x

dx)

\sf\large= {x}^{2} {e}^{x} - 2 \displaystyle\int \: x. {e}^{x} dx=x

2

e

x

−2∫x.e

x

dx

\sf\large= {x}^{2} {e}^{x} - 2\bigg(x {e}^{x} - \displaystyle \int1. {e}^{x} dx\bigg)=x

2

e

x

−2(xe

x

−∫1.e

x

dx)

\sf\large= {x}^{2} {e}^{x} - 2\bigg(x {e}^{x} - {e}^{x}\bigg )=x

2

e

x

−2(xe

x

−e

x

)

\sf\large= {x}^{2} {e}^{x} - 2x {e}^{x} + 2 {e}^{x}=x

2

e

x

−2xe

x

+2e

x