Question

Why integral t²(-e^-t) with limits t= 0, ∞ becomes zero.

Answer 1

$\bigstar$

Solution

$\boxed{\int_0^\infin t^2(-e^{-t})dt}$

$-\int_0^\infin t^2(e^{-t})dt$

integrate by parts: 

$∫ f g ′ = f g − ∫ f ′ g$

$f = t^ 2 , g ′ = e ^{− t}\\ ↓ steps \\ f ′ = 2 t , g = − e ^{− t}$

$= − t^ 2 e ^{− t} − ∫ − 2 t e ^{− t} d t$

Now solving:

${\displaystyle\int}-2t\mathrm{e}^{-t}\,\mathrm{d}t$

Apply linearity:

$-2{\displaystyle\int}t\mathrm{e}^{-t}\,\mathrm{d}t$

Now solving:

${\displaystyle\int}t\mathrm{e}^{-t}\,\mathrm{d}t$

Integrate by parts:

${\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}$

$f = t , g ′ = e ^{− t}\\ ↓ steps \\ f ′ = 1 , g = − e ^{− t} :$

$=-t\mathrm{e}^{-t}-{\displaystyle\int}-\mathrm{e}^{-t}\,\mathrm{d}t$

Now solving:

${\displaystyle\int}-\mathrm{e}^{-t}\,\mathrm{d}t$

$Substitute \space u = − t ⟶ \frac{ d u} {d t} = − 1 (steps) ⟶ d t = − d u :$

$={\displaystyle\int}\mathrm{e}^u\,\mathrm{d}u$

on solving

$=\mathrm{e}^{-t}$

Plug in solved integrals:

$-t\mathrm{e}^{-t}-{\displaystyle\int}-\mathrm{e}^{-t}\,\mathrm{d}t$

$=-t\mathrm{e}^{-t}-\mathrm{e}^{-t}$

Plug in solved integrals:

$-2{\displaystyle\int}t\mathrm{e}^{-t}\,\mathrm{d}t$

$=2t\mathrm{e}^{-t}+2\mathrm{e}^{-t}$

Plug in solved integrals:

$-t^2\mathrm{e}^{-t}-{\displaystyle\int}-2t\mathrm{e}^{-t}\,\mathrm{d}t$

$=-t^2\mathrm{e}^{-t}-2t\mathrm{e}^{-t}-2\mathrm{e}^{-t}$

Plug in solved integrals:

$-{\displaystyle\int}t^2\mathrm{e}^{-t}\,\mathrm{d}t$

$=t^2\mathrm{e}^{-t}+2t\mathrm{e}^{-t}+2\mathrm{e}^{-t}$

The problem is solved:

${\displaystyle\int}-t^2\mathrm{e}^{-t}\,\mathrm{d}t$

$=\left(t^2+2t+2\right)\mathrm{e}^{-t}+C$

After solving integral , we got

$\implies$ $\boxed{t^2+2t+2\over e^t}$ limit 0 to $\infin$

When we put limits

And solve

$\bigstar$

Limit t$_{\rightarrow \infin}$ [$t^2+2t+2\over e^t$ ]

Limit is $\infin\over\infin$

To solve this

We use L hospital rule , and

$\bigstar$

We will get the value , answer is

"0"