Question

St^2 e^-10t dt
Evaluate the intergral

Answer 1

It looks like you're talking about the indefinite integral

$\displaystyle \int t^2 e^{-10t} \, dt$

Integrate by parts:

$\displaystyle u\,dv = uv - \int v\,du$

Let

$u = t^2 \implies du = 2t \, dt$

$dv = e^{-10t} \, dt \implies v = -\dfrac1{10} e^{-10t}$

Then

$\displaystyle \int t^2 e^{-10t} \, dt = -\frac1{10} t^2 e^{-10t} + \frac15 \int t e^{-10t} \, dt$

Integrate by parts again, this time with

$u = t \implies du = dt$

$dv = e^{-10t} \, dt \implies v = -\dfrac1{10} e^{-10t}$

Then

$\displaystyle \int t e^{-10t} \, dt = -\frac1{10} t e^{-10t} + \frac1{10} \int e^{-10t} \, dt$

Putting everything together, we get

$\displaystyle \int t^2 e^{-10t} \, dt = -\frac1{10} t^2 e^{-10t} + \frac15 \int t e^{-10t} \, dt$

$\displaystyle \int t^2 e^{-10t} \, dt = -\frac1{10} t^2 e^{-10t} + \frac15 \left(-\frac1{10} t e^{-10t} + \frac1{10} \int e^{-10t} \, dt\right)$

$\displaystyle \int t^2 e^{-10t} \, dt = -\frac1{10} t^2 e^{-10t} - \frac1{50} t e^{-10t} + \frac1{50} \int e^{-10t} \, dt$

$\displaystyle \int t^2 e^{-10t} \, dt = -\frac1{10} t^2 e^{-10t} - \frac1{50} t e^{-10t} + \frac1{50} \left(-\frac1{10} e^{-10t}\right) + C$

$\displaystyle \int t^2 e^{-10t} \, dt = -\frac1{10} t^2 e^{-10t} - \frac1{50} t e^{-10t} - \frac1{500} e^{-10t} + C$

$\displaystyle \int t^2 e^{-10t} \, dt = \boxed{-\frac{e^{-10t}}{500} \left(50t^2 + 10t + 1\right) + C}$