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 = -\frac1{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 = -\frac1{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{-(e^(-10t))/(500) \left(50t^2 + 10t + 1\right) + C}`