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Anestetic [448]
2 years ago
7

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

Mathematics
1 answer:
Marina CMI [18]2 years ago
7 0

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}

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