What is the integral of t^2e^{-t/3} ?

The integral of t^2e^{-t/3} is -3e^{-t/3}t^2+6(-3e^{-t/3}t-9e^{-t/3})+C

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integral of t^2e^{-t/3}

\int t^{2} e^{- \frac{t}{3}} d t

- 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}) + C

Solution steps

\int t^{2} e^{- \frac{t}{3}} d t

Apply Integration By Parts

: - 3 e^{- \frac{t}{3}} t^{2} - \int - 6 e^{- \frac{t}{3}} t d t

= - 3 e^{- \frac{t}{3}} t^{2} - \int - 6 e^{- \frac{t}{3}} t d t

\int - 6 e^{- \frac{t}{3}} t d t = - 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}})

= - 3 e^{- \frac{t}{3}} t^{2} - (- 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}))

Simplify

= - 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}})

Add a constant to the solution

= - 3 e^{- \frac{t}{3}} t^{2} + 6 (- 3 e^{- \frac{t}{3}} t - 9 e^{- \frac{t}{3}}) + C

n = 7 \sum \infty \frac{1}{6 n}

y = x^{2} + 5 x, x = 8

\int (x^{2} - 2 x + 3) d x

\int \frac{3 x ^{3}}{x ^{2} + 1} d x

n = 0 \sum \infty \frac{1}{n}

What is the integral of t^2e^{-t/3} ?
The integral of t^2e^{-t/3} is -3e^{-t/3}t^2+6(-3e^{-t/3}t-9e^{-t/3})+C