integral of t^2e^{-4t} — step-by-step solution

⚠ t^2e^{-4t} is read as t^2e^(-4t): use ^(...) to group an exponent, use * to multiply

Before integrating, let's calculate:

t^{2} e^{(-4 t)}

\frac{t^{2}}{e^{(4 t)}}

Let's compute the integral:

\int(\frac{t^{2}}{e^{(4 t)}})dt

An antiderivative of

e^{(- 4 t)}

- \frac{1}{e^{(4 t)} \cdot 4}

. Repeat the formula

\int(t^{2} e^{(- 4 t)})dt = t^{2} \cdot (- \frac{1}{e^{(4 t)} \cdot 4}) - \int(2 t \cdot (- \frac{1}{e^{(4 t)} \cdot 4}))dt

Integrate by parts by differentiating the polynomial and integrating the exponential. Each derivative lowers the degree, down to zero

After 0 differentiations, the polynomial is

t^{2}

; its contribution to the antiderivative is

e^{(- 4 t)} \cdot (- \frac{t^{2}}{4})

After 1 differentiation, the polynomial is

2 t

; its contribution to the antiderivative is

e^{(- 4 t)} \cdot (- \frac{2 t}{16})

After 2 differentiations, the polynomial is

2

; its contribution to the antiderivative is

e^{(- 4 t)} \cdot (- \frac{2}{64})

The next derivative is zero: add the contributions over a common denominator, keeping the exponential aside:

- \frac{t^{2}}{4} - \frac{2 t}{16} - \frac{2}{64}

\frac{- 16 t^{2} - 4 \cdot 2 t - 2}{64}

- \frac{16 t^{2} + 8 t + 2}{64}

- \frac{8 t^{2} + 4 t + 1}{32}

Restore the exponential factor:

e^{(- 4 t)} \cdot \frac{- 8 t^{2} - 4 t - 1}{32}

\frac{e^{(- 4 t)} \cdot (- (8 t^{2} + 4 t + 1))}{32}

- \frac{e^{(- 4 t)} \cdot (8 t^{2} + 4 t + 1)}{32}

- \frac{8 t^{2} + 4 t + 1}{e^{(4 t)} \cdot 32}

▶Indefinite integrals are defined up to an additive constant C, so this yields:

- \frac{8 t^{2} + 4 t + 1}{32 e^{(4 t)}} + C