What is the integral from 0 to 1 of t^3e^{-2t} ?

The integral from 0 to 1 of t^3e^{-2t} is -(-6e^2+38)/(16e^2)

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integral from 0 to 1 of t^3e^{-2t}

\int_{0}^{1} t^{3} e^{- 2 t} d t

- \frac{- 6 e ^{2} + 38}{16 e ^{2}}

Decimal

0.05357 \dots

Solution steps

\int_{0}^{1} t^{3} e^{- 2 t} d t

Apply u-substitution

: \int_{0}^{- 2} \frac{e ^{u} u ^{3}}{16} d u

= \int_{0}^{- 2} \frac{e ^{u} u ^{3}}{16} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 2}^{0} \frac{e ^{u} u ^{3}}{16} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \frac{1}{16} \cdot \int_{- 2}^{0} e^{u} u^{3} d u

Apply Integration By Parts

: [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- 2}^{0}

= - \frac{1}{16} [u^{3} e^{u} - \int 3 u^{2} e^{u} d u]_{- 2}^{0}

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= - \frac{1}{16} [u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))]_{- 2}^{0}

Compute the boundaries:

- 6 + \frac{38}{e ^{2}}

= - \frac{1}{16} (- 6 + \frac{38}{e ^{2}})

Simplify

= - \frac{- 6 e ^{2} + 38}{16 e ^{2}}

x \to 0 lim (\frac{e ^{x} - 1}{x ^{2}})

y = x^{2} - 12, y = 4 x

\frac{d y}{d x} (2 + x) = 3 y

x \to 1 lim (\frac{x - 1}{3})

f (x) = 4 x^{2} tan (x)

What is the integral from 0 to 1 of t^3e^{-2t} ?
The integral from 0 to 1 of t^3e^{-2t} is -(-6e^2+38)/(16e^2)