What is the integral from 0 to 2 of x^2e^{-5x} ?

The integral from 0 to 2 of x^2e^{-5x} is (2e^{10}-122)/(125e^{10)}

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integral from 0 to 2 of x^2e^{-5x}

\int_{0}^{2} x^{2} e^{- 5 x} d x

\frac{2 e ^{10} - 122}{125 e ^{10}}

Decimal

0.01595 \dots

Solution steps

\int_{0}^{2} x^{2} e^{- 5 x} d x

Apply u-substitution

: \int_{0}^{- 10} - \frac{e ^{u} u ^{2}}{125} d u

= \int_{0}^{- 10} - \frac{e ^{u} u ^{2}}{125} d u

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

= - \int_{- 10}^{0} - \frac{e ^{u} u ^{2}}{125} d u

Take the constant out:

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

= - (- \frac{1}{125} \cdot \int_{- 10}^{0} e^{u} u^{2} d u)

Apply Integration By Parts

: [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 10}^{0}

= - (- \frac{1}{125} [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 10}^{0})

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

= - (- \frac{1}{125} [u^{2} e^{u} - 2 (e^{u} u - e^{u})]_{- 10}^{0})

Simplify

= \frac{1}{125} [e^{u} u^{2} - 2 (e^{u} u - e^{u})]_{- 10}^{0}

Compute the boundaries:

2 - \frac{122}{e ^{10}}

= \frac{1}{125} (2 - \frac{122}{e ^{10}})

Simplify

= \frac{2 e ^{10} - 122}{125 e ^{10}}

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\frac{\partial}{\partial y} (x^{2} - x y)

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e^{t} cos (2 t)

\frac{d}{d x} (e^{0 x})

What is the integral from 0 to 2 of x^2e^{-5x} ?
The integral from 0 to 2 of x^2e^{-5x} is (2e^{10}-122)/(125e^{10)}