What is the integral from 0 to 2 of x^2e^{-9x} ?

The integral from 0 to 2 of x^2e^{-9x} is (2e^{18}-362)/(729e^{18)}

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integral from 0 to 2 of x^2e^{-9x}

\int_{0}^{2} x^{2} e^{- 9 x} d x

\frac{2 e ^{18} - 362}{729 e ^{18}}

Decimal Notation

0.00274 \dots

Solution steps

\int_{0}^{2} x^{2} e^{- 9 x} d x

Apply u-substitution

: \int_{0}^{- 18} - \frac{e ^{u} u ^{2}}{729} d u

= \int_{0}^{- 18} - \frac{e ^{u} u ^{2}}{729} d u

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

= - \int_{- 18}^{0} - \frac{e ^{u} u ^{2}}{729} d u

Take the constant out:

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

= - (- \frac{1}{729} \cdot \int_{- 18}^{0} e^{u} u^{2} d u)

Apply Integration By Parts

: [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 18}^{0}

= - (- \frac{1}{729} [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 18}^{0})

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

= - (- \frac{1}{729} [u^{2} e^{u} - 2 (e^{u} u - e^{u})]_{- 18}^{0})

Simplify

= \frac{1}{729} [e^{u} u^{2} - 2 (e^{u} u - e^{u})]_{- 18}^{0}

Compute the boundaries:

2 - \frac{362}{e ^{18}}

= \frac{1}{729} (2 - \frac{362}{e ^{18}})

Simplify

= \frac{2 e ^{18} - 362}{729 e ^{18}}

\frac{d}{d x} (6 (sec (x) - tan (x))^{\frac{3}{2}})

x^{2} - 11, 10 x, [- 11, 0] y = 0

\frac{d}{d x} (- 1 0^{3 x^{2} - 4})

x \frac{d y}{d x} = e^{y + x}

What is the integral from 0 to 2 of x^2e^{-9x} ?
The integral from 0 to 2 of x^2e^{-9x} is (2e^{18}-362)/(729e^{18)}