What is the integral from 0 to a of r^2e^{-(2r)/a} ?

The integral from 0 to a of r^2e^{-(2r)/a} is ((2e^2-10)a^3)/(8e^2)

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integral from 0 to a of r^2e^{-(2r)/a}

\int_{0}^{a} r^{2} e^{- \frac{2 r}{a}} d r

\frac{( 2 e ^{2} - 10 ) a ^{3}}{8 e ^{2}}

Solution steps

\int_{0}^{a} r^{2} e^{- \frac{2 r}{a}} d r

Apply u-substitution

: \int_{0}^{- 2} - \frac{a ^{3} e ^{u} u ^{2}}{8} d u

= \int_{0}^{- 2} - \frac{a ^{3} e ^{u} u ^{2}}{8} d u

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

= - \int_{- 2}^{0} - \frac{a ^{3} e ^{u} u ^{2}}{8} d u

Take the constant out:

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

= - (- \frac{a ^{3}}{8} \cdot \int_{- 2}^{0} e^{u} u^{2} d u)

Apply Integration By Parts

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

= - (- \frac{a ^{3}}{8} [u^{2} e^{u} - \int 2 u e^{u} d u]_{- 2}^{0})

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

= - (- \frac{a ^{3}}{8} [u^{2} e^{u} - 2 (e^{u} u - e^{u})]_{- 2}^{0})

Simplify

= \frac{a ^{3}}{8} [e^{u} u^{2} - 2 (e^{u} u - e^{u})]_{- 2}^{0}

Compute the boundaries:

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

= \frac{a ^{3}}{8} (2 - \frac{10}{e ^{2}})

Simplify

= \frac{( 2 e ^{2} - 10 ) a ^{3}}{8 e ^{2}}

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What is the integral from 0 to a of r^2e^{-(2r)/a} ?
The integral from 0 to a of r^2e^{-(2r)/a} is ((2e^2-10)a^3)/(8e^2)