What is the integral from 0 to b of xe^{-ax} ?

The integral from 0 to b of xe^{-ax} is (-abe^{-ab}-e^{-ab}+1)/(a^2)

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integral from 0 to b of xe^{-ax}

\int_{0}^{b} x e^{- a x} d x

\frac{- ab e ^{- ab} - e ^{- ab} + 1}{a ^{2}}

Solution steps

\int_{0}^{b} x e^{- a x} d x

Apply u-substitution

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

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

Take the constant out:

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

= \frac{1}{a ^{2}} \cdot \int_{0}^{- ab} e^{u} u d u

Apply Integration By Parts

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

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

\int e^{u} d u = e^{u}

= \frac{1}{a ^{2}} [e^{u} u - e^{u}]_{0}^{- ab}

Simplify

\frac{1}{a ^{2}} [e^{u} u - e^{u}]_{0}^{- ab} : \frac{[ e ^{u} u - e ^{u} ] _{0}^{- ab}}{a ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{0}^{- ab}}{a ^{2}}

Compute the boundaries:

- ab e^{- ab} - e^{- ab} + 1

= \frac{- ab e ^{- ab} - e ^{- ab} + 1}{a ^{2}}

\int_{1}^{5} \frac{x}{12} d x

\int_{1}^{2} \frac{x}{6} d x

\int_{- 2}^{2} t d t

\int_{- 1}^{1} x \cdot e^{x} d x

\int_{1}^{2} \frac{x - 1}{2} d x

What is the integral from 0 to b of xe^{-ax} ?
The integral from 0 to b of xe^{-ax} is (-abe^{-ab}-e^{-ab}+1)/(a^2)