What is the integral from 0 to b of e^{ux}a(b-x) ?

The integral from 0 to b of e^{ux}a(b-x) is a((b(e^{bu}-1))/u-(be^{bu})/u-(-e^{bu}+1)/(u^2))

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integral from 0 to b of e^{ux}a(b-x)

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

a (\frac{b ( e ^{b u} - 1 )}{u} - \frac{b e ^{b u}}{u} - \frac{- e ^{b u} + 1}{u ^{2}})

Solution steps

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

Take the constant out:

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

= a \cdot \int_{0}^{b} (b - x) e^{ux} d x

Expand

(b - x) e^{ux} : b e^{ux} - e^{ux} x

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= a (\int_{0}^{b} b e^{ux} d x - \int_{0}^{b} e^{ux} x d x)

\int_{0}^{b} b e^{ux} d x = \frac{b ( e ^{b u} - 1 )}{u}

\int_{0}^{b} e^{ux} x d x = \frac{b e ^{b u}}{u} + \frac{- e ^{b u} + 1}{u ^{2}}

= a (\frac{b ( e ^{b u} - 1 )}{u} - (\frac{b e ^{b u}}{u} + \frac{- e ^{b u} + 1}{u ^{2}}))

Simplify

= a (\frac{b ( e ^{b u} - 1 )}{u} - \frac{b e ^{b u}}{u} - \frac{- e ^{b u} + 1}{u ^{2}})

\int \frac{1}{x e ^{l n (x)}} d x

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

\frac{d}{d y} (y^{\frac{1}{2}})

\frac{\partial}{\partial x} (1 (x - y) e^{1 y + 2 x^{2}})

f (x) = x

What is the integral from 0 to b of e^{ux}a(b-x) ?
The integral from 0 to b of e^{ux}a(b-x) is a((b(e^{bu}-1))/u-(be^{bu})/u-(-e^{bu}+1)/(u^2))