What is the integral from 0 to t of (t-w)e^{3w} ?

The integral from 0 to t of (t-w)e^{3w} is (t(e^{3t}-1))/3-1/9 (3te^{3t}-e^{3t}+1)

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integral from 0 to t of (t-w)e^{3w}

\int_{0}^{t} (t - w) e^{3 w} d w

\frac{t ( e ^{3 t} - 1 )}{3} - \frac{1}{9} (3 t e^{3 t} - e^{3 t} + 1)

Solution steps

\int_{0}^{t} (t - w) e^{3 w} d w

Expand

(t - w) e^{3 w} : t e^{3 w} - e^{3 w} w

= \int_{0}^{t} t e^{3 w} - e^{3 w} w d w

Apply the Sum Rule:

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

= \int_{0}^{t} t e^{3 w} d w - \int_{0}^{t} e^{3 w} w d w

\int_{0}^{t} t e^{3 w} d w = \frac{t ( e ^{3 t} - 1 )}{3}

\int_{0}^{t} e^{3 w} w d w = \frac{1}{9} (3 t e^{3 t} - e^{3 t} + 1)

= \frac{t ( e ^{3 t} - 1 )}{3} - \frac{1}{9} (3 t e^{3 t} - e^{3 t} + 1)

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What is the integral from 0 to t of (t-w)e^{3w} ?
The integral from 0 to t of (t-w)e^{3w} is (t(e^{3t}-1))/3-1/9 (3te^{3t}-e^{3t}+1)