What is the integral from 0 to t of (t-w)e^{6w} ?

The integral from 0 to t of (t-w)e^{6w} is (t(e^{6t}-1))/6-1/36 (6te^{6t}-e^{6t}+1)

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integral from 0 to t of (t-w)e^{6w}

\int_{0}^{t} (t - w) e^{6 w} d w

\frac{t ( e ^{6 t} - 1 )}{6} - \frac{1}{36} (6 t e^{6 t} - e^{6 t} + 1)

Solution steps

\int_{0}^{t} (t - w) e^{6 w} d w

Expand

(t - w) e^{6 w} : t e^{6 w} - e^{6 w} w

= \int_{0}^{t} t e^{6 w} - e^{6 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^{6 w} d w - \int_{0}^{t} e^{6 w} w d w

\int_{0}^{t} t e^{6 w} d w = \frac{t ( e ^{6 t} - 1 )}{6}

\int_{0}^{t} e^{6 w} w d w = \frac{1}{36} (6 t e^{6 t} - e^{6 t} + 1)

= \frac{t ( e ^{6 t} - 1 )}{6} - \frac{1}{36} (6 t e^{6 t} - e^{6 t} + 1)

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What is the integral from 0 to t of (t-w)e^{6w} ?
The integral from 0 to t of (t-w)e^{6w} is (t(e^{6t}-1))/6-1/36 (6te^{6t}-e^{6t}+1)