integral from 0 to t of (t-w)e^{2w}

解答

\int_{0}^{t} (t - w) e^{2 w} d w

\frac{t ( e ^{2 t} - 1 )}{2} - \frac{1}{4} (2 t e^{2 t} - e^{2 t} + 1)

求解步骤

\int_{0}^{t} (t - w) e^{2 w} d w

乘开

(t - w) e^{2 w} : t e^{2 w} - e^{2 w} w

= \int_{0}^{t} t e^{2 w} - e^{2 w} w d w

使用积分加法定则:

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

= \int_{0}^{t} t e^{2 w} d w - \int_{0}^{t} e^{2 w} w d w

\int_{0}^{t} t e^{2 w} d w = \frac{t ( e ^{2 t} - 1 )}{2}

\int_{0}^{t} e^{2 w} w d w = \frac{1}{4} (2 t e^{2 t} - e^{2 t} + 1)

= \frac{t ( e ^{2 t} - 1 )}{2} - \frac{1}{4} (2 t e^{2 t} - e^{2 t} + 1)

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