integral from 0 to 5 of (x^2-x+1)e^{2x}

解答

\int_{0}^{5} (x^{2} - x + 1) e^{2 x} d x

\frac{34 e ^{10} - 4}{4}

十进制

187223.95925 \dots

求解步骤

\int_{0}^{5} (x^{2} - x + 1) e^{2 x} d x

乘开

(x^{2} - x + 1) e^{2 x} : e^{2 x} x^{2} - e^{2 x} x + e^{2 x}

= \int_{0}^{5} e^{2 x} x^{2} - e^{2 x} x + e^{2 x} d x

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}^{5} e^{2 x} x^{2} d x - \int_{0}^{5} e^{2 x} x d x + \int_{0}^{5} e^{2 x} d x

\int_{0}^{5} e^{2 x} x^{2} d x = \frac{41 e ^{10} - 1}{4}

\int_{0}^{5} e^{2 x} x d x = \frac{9 e ^{10} + 1}{4}

\int_{0}^{5} e^{2 x} d x = \frac{e ^{10} - 1}{2}

= \frac{41 e ^{10} - 1}{4} - \frac{9 e ^{10} + 1}{4} + \frac{e ^{10} - 1}{2}

化简

\frac{41 e ^{10} - 1}{4} - \frac{9 e ^{10} + 1}{4} + \frac{e ^{10} - 1}{2} : \frac{34 e ^{10} - 4}{4}

= \frac{34 e ^{10} - 4}{4}

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