integral from 0 to 4 of xln(5-x)

解答

\int_{0}^{4} x ln (5 - x) d x

- 14 + \frac{25}{2} ln (5)

十进制

6.11797 \dots

求解步骤

\int_{0}^{4} x ln (5 - x) d x

使用分布积分法

= [\frac{1}{2} x^{2} ln (5 - x) - \int - \frac{x ^{2}}{2 ( 5 - x )} d x]_{0}^{4}

\int - \frac{x ^{2}}{2 ( 5 - x )} d x = - \frac{1}{2} (- \frac{x ^{2}}{2} - 5 x - 25 ln ∣ - x + 5 ∣)

= [\frac{1}{2} x^{2} ln (5 - x) - (- \frac{1}{2} (- \frac{x ^{2}}{2} - 5 x - 25 ln ∣ - x + 5 ∣))]_{0}^{4}

化简

= [\frac{1}{2} x^{2} ln (5 - x) + \frac{1}{2} (- \frac{x ^{2}}{2} - 5 x - 25 ln ∣ - x + 5 ∣)]_{0}^{4}

带入上下限计算后得:

- 14 + \frac{25}{2} ln (5)

= - 14 + \frac{25}{2} ln (5)

\int_{0}^{5} x e^{5 x} d x

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

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