integral from a to b of ln(r^2)r

解答

\int_{a}^{b} ln (r^{2}) r d r

b^{2} ln (b) + \frac{- b ^{2} + a ^{2}}{2} - a^{2} ln (a)

求解步骤

\int_{a}^{b} ln (r^{2}) r d r

使用分布积分法

= [\frac{1}{2} r^{2} ln (r^{2}) - \int r d r]_{a}^{b}

\int r d r = \frac{r ^{2}}{2}

= [\frac{1}{2} r^{2} ln (r^{2}) - \frac{r ^{2}}{2}]_{a}^{b}

\frac{1}{2} r^{2} ln (r^{2}) = r^{2} ln (r)

= [r^{2} ln (r) - \frac{r ^{2}}{2}]_{a}^{b}

带入上下限计算后得:

b^{2} ln (b) - \frac{b ^{2}}{2} - a^{2} ln (a) + \frac{a ^{2}}{2}

= b^{2} ln (b) - \frac{b ^{2}}{2} - a^{2} ln (a) + \frac{a ^{2}}{2}

化简

= b^{2} ln (b) + \frac{- b ^{2} + a ^{2}}{2} - a^{2} ln (a)

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