integral from a to b of (ax+b)^p

解答

\int_{a}^{b} (a x + b)^{p} d x

\frac{( b + ab ) ^{p + 1} - ( b + a ^{2} ) ^{p + 1}}{a ( p + 1 )}

求解步骤

\int_{a}^{b} (a x + b)^{p} d x

使用换元积分法

= \int_{b + a^{2}}^{b + ab} \frac{u ^{p}}{a} d u

提出常数:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{a} \cdot \int_{b + a^{2}}^{b + ab} u^{p} d u

使用幂法则

= \frac{1}{a} [\frac{u ^{p + 1}}{p + 1}]_{b + a^{2}}^{b + ab}

化简

\frac{1}{a} [\frac{u ^{p + 1}}{p + 1}]_{b + a^{2}}^{b + ab} : \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{b + a^{2}}^{b + ab}}{a}

= \frac{[ \frac{u ^{p + 1}}{p + 1} ] _{b + a^{2}}^{b + ab}}{a}

带入上下限计算后得:

\frac{( b + ab ) ^{p + 1}}{p + 1} - \frac{( b + a ^{2} ) ^{p + 1}}{p + 1}

= \frac{\frac{( b + ab ) ^{p + 1}}{p + 1} - \frac{( b + a ^{2} ) ^{p + 1}}{p + 1}}{a}

化简

= \frac{( b + ab ) ^{p + 1} - ( b + a ^{2} ) ^{p + 1}}{a ( p + 1 )}

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