integral from 0 to 1 of x(a+bx^2)

解

\int_{0}^{1} x (a + b x^{2}) d x

\frac{1}{4} (2 a + b)

解答ステップ

\int_{0}^{1} x (a + b x^{2}) d x

部分積分を適用する

= [\frac{1}{2} x^{2} (a + b x^{2}) - \int b x^{3} d x]_{0}^{1}

\int b x^{3} d x = \frac{b x ^{4}}{4}

= [\frac{1}{2} x^{2} (a + b x^{2}) - \frac{b x ^{4}}{4}]_{0}^{1}

簡素化

[\frac{1}{2} x^{2} (a + b x^{2}) - \frac{b x ^{4}}{4}]_{0}^{1} : [\frac{1}{4} (2 a x^{2} + b x^{4})]_{0}^{1}

= [\frac{1}{4} (2 a x^{2} + b x^{4})]_{0}^{1}

限度を計算する:

\frac{1}{4} (2 a + b)

= \frac{1}{4} (2 a + b)