integral from 0 to 1 of x^3(2+x^4)

解

\int_{0}^{1} x^{3} (2 + x^{4}) d x

\frac{5}{8}

十進法表記

0.625

解答ステップ

\int_{0}^{1} x^{3} (2 + x^{4}) d x

部分積分を適用する

= [\frac{1}{4} x^{4} (2 + x^{4}) - \int x^{7} d x]_{0}^{1}

\int x^{7} d x = \frac{x ^{8}}{8}

= [\frac{1}{4} x^{4} (2 + x^{4}) - \frac{x ^{8}}{8}]_{0}^{1}

簡素化

[\frac{1}{4} x^{4} (2 + x^{4}) - \frac{x ^{8}}{8}]_{0}^{1} : [\frac{1}{8} (4 x^{4} + x^{8})]_{0}^{1}

= [\frac{1}{8} (4 x^{4} + x^{8})]_{0}^{1}

限度を計算する:

\frac{5}{8}

= \frac{5}{8}