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人気のある微分積分 >

integral from-1 to 0 of x^2sqrt(x+1)

解

\int_{- 1}^{0} x^{2} x + 1 d x

解

\frac{16}{105}

+1

十進法表記

0.15238 \dots

解答ステップ

\int_{- 1}^{0} x^{2} x + 1 d x

u置換積分法を適用する

= \int_{0}^{1} (u - 1)^{2} u d u

拡張

(u - 1)^{2} u : u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u

= \int_{0}^{1} u^{\frac{5}{2}} - 2 u^{\frac{3}{2}} + u d u

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} u^{\frac{5}{2}} d u - \int_{0}^{1} 2 u^{\frac{3}{2}} d u + \int_{0}^{1} u d u

\int_{0}^{1} u^{\frac{5}{2}} d u = \frac{2}{7}

\int_{0}^{1} 2 u^{\frac{3}{2}} d u = \frac{4}{5}

\int_{0}^{1} u d u = \frac{2}{3}

= \frac{2}{7} - \frac{4}{5} + \frac{2}{3}

簡素化

\frac{2}{7} - \frac{4}{5} + \frac{2}{3} : \frac{16}{105}

= \frac{16}{105}

グラフ

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