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人気のある微分積分 >

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

解

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

解

\frac{562}{105}

+1

十進法表記

5.35238 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{4} (u - 2)^{2} u d u

拡張

(u - 2)^{2} u : u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 u

= \int_{1}^{4} u^{\frac{5}{2}} - 4 u^{\frac{3}{2}} + 4 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_{1}^{4} u^{\frac{5}{2}} d u - \int_{1}^{4} 4 u^{\frac{3}{2}} d u + \int_{1}^{4} 4 u d u

\int_{1}^{4} u^{\frac{5}{2}} d u = \frac{254}{7}

\int_{1}^{4} 4 u^{\frac{3}{2}} d u = \frac{248}{5}

\int_{1}^{4} 4 u d u = \frac{56}{3}

= \frac{254}{7} - \frac{248}{5} + \frac{56}{3}

簡素化

\frac{254}{7} - \frac{248}{5} + \frac{56}{3} : \frac{562}{105}

= \frac{562}{105}

グラフ

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