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人気のある微分積分 >

integral from 1 to 2 of xsqrt((x-1))

解

\int_{1}^{2} x (x - 1) d x

解

\frac{16}{15}

+1

十進法表記

1.06666 \dots

解答ステップ

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

u置換積分法を適用する

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

拡張

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

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

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

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

= \frac{2}{5} + \frac{2}{3}

簡素化

\frac{2}{5} + \frac{2}{3} : \frac{16}{15}

= \frac{16}{15}

グラフ

人気の例

integral from-1 to 3 of 2x^3

\int_{- 1}^{3} 2 x^{3} d x

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

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

integral from 0 to 0.3 of 3e^{-x^2}

\int_{0}^{0.3} 3 e^{- x^{2}} d x

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

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

integral from 1 to 2 of x/((x^2+1)^2)

\int_{1}^{2} \frac{x}{( x ^{2} + 1 ) ^{2}} d x