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人気のある微分積分 >

integral from 1 to 2 of x(sqrt(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 3 to 4 of-xln(x-3)

\int_{3}^{4} - x ln (x - 3) d x

integral from 0 to pi/2 of 1-sin^2(x)

\int_{0}^{\frac{π}{2}} 1 - sin^{2} (x) d x

integral from 0 to 2 of sqrt(3x)

\int_{0}^{2} 3 x d x

integral from 1 to 5 of (3x)/(x^3)

\int_{1}^{5} \frac{3 x}{x ^{3}} d x

integral from 0 to 1 of x^2-x^5

\int_{0}^{1} x^{2} - x^{5} d x