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人気のある微分積分 >

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

解

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

解

- \frac{1}{3} + \frac{π}{4}

+1

十進法表記

0.45206 \dots

解答ステップ

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

平方完成する

2 x - x^{2} : - (x - 1)^{2} + 1

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

u置換積分法を適用する

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

拡張

(u + 1) - u^{2} + 1 : u - u^{2} + 1 + - u^{2} + 1

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

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

\int_{- 1}^{0} - u^{2} + 1 d u = \frac{π}{4}

= - \frac{1}{3} + \frac{π}{4}

グラフ

人気の例

integral from-5 to 5 of (-0.4x^2+10)

\int_{- 5}^{5} (- 0.4 x^{2} + 10) d x

integral from 2x to 2 of e^{y^2}

\int_{2 x}^{2} e^{y^{2}} d y

integral from 0 to 3 of sqrt(1+64x^2)

\int_{0}^{3} 1 + 64 x^{2} d x

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

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

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

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