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人気のある微分積分 >

integral from 0 to 1 of x^3*sqrt(1-x^2)

解

\int_{0}^{1} x^{3} \cdot 1 - x^{2} d x

解

\frac{2}{15}

+1

十進法表記

0.13333 \dots

解答ステップ

\int_{0}^{1} x^{3} 1 - x^{2} d x

u置換積分法を適用する

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

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

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

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

拡張

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

= - (- \frac{1}{2} \cdot \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

= - (- \frac{1}{2} (- \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{1}{2} (- \frac{2}{5} + \frac{2}{3}))

簡素化

- (- \frac{1}{2} (- \frac{2}{5} + \frac{2}{3})) : \frac{2}{15}

= \frac{2}{15}

グラフ

人気の例

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

\int_{1}^{5} (x^{3} + 1) d x

integral from 0 to 1 of x^a

\int_{0}^{1} x^{a} d x

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

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

integral from pi/3 to (2pi)/3 of sin(x)

\int_{\frac{π}{3}}^{\frac{2 π}{3}} sin (x) d x

integral from 0 to 4 of xsin(3x)

\int_{0}^{4} x sin (3 x) d x