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人気のある微分積分 >

integral from 0 to 7 of 4x\sqrt[3]{x+1}

解

\int_{0}^{7} 4 x 3 x + 1 d x

解

\frac{1209}{7}

+1

十進法表記

172.71428 \dots

解答ステップ

\int_{0}^{7} 4 x 3 x + 1 d x

定数を除く:

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

= 4 \cdot \int_{0}^{7} x 3 x + 1 d x

u置換積分法を適用する

= 4 \cdot \int_{1}^{8} (u - 1) 3 u d u

拡張

(u - 1) 3 u : u^{\frac{4}{3}} - 3 u

= 4 \cdot \int_{1}^{8} u^{\frac{4}{3}} - 3 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

= 4 (\int_{1}^{8} u^{\frac{4}{3}} d u - \int_{1}^{8} 3 u d u)

\int_{1}^{8} u^{\frac{4}{3}} d u = \frac{381}{7}

\int_{1}^{8} 3 u d u = \frac{45}{4}

= 4 (\frac{381}{7} - \frac{45}{4})

簡素化

4 (\frac{381}{7} - \frac{45}{4}) : \frac{1209}{7}

= \frac{1209}{7}

グラフ

人気の例

integral from 0 to 3 of (x^2-5x+9)

\int_{0}^{3} (x^{2} - 5 x + 9) d x

integral from 2 to 4 of x^2+5x+7+1

\int_{2}^{4} x^{2} + 5 x + 7 + 1 d x

integral from 2 to 7 of sqrt(3x-1)

\int_{2}^{7} 3 x - 1 d x

integral from-1 to 2 of (2x^4-5x^2)/(2x)

\int_{- 1}^{2} \frac{2 x ^{4} - 5 x ^{2}}{2 x} d x

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

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