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人気のある微分積分 >

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

解

\int_{0}^{2} t^{3} 3 t^{2} + 1 d t

解

\frac{2 ( 221 13 + 1 )}{135}

+1

十進法表記

11.81965 \dots

解答ステップ

\int_{0}^{2} t^{3} 3 t^{2} + 1 d t

u置換積分法を適用する

= \int_{1}^{13} \frac{u ( u - 1 )}{18} d u

定数を除く:

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

= \frac{1}{18} \cdot \int_{1}^{13} u (u - 1) d u

拡張

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

= \frac{1}{18} \cdot \int_{1}^{13} 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}{18} (\int_{1}^{13} u^{\frac{3}{2}} d u - \int_{1}^{13} u d u)

\int_{1}^{13} u^{\frac{3}{2}} d u = \frac{338 13 - 2}{5}

\int_{1}^{13} u d u = \frac{26 13 - 2}{3}

= \frac{1}{18} (\frac{338 13 - 2}{5} - \frac{26 13 - 2}{3})

簡素化

\frac{1}{18} (\frac{338 13 - 2}{5} - \frac{26 13 - 2}{3}) : \frac{2 ( 221 13 + 1 )}{135}

= \frac{2 ( 221 13 + 1 )}{135}

グラフ

人気の例

integral from 0 to 1 of 16x^3

\int_{0}^{1} 16 x^{3} d x

integral from 6 to infinity of 2x^{-3}

\int_{6}^{\infty} 2 x^{- 3} d x

integral from 0 to 5 of 0.5e^{-0.5x}

\int_{0}^{5} 0.5 e^{- 0.5 x} d x

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

integral from 0 to 5 of 7/3 x

\int_{0}^{5} \frac{7}{3} x d x