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人気のある微分積分 >

integral from 0 to 1 of 2t*sqrt(1+4t^2)

解

\int_{0}^{1} 2 t \cdot 1 + 4 t^{2} d t

解

\frac{5 5 - 1}{6}

+1

十進法表記

1.69672 \dots

解答ステップ

\int_{0}^{1} 2 t 1 + 4 t^{2} d t

定数を除く:

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

= 2 \cdot \int_{0}^{1} t 1 + 4 t^{2} d t

u置換積分法を適用する

= 2 \cdot \int_{1}^{5} \frac{u}{8} d u

定数を除く:

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

= 2 \cdot \frac{1}{8} \cdot \int_{1}^{5} u d u

べき乗則を適用する

= 2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

簡素化

2 \cdot \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5} : \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

= \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{5}

限度を計算する:

\frac{10 5}{3} - \frac{2}{3}

= \frac{1}{4} (\frac{10 5}{3} - \frac{2}{3})

簡素化

= \frac{5 5 - 1}{6}

グラフ

人気の例

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

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\int_{0}^{1} (x^{4})^{2} - (x^{7})^{2} d x

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

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integral from 2 to 1 of (ln(4x))^2

\int_{2}^{1} (ln (4 x))^{2} d x

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\int_{0}^{π} 5 sin (t) d t