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人気のある微分積分 >

integral from 0 to 3 of xsqrt(1+4x^2)

解

\int_{0}^{3} x 1 + 4 x^{2} d x

解

\frac{37 37 - 1}{12}

+1

十進法表記

18.67185 \dots

解答ステップ

\int_{0}^{3} x 1 + 4 x^{2} d x

u置換積分法を適用する

= \int_{1}^{37} \frac{u}{8} d u

定数を除く:

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

= \frac{1}{8} \cdot \int_{1}^{37} u d u

べき乗則を適用する

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{37}

限度を計算する:

\frac{74 37}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{74 37}{3} - \frac{2}{3})

簡素化

= \frac{37 37 - 1}{12}

グラフ

人気の例

integral from 0 to 2 of 3x^2-2x+1

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

integral from 0 to 6 of (x^2+1)e^{3x}

\int_{0}^{6} (x^{2} + 1) e^{3 x} d x

integral from 3 to x of 9x

\int_{3}^{x} 9 x d x

integral from 0 to 2 of (2x^2)^2

\int_{0}^{2} (2 x^{2})^{2} d x

integral from 1 to 5 of (x-arctan(x))

\int_{1}^{5} (x - arctan (x)) d x