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人気のある微分積分 >

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

解

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

解

\frac{17 17 - 1}{12}

+1

十進法表記

5.75773 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{17} \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}^{17} u d u

べき乗則を適用する

= \frac{1}{8} [\frac{2}{3} u^{\frac{3}{2}}]_{1}^{17}

限度を計算する:

\frac{34 17}{3} - \frac{2}{3}

= \frac{1}{8} (\frac{34 17}{3} - \frac{2}{3})

簡素化

= \frac{17 17 - 1}{12}

グラフ

人気の例

integral from 2 to infinity of 5/(4+x^2)

\int_{2}^{\infty} \frac{5}{4 + x ^{2}} d x

integral from 0 to 2 of (x^3)^2

\int_{0}^{2} (x^{3})^{2} d x

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

\int_{- 1}^{2} \frac{x ^{2}}{x + 2} d x

integral from 0 to 3 of 6x-x^2

\int_{0}^{3} 6 x - x^{2} d x

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

\int_{1}^{2} x^{2} + \frac{1}{x ^{4}} d x