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人気のある微分積分 >

integral from 0 to 2 of x/(sqrt(4x+1))

解

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

解

\frac{5}{6}

+1

十進法表記

0.83333 \dots

解答ステップ

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

u置換積分法を適用する

= \int_{1}^{3} \frac{u ^{2} - 1}{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}^{3} u^{2} - 1 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}{8} (\int_{1}^{3} u^{2} d u - \int_{1}^{3} 1 d u)

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

\int_{1}^{3} 1 d u = 2

= \frac{1}{8} (\frac{26}{3} - 2)

簡素化

\frac{1}{8} (\frac{26}{3} - 2) : \frac{5}{6}

= \frac{5}{6}

グラフ

人気の例

integral from 0 to 2 of x^3e^{4x}

\int_{0}^{2} x^{3} e^{4 x} d x

integral from 0 to pi of t^2cos(3t)

\int_{0}^{π} t^{2} cos (3 t) d t

integral from 1 to e of (12+ln(x))^2

\int_{1}^{e} (12 + ln (x))^{2} d x

integral from 0 to pi of (x-sin^2(x))

\int_{0}^{π} (x - sin^{2} (x)) d x

integral from-8 to 8 of 1/(x^{1/3)}

\int_{- 8}^{8} \frac{1}{x ^{\frac{1}{3}}} d x