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人気のある微分積分 >

integral from 0 to 1 of 2ksqrt(x)

解

\int_{0}^{1} 2 k x d x

解

\frac{4 k}{3}

解答ステップ

\int_{0}^{1} 2 k x d x

定数を除く:

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

= 2 k \cdot \int_{0}^{1} x d x

べき乗則を適用する

= 2 k [\frac{2}{3} x^{\frac{3}{2}}]_{0}^{1}

限度を計算する:

\frac{2}{3}

= 2 k \frac{2}{3}

簡素化

= \frac{4 k}{3}

人気の例

integral from 0 to 2 of (4t-1)^{10}

\int_{0}^{2} (4 t - 1)^{10} d t

integral from 3 to 4 of (x-3)(4-x)

\int_{3}^{4} (x - 3) (4 - x) d x

integral from 0 to pi of sqrt(2-2cos(t))

\int_{0}^{π} 2 - 2 cos (t) d t

integral from 0 to 2 of x/(1+10x^2)

\int_{0}^{2} \frac{x}{1 + 10 x ^{2}} d x

integral from 1 to 8 of (1/x)

\int_{1}^{8} (\frac{1}{x}) d x