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人気のある微分積分 >

integral from 0 to 3 of pisqrt(4x+5)

解

\int_{0}^{3} π 4 x + 5 d x

解

\frac{π ( 17 17 - 5 5 )}{6}

+1

十進法表記

30.84648 \dots

解答ステップ

\int_{0}^{3} π 4 x + 5 d x

定数を除く:

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

= π \cdot \int_{0}^{3} 4 x + 5 d x

u置換積分法を適用する

= π \cdot \int_{5}^{17} \frac{u}{4} d u

定数を除く:

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

= π \frac{1}{4} \cdot \int_{5}^{17} u d u

べき乗則を適用する

= π \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{17}

簡素化

π \frac{1}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{17} : \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{17}

= \frac{π}{4} [\frac{2}{3} u^{\frac{3}{2}}]_{5}^{17}

限度を計算する:

\frac{34 17 - 10 5}{3}

= \frac{π}{4} \cdot \frac{34 17 - 10 5}{3}

簡素化

= \frac{π ( 17 17 - 5 5 )}{6}

グラフ

人気の例

integral from 0 to 1 of x/(sqrt(1+3x)+1)

\int_{0}^{1} \frac{x}{1 + 3 x + 1} d x

integral from 1 to infinity of 2/(5x-1)

\int_{1}^{\infty} \frac{2}{5 x - 1} d x

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

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

integral from 0 to pi of |cos(t)|

\int_{0}^{π} ∣ cos (t) ∣ d t

integral from-5 to 0 of 1/(x^2-16)

\int_{- 5}^{0} \frac{1}{x ^{2} - 16} d x