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人気のある微分積分 >

integral from 1 to 2 of (v^4+3v^5)/(v^3)

解

\int_{1}^{2} \frac{v ^{4} + 3 v ^{5}}{v ^{3}} d v

解

\frac{17}{2}

+1

十進法表記

8.5

解答ステップ

\int_{1}^{2} \frac{v ^{4} + 3 v ^{5}}{v ^{3}} d v

簡素化

\frac{v ^{4} + 3 v ^{5}}{v ^{3}} : v (3 v + 1)

= \int_{1}^{2} v (3 v + 1) d v

拡張

v (3 v + 1) : 3 v^{2} + v

= \int_{1}^{2} 3 v^{2} + v d v

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

= \int_{1}^{2} 3 v^{2} d v + \int_{1}^{2} v d v

\int_{1}^{2} 3 v^{2} d v = 7

\int_{1}^{2} v d v = \frac{3}{2}

= 7 + \frac{3}{2}

簡素化

7 + \frac{3}{2} : \frac{17}{2}

= \frac{17}{2}

グラフ

人気の例

integral from 0 to pi of 1/2 sin^2(x)

\int_{0}^{π} \frac{1}{2} sin^{2} (x) d x

integral from-1 to 3 of 4^x

\int_{- 1}^{3} 4^{x} d x

integral from 1 to 2 of x^4-x

\int_{1}^{2} x^{4} - x d x

integral from 0 to t of μe^{t-μ}

\int_{0}^{t} μ e^{t - μ} d μ

integral from 0 to 1 of 5/(4x+1)

\int_{0}^{1} \frac{5}{4 x + 1} d x