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人気のある微分積分 >

integral from 0 to 3 of x^5sqrt(2x^3+1)

解

\int_{0}^{3} x^{5} 2 x^{3} + 1 d x

解

\frac{4400 55 + 1}{45}

+1

十進法表記

725.16162 \dots

解答ステップ

\int_{0}^{3} x^{5} 2 x^{3} + 1 d x

u置換積分法を適用する

= \int_{1}^{55} \frac{u ( u - 1 )}{12} d u

定数を除く:

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

= \frac{1}{12} \cdot \int_{1}^{55} u (u - 1) d u

拡張

u (u - 1) : u^{\frac{3}{2}} - u

= \frac{1}{12} \cdot \int_{1}^{55} u^{\frac{3}{2}} - u 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}{12} (\int_{1}^{55} u^{\frac{3}{2}} d u - \int_{1}^{55} u d u)

\int_{1}^{55} u^{\frac{3}{2}} d u = 121055 - \frac{2}{5}

\int_{1}^{55} u d u = \frac{110 55 - 2}{3}

= \frac{1}{12} (121055 - \frac{2}{5} - \frac{110 55 - 2}{3})

簡素化

\frac{1}{12} (121055 - \frac{2}{5} - \frac{110 55 - 2}{3}) : \frac{4400 55 + 1}{45}

= \frac{4400 55 + 1}{45}

グラフ

人気の例

derivative of 5^{2x+1}

\frac{d}{d x} (5^{2 x + 1})

y^{''}-2y^'+y=((e^t))/(1+t^2)

y^{^{''}} - 2 y^{^{'}} + y = \frac{( e ^{t} )}{1 + t ^{2}}

limit as x approaches 0-of 12e^{-2/x}

x \to 0 - lim (12 e^{- \frac{2}{x}})

tangent f(x)=(x^2-1)/(x^2+x+1)

t an g e n t f (x) = \frac{x ^{2} - 1}{x ^{2} + x + 1}

limit as x approaches-2 of x^2-x+4

x \to - 2 lim (x^{2} - x + 4)