ja

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

人気のある微分積分 >

integral from 1 to 3 of sqrt(1+(x^3-1/(4x^3))^2)

解

\int_{1}^{3} 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

解

\frac{181}{9}

+1

十進法表記

20.11111 \dots

解答ステップ

\int_{1}^{3} 1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} d x

簡素化

1 + (x^{3} - \frac{1}{4 x ^{3}})^{2} : \frac{8 x ^{6} + 1 + 16 x ^{12}}{4 x ^{3}}

= \int_{1}^{3} \frac{8 x ^{6} + 1 + 16 x ^{12}}{4 x ^{3}} d x

定数を除く:

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

= \frac{1}{4} \cdot \int_{1}^{3} \frac{8 x ^{6} + 1 + 16 x ^{12}}{x ^{3}} d x

u置換積分法を適用する

= \frac{1}{4} \cdot \int_{1}^{729} \frac{4 u + 1}{6 u ^{\frac{4}{3}}} d u

定数を除く:

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

= \frac{1}{4} \cdot \frac{1}{6} \cdot \int_{1}^{729} \frac{4 u + 1}{u ^{\frac{4}{3}}} d u

拡張

\frac{4 u + 1}{u ^{\frac{4}{3}}} : \frac{4}{u ^{\frac{1}{3}}} + \frac{1}{u ^{\frac{4}{3}}}

= \frac{1}{4} \cdot \frac{1}{6} \cdot \int_{1}^{729} \frac{4}{u ^{\frac{1}{3}}} + \frac{1}{u ^{\frac{4}{3}}} 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}{4} \cdot \frac{1}{6} (\int_{1}^{729} \frac{4}{u ^{\frac{1}{3}}} d u + \int_{1}^{729} \frac{1}{u ^{\frac{4}{3}}} d u)

\int_{1}^{729} \frac{4}{u ^{\frac{1}{3}}} d u = 480

\int_{1}^{729} \frac{1}{u ^{\frac{4}{3}}} d u = \frac{8}{3}

= \frac{1}{4} \cdot \frac{1}{6} (480 + \frac{8}{3})

簡素化

\frac{1}{4} \cdot \frac{1}{6} (480 + \frac{8}{3}) : \frac{181}{9}

= \frac{181}{9}

グラフ

人気の例

(\partial)/(\partial v)(sqrt(uv))

\frac{\partial}{\partial v} (uv)

(\partial)/(\partial z)(2x^4yz^3)

\frac{\partial}{\partial z} (2 x^{4} y z^{3})

derivative of 3/(4x^8)

\frac{d}{d x} (\frac{3}{4 x ^{8}})

xy^'+y=6xy^2,y(1)=-4

x y^{^{'}} + y = 6 x y^{2}, y (1) = - 4

derivative 1-2sin^3(x)

d er i v a t i v e 1 - 2 sin^{3} (x)