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 0 to 7 of sin(8x^2)

解

\int_{0}^{7} sin (8 x^{2}) d x

解

\frac{1}{2 2} (\frac{π}{2} § (\frac{2}{π} \cdot 142) - \frac{π}{2} § (\frac{2}{π} \cdot 0))

解答ステップ

\int_{0}^{7} sin (8 x^{2}) d x

置換積分法を適用する

= \int_{0}^{14} \frac{1}{2} sin (2 u^{2}) d u

定数を除く:

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

= \frac{1}{2} \cdot \int_{0}^{14} sin (2 u^{2}) d u

置換積分法を適用する

= \frac{1}{2} \cdot \int_{0}^{28} \frac{1}{2} sin (\frac{v ^{2}}{2}) d v

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{28} sin (\frac{v ^{2}}{2}) d v

指数の規則を適用する:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{v ^{2}}{2} = (\frac{v}{2})^{2}

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{28} sin ((\frac{v}{2})^{2}) d v

u置換積分法を適用する

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int_{0}^{142} sin (w^{2}) 2 d w

定数を除く:

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

= \frac{1}{2} \cdot \frac{1}{2} 2 \cdot \int_{0}^{142} sin (w^{2}) d w

非初等積分を使用する:

\int sin (x^{2}) d x = \frac{π}{2} § (\frac{2}{π} x)

= \frac{1}{2} \cdot \frac{1}{2} 2 [\frac{π}{2} § (\frac{2}{π} w)]_{0}^{142}

簡素化

\frac{1}{2} \cdot \frac{1}{2} 2 [\frac{π}{2} § (\frac{2}{π} w)]_{0}^{142} : \frac{1}{2 2} [\frac{π}{2} § (\frac{2}{π} w)]_{0}^{142}

= \frac{1}{2 2} [\frac{π}{2} § (\frac{2}{π} w)]_{0}^{142}

限度を計算する:

\frac{π}{2} § (\frac{2}{π} \cdot 142) - \frac{π}{2} § (\frac{2}{π} \cdot 0)

= \frac{1}{2 2} (\frac{π}{2} § (\frac{2}{π} \cdot 142) - \frac{π}{2} § (\frac{2}{π} \cdot 0))

人気の例

integral from 1 to 5 of x^2+x+2

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

integral from 1 to 4 of 4/(1-3x)

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

integral from 0 to 1 of 2te^{t^2-1}+3t^7

\int_{0}^{1} 2 t e^{t^{2} - 1} + 3 t^{7} d t

integral from-1 to 3 of x^2+2x+1

\int_{- 1}^{3} x^{2} + 2 x + 1 d x

integral from 1 to 4 of (u-8)/(sqrt(u))

\int_{1}^{4} \frac{u - 8}{u} d u