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 1 of (3x^2-2x)^2

解

\int_{0}^{1} (3 x^{2} - 2 x)^{2} d x

解

\frac{2}{15}

+1

十進法表記

0.13333 \dots

解答ステップ

\int_{0}^{1} (3 x^{2} - 2 x)^{2} d x

平方完成する

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

= \int_{0}^{1} (3 (x - \frac{1}{3})^{2} - \frac{1}{3})^{2} d x

u置換積分法を適用する

= \int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} - 2 u^{2} + \frac{1}{9} 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

= \int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} d u - \int_{- \frac{1}{3}}^{\frac{2}{3}} 2 u^{2} d u + \int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{9} d u

\int_{- \frac{1}{3}}^{\frac{2}{3}} 9 u^{4} d u = \frac{11}{45}

\int_{- \frac{1}{3}}^{\frac{2}{3}} 2 u^{2} d u = \frac{2}{9}

\int_{- \frac{1}{3}}^{\frac{2}{3}} \frac{1}{9} d u = \frac{1}{9}

= \frac{11}{45} - \frac{2}{9} + \frac{1}{9}

簡素化

\frac{11}{45} - \frac{2}{9} + \frac{1}{9} : \frac{2}{15}

= \frac{2}{15}

グラフ

人気の例

integral from 0 to t of 1/(1-x)

\int_{0}^{t} \frac{1}{1 - x} d x

integral from 0 to 2 of 1/(x^2+4x+8)

\int_{0}^{2} \frac{1}{x ^{2} + 4 x + 8} d x

integral from 0 to 1 of (6x^2)/(1-x^3)

\int_{0}^{1} \frac{6 x ^{2}}{1 - x ^{3}} d x

integral from 0 to 1/2 of 1/(1+4y^2)

\int_{0}^{\frac{1}{2}} \frac{1}{1 + 4 y ^{2}} d y

integral from 0 to pi/3 of 2cos(x)-1

\int_{0}^{\frac{π}{3}} 2 cos (x) - 1 d x