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 4 of xsqrt(4x-x^2)

解

\int_{0}^{4} x 4 x - x^{2} d x

解

4 π

+1

十進法表記

12.56637 \dots

解答ステップ

\int_{0}^{4} x 4 x - x^{2} d x

平方完成する

4 x - x^{2} : - (x - 2)^{2} + 4

= \int_{0}^{4} x - (x - 2)^{2} + 4 d x

u置換積分法を適用する

= \int_{- 2}^{2} (u + 2) - u^{2} + 4 d u

拡張

(u + 2) - u^{2} + 4 : u - u^{2} + 4 + 2 - u^{2} + 4

= \int_{- 2}^{2} u - u^{2} + 4 + 2 - u^{2} + 4 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_{- 2}^{2} u - u^{2} + 4 d u + \int_{- 2}^{2} 2 - u^{2} + 4 d u

\int_{- 2}^{2} u - u^{2} + 4 d u = 0

\int_{- 2}^{2} 2 - u^{2} + 4 d u = 4 π

= 0 + 4 π

簡素化

= 4 π

グラフ

人気の例

integral from-pi to 0 of 1/pi

\int_{- π}^{0} \frac{1}{π}

integral from-2 to 2 of 5t^4

\int_{- 2}^{2} 5 t^{4} d t

integral from 0 to 1 of (1-x^2)cos(npix)

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

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

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

integral from 0 to pi of (3+cos(x))

\int_{0}^{π} (3 + cos (x)) d x