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 5 of (25-x^2)^{1/2}

解

\int_{0}^{5} (25 - x^{2})^{\frac{1}{2}} d x

解

\frac{25 π}{4}

+1

十進法表記

19.63495 \dots

解答ステップ

\int_{0}^{5} (25 - x^{2})^{\frac{1}{2}} d x

三角関数による置換を適用する

= \int_{0}^{\frac{π}{2}} 25 cos^{2} (u) d u

定数を除く:

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

= 25 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (u) d u

三角関数の公式を使用して書き換える

= 25 \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

定数を除く:

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

= 25 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{2}} 1 + cos (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

= 25 \cdot \frac{1}{2} (\int_{0}^{\frac{π}{2}} 1 d u + \int_{0}^{\frac{π}{2}} cos (2 u) d u)

\int_{0}^{\frac{π}{2}} 1 d u = \frac{π}{2}

\int_{0}^{\frac{π}{2}} cos (2 u) d u = 0

= 25 \cdot \frac{1}{2} (\frac{π}{2} + 0)

簡素化

25 \cdot \frac{1}{2} (\frac{π}{2} + 0) : \frac{25 π}{4}

= \frac{25 π}{4}

グラフ

人気の例

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

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

integral from 0 to 4 of (2+x)

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

integral from 1 to 2 of sqrt(x^2)

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

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

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

integral from 1 to 3 of-2x

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