zs

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

受欢迎的微积分 >

integral from 0 to 2.5 of sqrt(25-x^2)

解答

\int_{0}^{2.5} 25 - x^{2} d x

解答

\frac{25 ( 2 π + 3 3 )}{24}

+1

十进制

11.95764 \dots

求解步骤

\int_{0}^{2.5} 25 - x^{2} d x

使用三角换元法

= \int_{0}^{\frac{π}{6}} 25 cos^{2} (u) d u

提出常数:

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

= 25 \cdot \int_{0}^{\frac{π}{6}} cos^{2} (u) d u

使用三角恒等式改写

= 25 \cdot \int_{0}^{\frac{π}{6}} \frac{1 + cos ( 2 u )}{2} d u

提出常数:

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

= 25 \cdot \frac{1}{2} \cdot \int_{0}^{\frac{π}{6}} 1 + cos (2 u) d u

使用积分加法定则:

\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}{6}} 1 d u + \int_{0}^{\frac{π}{6}} cos (2 u) d u)

\int_{0}^{\frac{π}{6}} 1 d u = \frac{π}{6}

\int_{0}^{\frac{π}{6}} cos (2 u) d u = \frac{3}{4}

= 25 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4})

化简

25 \cdot \frac{1}{2} (\frac{π}{6} + \frac{3}{4}) : \frac{25 ( 2 π + 3 3 )}{24}

= \frac{25 ( 2 π + 3 3 )}{24}

作图

流行的例子

integral of 1/(sqrt(x^2+14))

\int \frac{1}{x ^{2} + 14} d x

化简 sin(x+y)

s im pl i f y sin (x + y)

integral of sin(bt)

\int sin (b t) d t

derivative of x/((2x+3^3))

\frac{d}{d x} (\frac{x}{( 2 x + 3 ) ^{3}})

integral of x^2sqrt(8+x^3)

\int x^{2} 8 + x^{3} d x