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

解答

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

解答

π

+1

十进制

3.14159 \dots

求解步骤

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

对

3 x - x^{2}

配方:

- (x - \frac{3}{2})^{2} + \frac{9}{4}

= \int_{0}^{3} \frac{1}{- ( x - \frac{3}{2} ) ^{2} + \frac{9}{4}} d x

使用换元积分法

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

提出常数:

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

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

使用三角换元法

= 2 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1}{2} d v

提出常数:

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

= 2 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d v

常数积分:

\int a d x = a x

= 2 \cdot \frac{1}{2} [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}}

化简

2 \cdot \frac{1}{2} [1 \cdot v]_{- \frac{π}{2}}^{\frac{π}{2}} : [v]_{- \frac{π}{2}}^{\frac{π}{2}}

= [v]_{- \frac{π}{2}}^{\frac{π}{2}}

带入上下限计算后得:

π

= π

作图

流行的例子

integral from 0 to x^3 of e^{y/x}

\int_{0}^{x^{3}} e^{\frac{y}{x}} d y

integral from 0 to 1 of x^2sqrt(9+4x^2)

\int_{0}^{1} x^{2} 9 + 4 x^{2} d x

integral from 0 to 2 of 1/(e^x+e^{-x)}

\int_{0}^{2} \frac{1}{e ^{x} + e ^{- x}} d x

integral from-4 to 4 of (-1/5 x^3+5x)

\int_{- 4}^{4} (- \frac{1}{5} x^{3} + 5 x) d x

integral from 1 to 10 of (1/x)

\int_{1}^{10} (\frac{1}{x}) d x