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 1 of x/(sqrt(x(1-x)))

解答

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

解答

\frac{π}{2}

+1

十进制

1.57079 \dots

求解步骤

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

使用指数法则:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

x (1 - x) = x (1 - x)

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

化简

\frac{x}{x 1 - x} : \frac{x}{1 - x}

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

化简

\frac{x}{1 - x} : \frac{x}{1 - x}

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

使用换元积分法

= \int_{1}^{0} - 2 - u^{2} + 1 d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{0}^{1} - 2 - u^{2} + 1 d u

提出常数:

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

= - (- 2 \cdot \int_{0}^{1} - u^{2} + 1 d u)

使用三角换元法

= - (- 2 \cdot \int_{0}^{\frac{π}{2}} cos^{2} (v) d v)

使用三角恒等式改写

= - (- 2 \cdot \int_{0}^{\frac{π}{2}} \frac{1 + cos ( 2 v )}{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_{0}^{\frac{π}{2}} 1 + cos (2 v) d v)

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

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

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

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

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

化简

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

= \frac{π}{2}

作图

流行的例子

integral from 1 to 3 of (x^2+x+4)

\int_{1}^{3} (x^{2} + x + 4) d x

integral from 1 to 9 of 3x^2+xsqrt(x)+1

\int_{1}^{9} 3 x^{2} + x x + 1 d x

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

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

integral from 6 to 7 of 1/((x-6)^{5/4)}

\int_{6}^{7} \frac{1}{( x - 6 ) ^{\frac{5}{4}}} d x

integral from 0 to pi/2 of xcos(9x)

\int_{0}^{\frac{π}{2}} x cos (9 x) d x