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 sqrt(x)sqrt(2-x)

解答

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

解答

\frac{π}{4}

+1

十进制

0.78539 \dots

求解步骤

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

使用指数法则:

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

x 2 - x = x (2 - x)

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

对

x (2 - x)

配方:

- (x - 1)^{2} + 1

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

使用换元积分法

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

使用三角换元法

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

使用三角恒等式改写

= \int_{- \frac{π}{2}}^{0} \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

= \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{0} 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

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

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

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

= \frac{1}{2} (\frac{π}{2} + 0)

化简

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

= \frac{π}{4}

作图

流行的例子

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

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

integral from 3 to 4 of 1/(25-x^2)

\int_{3}^{4} \frac{1}{25 - x ^{2}} d x

integral from 1 to 4 of-2x+8

\int_{1}^{4} - 2 x + 8 d x

integral from 3 to 8 of 1/(sqrt(1+x)-1)

\int_{3}^{8} \frac{1}{1 + x - 1} d x

integral from 2 to 5 of (x^2-2x+3)

\int_{2}^{5} (x^{2} - 2 x + 3) d x