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 1 to 2 of sqrt(4-x^2)

解答

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

解答

2 (\frac{π}{3} - \frac{3}{4})

+1

十进制

1.22836 \dots

求解步骤

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

使用三角换元法

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

提出常数:

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

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

使用三角恒等式改写

= 4 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} \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

= 4 \cdot \frac{1}{2} \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} 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

= 4 \cdot \frac{1}{2} (\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u + \int_{\frac{π}{6}}^{\frac{π}{2}} cos (2 u) d u)

\int_{\frac{π}{6}}^{\frac{π}{2}} 1 d u = \frac{π}{3}

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

= 4 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

化简

4 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 2 (\frac{π}{3} - \frac{3}{4})

= 2 (\frac{π}{3} - \frac{3}{4})

作图

流行的例子

derivative of arctan(sin(3x))

\frac{d}{d x} (arctan (sin (3 x)))

d/(dt)(((8t)^2)/(10))

\frac{d}{d t} (\frac{( 8 t ) ^{2}}{10})

sum from n=1 to infinity of x/(2^n)

n = 1 \sum \infty \frac{x}{2 ^{n}}

(dy)/(dx)= x/(ye^{y+x^2)}

\frac{d y}{d x} = \frac{x}{y e ^{y + x^{2}}}

(\partial)/(\partial x)(sin(7x^5y-7xy^4))

\frac{\partial}{\partial x} (sin (7 x^{5} y - 7 x y^{4}))