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 4 to 8 of sqrt(64-x^2)

解答

\int_{4}^{8} 64 - x^{2} d x

解答

32 (\frac{π}{3} - \frac{3}{4})

+1

十进制

19.65391 \dots

求解步骤

\int_{4}^{8} 64 - x^{2} d x

使用三角换元法

= \int_{\frac{π}{6}}^{\frac{π}{2}} 64 cos^{2} (u) d u

提出常数:

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

= 64 \cdot \int_{\frac{π}{6}}^{\frac{π}{2}} cos^{2} (u) d u

使用三角恒等式改写

= 64 \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

= 64 \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

= 64 \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}

= 64 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4})

化简

64 \cdot \frac{1}{2} (\frac{π}{3} - \frac{3}{4}) : 32 (\frac{π}{3} - \frac{3}{4})

= 32 (\frac{π}{3} - \frac{3}{4})

作图

流行的例子

integral from 0 to 2pi of cos^2(nx)

\int_{0}^{2 π} cos^{2} (n x) d x

integral from-2 to 2 of (x+4)sqrt(4-x^2)

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

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

\int_{1}^{e} \frac{x ^{3}}{1 + x ^{2}} d x

integral from 0 to 2 of x^3-8

\int_{0}^{2} x^{3} - 8 d x

integral from 0 to pi of sin(x)sin(n)x

\int_{0}^{π} sin (x) sin (n) x d x