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受欢迎的微积分 >

integral from 0 to 2 of rsqrt(16-r^2)

解答

\int_{0}^{2} r 16 - r^{2} d r

解答

\frac{8 ( 8 - 3 3 )}{3}

+1

十进制

7.47692 \dots

求解步骤

\int_{0}^{2} r 16 - r^{2} d r

使用换元积分法

= \int_{16}^{12} - \frac{u}{2} d u

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

= - \int_{12}^{16} - \frac{u}{2} d u

提出常数:

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

= - (- \frac{1}{2} \cdot \int_{12}^{16} u d u)

使用幂法则

= - (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16})

化简

= \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{12}^{16}

带入上下限计算后得:

\frac{128}{3} - 163

= \frac{1}{2} (\frac{128}{3} - 163)

化简

= \frac{8 ( 8 - 3 3 )}{3}

作图

流行的例子

integral from 1 to 4 of (3sqrt(x))

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

integral from 4 to 0 of (2x^5+3x^4-2x^3)

\int_{4}^{0} (2 x^{5} + 3 x^{4} - 2 x^{3}) d x

integral from-1 to 1 of (8x^3-6x+1)

\int_{- 1}^{1} (8 x^{3} - 6 x + 1) d x

integral from 1 to 3 of 1/(x^2-16)

\int_{1}^{3} \frac{1}{x ^{2} - 16} d x

integral from 0 to 3 of cos(x^2)

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