integral from 0 to 3 of 2xsqrt(9-x^2)

解答

\int_{0}^{3} 2 x 9 - x^{2} d x

18

求解步骤

\int_{0}^{3} 2 x 9 - x^{2} d x

提出常数:

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

= 2 \cdot \int_{0}^{3} x 9 - x^{2} d x

使用换元积分法

= 2 \cdot \int_{9}^{0} - \frac{u}{2} d u

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

= 2 (- \int_{0}^{9} - \frac{u}{2} d u)

提出常数:

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

= 2 (- (- \frac{1}{2} \cdot \int_{0}^{9} u d u))

使用幂法则

= 2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}))

化简

2 (- (- \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9})) : [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

= [\frac{2}{3} u^{\frac{3}{2}}]_{0}^{9}

带入上下限计算后得:

18

= 18

\int_{0}^{\frac{2 π}{3}} sin (x) d x

\int_{0}^{3} 3^{x} + 9 - x^{2} d x

\int_{2 + c o s (x)}^{x^{2} + 1} \frac{1}{x} d x

\int_{0}^{4} x x^{2} + 3 d x

\int_{0}^{1} (x^{2} + 4) e^{3 x} d x