integral from 0 to 1 of 15xsqrt(1-x)

解答

\int_{0}^{1} 15 x 1 - x d x

4

求解步骤

\int_{0}^{1} 15 x 1 - x d x

提出常数:

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

= 15 \cdot \int_{0}^{1} x 1 - x d x

使用换元积分法

= 15 \cdot \int_{1}^{0} - (- u + 1) u d u

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

= 15 (- \int_{0}^{1} - (- u + 1) u d u)

提出常数:

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

= 15 (- (- \int_{0}^{1} (- u + 1) u d u))

乘开

(- u + 1) u : - u^{\frac{3}{2}} + u

= 15 (- (- \int_{0}^{1} - u^{\frac{3}{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

= 15 (- (- (- \int_{0}^{1} u^{\frac{3}{2}} d u + \int_{0}^{1} u d u)))

\int_{0}^{1} u^{\frac{3}{2}} d u = \frac{2}{5}

\int_{0}^{1} u d u = \frac{2}{3}

= 15 (- (- (- \frac{2}{5} + \frac{2}{3})))

化简

15 (- (- (- \frac{2}{5} + \frac{2}{3}))) : 4

= 4

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