integral from 0 to sqrt(5 of)6x*f(x^2+1)

解答

\int_{0}^{5} 6 x \cdot f (x^{2} + 1) d x

\frac{105}{2} f

求解步骤

\int_{0}^{5} 6 x f (x^{2} + 1) d x

提出常数:

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

= 6 f \cdot \int_{0}^{5} (1 + x^{2}) x d x

乘开

(1 + x^{2}) x : x + x^{3}

= 6 f \cdot \int_{0}^{5} x + x^{3} d x

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

= 6 f (\int_{0}^{5} x d x + \int_{0}^{5} x^{3} d x)

\int_{0}^{5} x d x = \frac{5}{2}

\int_{0}^{5} x^{3} d x = \frac{25}{4}

= 6 f (\frac{5}{2} + \frac{25}{4})

化简

6 f (\frac{5}{2} + \frac{25}{4}) : \frac{105}{2} f

= \frac{105}{2} f

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