integral from 0 to 1 of 4xf(x^2+1)

解答

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

3 f

求解步骤

\int_{0}^{1} 4 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

= 4 f \cdot \int_{0}^{1} (1 + x^{2}) x d x

乘开

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

= 4 f \cdot \int_{0}^{1} 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

= 4 f (\int_{0}^{1} x d x + \int_{0}^{1} x^{3} d x)

\int_{0}^{1} x d x = \frac{1}{2}

\int_{0}^{1} x^{3} d x = \frac{1}{4}

= 4 f (\frac{1}{2} + \frac{1}{4})

化简

4 f (\frac{1}{2} + \frac{1}{4}) : 3 f

= 3 f

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