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

解答

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

\frac{28}{3}

十进制

9.33333 \dots

求解步骤

\int_{0}^{4} x^{2} - 2 x + 1 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

= \int_{0}^{4} x^{2} d x - \int_{0}^{4} 2 x d x + \int_{0}^{4} 1 d x

\int_{0}^{4} x^{2} d x = \frac{64}{3}

\int_{0}^{4} 2 x d x = 16

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

= \frac{64}{3} - 16 + 4

化简

\frac{64}{3} - 16 + 4 : \frac{28}{3}

= \frac{28}{3}

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