integral from 0 to 4 of x^2-8x+16

解答

\int_{0}^{4} x^{2} - 8 x + 16 d x

\frac{64}{3}

十进制

21.33333 \dots

求解步骤

\int_{0}^{4} x^{2} - 8 x + 16 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} 8 x d x + \int_{0}^{4} 16 d x

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

\int_{0}^{4} 8 x d x = 64

\int_{0}^{4} 16 d x = 64

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

化简

= \frac{64}{3}

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