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

解答

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

- \frac{4}{3}

十进制

- 1.33333 \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{4}{3}

= - \frac{4}{3}

\int_{1}^{e} \frac{x ^{7} + 1}{x} d x

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