integral from 0 to-1 of (x^3+8)/(x+2)

解答

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

- \frac{16}{3}

十进制

- 5.33333 \dots

求解步骤

\int_{0}^{- 1} \frac{x ^{3} + 8}{x + 2} d x

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{0} \frac{x ^{3} + 8}{x + 2} d x

化简

\frac{x ^{3} + 8}{x + 2} : x^{2} - 2 x + 4

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

\int_{- 1}^{0} x^{2} d x = \frac{1}{3}

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

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

= - (\frac{1}{3} - (- 1) + 4)

化简

- (\frac{1}{3} - (- 1) + 4) : - \frac{16}{3}

= - \frac{16}{3}

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p a r i t y \frac{tan ( x )}{x}

\int x - x + 2 d x