integral from-2 to 6 of 9-(x/2)^2-(6-x)

解答

\int_{- 2}^{6} 9 - (\frac{x}{2})^{2} - (6 - x) d x

\frac{64}{3}

十进制

21.33333 \dots

求解步骤

\int_{- 2}^{6} 9 - (\frac{x}{2})^{2} - (6 - x) d x

化简

9 - (\frac{x}{2})^{2} - (6 - x) : - \frac{x ^{2}}{4} + x + 3

= \int_{- 2}^{6} - \frac{x ^{2}}{4} + 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

= - \int_{- 2}^{6} \frac{x ^{2}}{4} d x + \int_{- 2}^{6} x d x + \int_{- 2}^{6} 3 d x

\int_{- 2}^{6} \frac{x ^{2}}{4} d x = \frac{56}{3}

\int_{- 2}^{6} x d x = 16

\int_{- 2}^{6} 3 d x = 24

= - \frac{56}{3} + 16 + 24

化简

- \frac{56}{3} + 16 + 24 : \frac{64}{3}

= \frac{64}{3}

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

\int_{1}^{6} 9^{x} d x

\int_{0}^{0.2} 25 e^{- 5 y} d y

\int_{- 1}^{0} e^{x + i x + i} d x

\int_{0}^{2} (6 x^{2} - 5) d x