integral from-1 to 9 of (9-x^2+8x)

解答

\int_{- 1}^{9} (9 - x^{2} + 8 x) d x

\frac{500}{3}

十进制

166.66666 \dots

求解步骤

\int_{- 1}^{9} 9 - x^{2} + 8 x 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}^{9} 9 d x - \int_{- 1}^{9} x^{2} d x + \int_{- 1}^{9} 8 x d x

\int_{- 1}^{9} 9 d x = 90

\int_{- 1}^{9} x^{2} d x = \frac{730}{3}

\int_{- 1}^{9} 8 x d x = 320

= 90 - \frac{730}{3} + 320

化简

90 - \frac{730}{3} + 320 : \frac{500}{3}

= \frac{500}{3}

\int \frac{x}{49 + x ^{4}} d x

d er i v a t i v e \frac{2}{t} - \frac{4}{t ^{2}}

\int_{0}^{5} 40 e^{- 0.3 t} d t

y^{^{'}} = 7 x e^{y}, y (0) = 0

x \to 0 lim (\frac{a ^{x} - 1}{2 x})