integral from-9 to 4 of |x+5|

解答

\int_{- 9}^{4} ∣ x + 5 ∣ d x

\frac{97}{2}

十进制

48.5

求解步骤

\int_{- 9}^{4} ∣ x + 5 ∣ d x

拿掉绝对值

= \int_{- 9}^{- 5} - x - 5 d x + \int_{- 5}^{4} x + 5 d x

\int_{- 9}^{- 5} - x - 5 d x = 8

\int_{- 5}^{4} x + 5 d x = \frac{81}{2}

= 8 + \frac{81}{2}

化简

8 + \frac{81}{2} : \frac{97}{2}

= \frac{97}{2}

\int \frac{x}{x ^{2} - 6 x + 8} d x

x^{^{'}} = x^{2}, x (0) = 1

n = 0 \sum \infty (\frac{1}{2})^{2 n + 1}

d er i v a t i v e \frac{1}{x + \frac{1}{x}}

x \to + (- 0) lim (f (x))