integral from 0 to 8 of |x^2-6x+5|

解答

\int_{0}^{8} x^{2} - 6 x + 5 d x

40

求解步骤

\int_{0}^{8} x^{2} - 6 x + 5 d x

拿掉绝对值

= \int_{0}^{1} x^{2} - 6 x + 5 d x + \int_{1}^{5} - x^{2} + 6 x - 5 d x + \int_{5}^{8} x^{2} - 6 x + 5 d x

\int_{0}^{1} x^{2} - 6 x + 5 d x = \frac{7}{3}

\int_{1}^{5} - x^{2} + 6 x - 5 d x = \frac{32}{3}

\int_{5}^{8} x^{2} - 6 x + 5 d x = 27

= \frac{7}{3} + \frac{32}{3} + 27

化简

\frac{7}{3} + \frac{32}{3} + 27 : 40

= 40

\int_{- 1}^{2} x^{3} - x^{2} - 2 x d x

\int_{0}^{2 x} x y^{3} d y

\int_{0}^{y} (x + y)^{2} d x

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

\int_{2}^{4} \frac{1}{x ( ln ( x ) ) ^{3}} d x