urgent; Evaluate : integrate limit 0 to 8 modulus (x-5) dx - Maths - Integrals

Question

urgent; Evaluate : integrate limit 0 to 8 modulus (x-5) dx

Duke Biswass · Accepted Answer

x-5 is negative from 0 to 5 and positive from 5 to 8

So the integral becomes

$\int_{0}^{5} (5 - x) ⅆ x + \int_{5}^{8} x - 5 ⅆ x$

= ${(5 x - \frac{x^{2}}{2})}_{0}^{5} + {(\frac{x^{2}}{2} - 5 x)}_{5}^{8}$

= $25 - \frac{25}{2} + (\frac{64}{2} - 40 - \frac{25}{2} + 25)$

= $\frac{25}{2} + \frac{9}{2} = \frac{34}{2} = 17$

urgent; Evaluate : integrate limit 0 to 8 modulus (x-5) dx .