Chapter 5 - The Integral - 5.4 The Fundamental Theorem of Calculus, Part I - Exercises - Page 258: 34

$\frac{13}{2}.$

We have $$ \int_{0}^{5}|3-x| d x= \int_{0}^{3}|3-x| d x+ \int_{3}^{5}|3-x| d x\\ = \int_{0}^{3}3-x \ d x- \int_{3}^{5}3-x \ d x=3x-\frac{1}{2}x^2|_{0}^3-(3x-\frac{1}{2}x^2)|_{3}^5\\ =9-\frac{9}{2}-15+\frac{25}{2}+9-\frac{9}{2}=\frac{13}{2}. $$