Explicație pas cu pas:
[x] este partea întreagă a lui x
a)
[tex]\int_{1}^{3} x[x] dx = \int_{1}^{2} x\cdot 1 dx + \int_{2}^{3} x\cdot 2 dx \\ = \int_{1}^{2} xdx + 2\int_{2}^{3} x dx = \left(\frac{ {x}^{2} }{2} \right)|_{1}^{2} + \left( \frac{ 2{x}^{2} }{2} \right)|_{2}^{3} \\ = \left(2 - \frac{1}{2} \right) + (9 - 4) = \frac{3}{2} + 5 = \frac{13}{2}[/tex]
b)
[tex]\int_{ - 1}^{3} x {3}^{[x]} dx = \int_{- 1}^{0} x\cdot {3}^{- 1} dx + \int_{0}^{1} x\cdot {3}^{0} dx + \int_{1}^{2} x\cdot {3}^{1} dx + \int_{2}^{3} x\cdot {3}^{2} dx \\ = \left(\frac{ {x}^{2} }{6} \right)|_{- 1}^{0} + \left(\frac{ {x}^{2} }{2} \right)|_{0}^{1} + \left(\frac{3 {x}^{2} }{2} \right)|_{1}^{2} + \left(\frac{9 {x}^{2} }{2} \right)|_{2}^{3} \\ = - \frac{1}{6} + \frac{1}{2} + \frac{9}{2} + \frac{45}{2} = \frac{82}{3} [/tex]
