Notifications
Clear all
0
07/02/2022 11:50 am
Topic starter
Let [t] denote the greatest integer ≤ t. Then the value of 8. \int_{-\frac{1}{2}}^1([2x]+ |x|) dx is .......
1 Answer
0
07/02/2022 11:58 am
I = \int_{-\frac{1}{2}}^1([2x]+ |x|) dx
= \int_{-\frac{1}{2}}^1[2x]dx + \int_{-\frac{1}{2}}^1|x|dx
= 0 + \int_{-\frac{1}{2}}^1(-x)dx + \int_0^1x dx
= \Big(-\frac{x^2}{2}\Big)^0_{\frac{1}{2}} + \Big(\frac{x^2}{2}\Big)^1_0
= \Big(0 + \frac{1}{8}\Big) + \frac{1}{2}
= \frac{5}{8}
8I = 5