The value of r=120 ( | ( 0r x| x|dx ) | ) is .
Answer
210
Solution
Let Ir = 0r x| x|dx, r a positive integer.
| (r - x)| = | x|, by 0r f(x)dx = 0r f(r - x)dx:
Ir = 0r (r - x)| x|dx
Adding: 2Ir = r 0r | x|dx
| x| has period 1 0r | x|dx = r 01 x dx = r 2
2Ir = 2r2 Ir = r2
| Ir | = r2 = r
r=120 r = 20 212 = 210
Answer: 210