find area

ne one?

See, using the property of mod, we can say that the graph changes its concavity at x=-1/2 in the max function and x= 1/2 in the min function.
to find area, break the limits at these points. And the graph will be that of x²+x and x²-x in the respective interval.

IS it that easy ?
I dont think so...

err...I made a mistake

if it would have had been that simple i wouldnt have uploaded it...

anyone pls try.....
i think its a really good question....

anant sir....nishant sir pls help me out.........

hey arshad why there is only one function inside max and min

there is only one function but check the limits...the limits are diff.

I ma getting g(x) as ..

Plzz tell if its correct or not...I need some expert's opinion

Eureka, your g(x) is not correct.
You see, that for -2 ≤ x < 0, g(x) is same as max f(t) for -2 ≤ t ≤ x. Notice that for t < 0, f(t) is an upward opening parabola having roots at t = -1 and 0. So the maximum of f(t) always occurs at t = -2. So g(x) = 2 for -2 ≤ x < 0.

Once you go to the positive side, f(t) is an upward opening parabola with roots at t =0, 1. So g(x) = f (x) as long as 0 ≤ x ≤ 1/2, after which the minimum is at x=1/2. So g(x) gets latched to f(1/2). As such

g(x) = 2 for -2 ≤ x < 0
= x² - x for 0 ≤ x ≤ 1/2
= -1/4 for 1/2 < x ≤ 3

So the required area = 113/24 sq units