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y=0 wtw /x-3/-2 =0
x-3=2 lub x-3=-2
x=5 lub x=1
|D|=a∫b [f(x)-g(x)]dx
|D|=₁∫⁵0-(|x-3|-2) dx=₁∫³x-1 dx+₃∫⁵-x+5x dx=
=(x²/2-x)|₁³+(-x²/2+5x)|₃⁵
=(9/2-3)-(½-1)+(-25/2+25)-(-9/2+15)=4-2-8+25-15=29-25=4
Pole obszaru wynosi 4.