UPSC NDA Math Model Paper 2

The given function can be summarized as follows:
Since the given function is a multi-valued function, let us separate the given definite integral into parts where the expressions of the function are different:

3

∫

−2

f(x)dx=

0

∫

−2

f(x)dx+

2

∫

0

f(x)dx +

3

∫

2

f(x)dx
=

0

∫

−2

(x+2)dx+

2

∫

0

(2−x)dx +

3

∫

2

(x−2)dx
=[

x2

2

+2x]−20+[2x−

x2

2

]02 +[

x2

2

−2x]23
=[0−(2−4)]+[4−2−0)] +[

9

2

−6−(2−4)]
=2+2+

9

2

−4
=4.5