NIMCET 2018 Question Paper with solutions

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Question : 48 of 120

Marks: +1, -0

Let f : R → R be defined by f (x) =

\{\table x+2, \if x < 0 ; {|x-2|} , \if x ≥ 0

. Find

∫↖{3}↙{-2}

f (x) dx

0.5
2.5
4.5
6.5

Solution:

Given,
f (x) =

\{\table x+2,x < 0;-(x-2) , 0 ≤ x <2;x - 2 , x > 2

f (x) =

\{\table x+2 , x < 0;2-x , 0 ≤ x <2 ; x-2 , x ≥ 2

∴

∫↖{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
=

[x^2/2+2x]^0_{-2}

+

[2x-x^2/2]^3_0

+

[x^2/2-2x]^3_2

=

[0-(4/2-4)]

+

[(4-4/2)-0]

+

[(9/2-6)-(4/2-4)]

= 2 + 2 -

3/2

+ 2 = 6 -

3/2

=

9/2

= 4.5

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