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Question

Question: \(\int _ { - 2 } ^ { 2 } | x | d x =\)...

∫−22∣x∣dx=\int _ { - 2 } ^ { 2 } | x | d x =

A

0

B

1

C

2

D

4

Answer

4

Explanation

Solution

I=∫−22∣x∣dxI = \int _ { - 2 } ^ { 2 } | x | d x =−[x22]−20+[x22]02= - \left[ \frac { x ^ { 2 } } { 2 } \right] _ { - 2 } ^ { 0 } + \left[ \frac { x ^ { 2 } } { 2 } \right] _ { 0 } ^ { 2 }

=−(−2)+(2)=4= - ( - 2 ) + ( 2 ) = 4.