The value of ∫2−1f(x)dx,wheref(x)=(x+1|+(x|+(x−1| is
A
92
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
192
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
152
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
212
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B192 We can redefine f as f(x)=⎛⎜⎝2−x,if−1<x≤0x+2,if0<x≤13x,if1<x≤2∴∫2−1f(x)dx=∫0−1f(x)dx+∫10f(x)dx+∫21f(x)dx=∫0−1(2−x)dx+∫10(x+2)dx+∫213xdx=(2x−x22)0−1+(x22+2x)10+3⋅(x22)21=(2+12)+(12+2)+3⋅(2−12)=192