Absolute Value Integrals PYQs
Practice Class 12 CBSE Board Previous Year Questions (2008-2026)
Q1
2013 Board
00:00
Evaluate $\int_1^3 [|x-1| + |x-2| + |x-3|] \, dx$. 4 Marks
Break the integral into $[1, 2]$ and $[2, 3]$.
In $[1, 2]$:
$|x-1| = x-1, |x-2| = 2-x, |x-3| = 3-x$.
Sum $= (x-1) + (2-x) + (3-x) = 4-x$.
In $[2, 3]$:
$|x-1| = x-1, |x-2| = x-2, |x-3| = 3-x$.
Sum $= (x-1) + (x-2) + (3-x) = x$.
Integral $= \int_1^2 (4-x) \, dx + \int_2^3 x \, dx$.
$= [4x - x^2/2]_1^2 + [x^2/2]_2^3$.
$= (8-2) - (4-0.5) + (4.5-2) = 6 - 3.5 + 2.5 = 5$.
5
Q2
2013 Board
00:00
Evaluate $\int_0^4 [|x| + |x-2| + |x-4|] \, dx$. 4 Marks
Break the integral at $x=2$.
In $[0, 2]$: $x + (2-x) + (4-x) = 6-x$.
In $[2, 4]$: $x + (x-2) + (4-x) = x+2$.
Integral $= \int_0^2 (6-x) \, dx + \int_2^4 (x+2) \, dx$.
$= [6x - x^2/2]_0^2 + [x^2/2 + 2x]_2^4$.
$= (12-2) - 0 + (8+8) - (2+4) = 10 + 16 - 6 = 20$.
20
Q3
2013 Board
00:00
Evaluate $\int_2^5 [|x-2| + |x-3| + |x-5|] \, dx$. 4 Marks
Break the integral at $x=3$.
In $[2, 3]$: $(x-2) + (3-x) + (5-x) = 6-x$.
In $[3, 5]$: $(x-2) + (x-3) + (5-x) = x$.
Integral $= \int_2^3 (6-x) \, dx + \int_3^5 x \, dx$.
$= [6x - x^2/2]_2^3 + [x^2/2]_3^5$.
$= (18-4.5) - (12-2) + (12.5 - 4.5) = 13.5 - 10 + 8 = 11.5$.
11.5