JEE Main 28 Jan 2025 Shift 1 Paper

The region defined is symmetric about the y-axis. We find the area for x≥0 and double the result.
The boundary functions are y=2x+1 and y=x2+1 within 0≤x≤3. They intersect at x=0 and x=2.
The total area is calculated by integrating the lower function in each interval:
Ahalf=

2

∫

0

(x2+1)‌dx+

3

∫

2

(2x+1)‌dx
Evaluate the integrals:
[‌

x3

3

+x]02=‌

14

3

[x2+x]23=6
Sum the half-area components:
A‌half ‌=‌

14

3

+6=‌

32

3

Double for the total area:
A‌total ‌=2×‌

32

3

=‌

64

3