JEE Main 2025 Apr 7 Shift 2, Mathematics Q7: Area as Definite Integral

The two upper-bound lines $y=x+7$ and $y=11-3x$ intersect where $x+7=11-3x$, giving $x=1,y=8$; $y=x+7$ is the lower of the two for $x\le 1$ and $y=11-3x$ is lower for $x\ge 1$.

The parabola $y=1+x^2$ meets $y=x+7$ at $x^2-x-6=0\Rightarrow x=-2$ or $x=3$; the relevant left boundary is $x=-2$. It meets $y=11-3x$ at $x^2+3x-10=0\Rightarrow x=-5$ or $x=2$; the relevant right boundary is $x=2$.

So the region runs from $x=-2$ to $x=1$ bounded above by $x+7$, and from $x=1$ to $x=2$ bounded above by $11-3x$, with $1+x^2$ as the lower bound throughout:

$A={\int }_{-2}^1[(x+7)-(1+x^2)]dx+{\int }_1^2[(11-3x)-(1+x^2)]dx$

$={\left[\frac{x^2}{2}+6x-\frac{x^3}{3}\right]}_{-2}^1+{\left[10x-\frac{3x^2}{2}-\frac{x^3}{3}\right]}_1^2=\frac{27}{2}+\frac{19}{6}=\frac{50}{3}$.

$\therefore 3A=50$.