JEE Main · 2025 · 04 Apr 2025 · Shift 1 · Area Under Curves

The boundaries $y=|x-5|$ and $y=4\sqrt{x}$ meet where $4\sqrt{x}=|x-5|$: for $x<5$ this gives $x=1$, and for $x\ge 5$ it gives $x=25$. The area is $A={\int }_1^{25}4\sqrt{x}dx-{\int }_1^{25}|x-5|dx$. The first integral is $\frac{8}{3}\left(25^{3/2}-1\right)=\frac{8}{3}(124)=\frac{992}{3}$. The region under $|x-5|$ is two triangles of areas $\frac{1}{2}(4)(4)=8$ and $\frac{1}{2}(20)(20)=200$, together $208$. Hence $A=\frac{992}{3}-208=\frac{368}{3}$, and $3A=368$.