JEE Main 2026 — 6 April, Evening Shift — Question 38

Find intersection points of $y = x^2 - 8x$ and $y = -x$: $x^2 - 8x = -x \Rightarrow x^2 - 7x = 0 \Rightarrow x(x-7)=0 \Rightarrow x=0,7$. The region is bounded between these curves for $0 \le x \le 7$. Upper curve: $y = -x$, lower curve: $y = x^2 - 8x$. Area = $\int_0^7 [(-x) - (x^2 - 8x)] \, dx = \int_0^7 (7x - x^2) \, dx$. Evaluate: $\int_0^7 7x \, dx = \left[ \frac{7x^2}{2} \right]_0^7 = \frac{7 \cdot 49}{2} = \frac{343}{2}$. $\int_0^7 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^7 = \frac{343}{3}$. Area = $\frac{343}{2} - \frac{343}{3} = 343 \left( \frac{1}{2} - \frac{1}{3} \right) = 343 \cdot \frac{1}{6} = \frac{343}{6}$.