See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Let   2022 1 0 I x f(x) dx    …(i) let x f(t) dx f '(t)dt            2022 0 1 1 2022 0 I f(t) f f(t) f '(t)dt f f(0) 0 f(1) 0 I f(t) t f '(t)dt ...(ii)           Now adding equation (1) and (2)     2022 1 0 2I x f(x) 1 f '(x) dx 1 p I 2023 q p q 2024          For More Material Join: @JEEAdvanced_2024