See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

1 1 x 2  , 2 3 x 2  Now,  3 5 2 2 4 f x 1 x 2x 4x 6x ..... 3 3 5                =    d 1 x x f x dx   (1 – x2)f(x) = 1 + xf(x)  f(0) = 0 and  1 2 sin x f x 1 x    Now, area = 3 1 2 2 2 1 2 sin x dx 24 1 x       a + b – 19 = 7