Please solve the question in the image attached Answer of this question is Option 2

hello manjula, this ques can be easily solved  by drawing a graph of the 2 functions. further note that [x+1] can be written as [x] + 1 where [.] represents gif. (this follows from std properties of gif)

once you draw the graph, you ll notice that y= x^2 and y= [x] + 1 tend to meet at x=1. also, y= [x] + 1 is greater than x^2 when x is less than 1 but greater than 0.

so area bounded A= ∫ from 0 to 1 ([x] + 1 – x^2) dx

note that in x belonging to (0, 1), [x]= 0

so A= ∫ from 0 to 1 (1 – x^2) dx

= x – x^3/3 from 0 to 1

= 1 – 1/3

= 2/3