 Hello and welcome to the session I am Deepika here. Let's discuss a question which says evaluate the Pauline definite integral from 0 to 1 dx upon under root of 1 minus x square. Now we know that by second fundamental theorem of integral calculus, we have integral of A to B fx dx where a continuous function defined on the closed interval AB is equal to derivative that is the integral from A to B fx dx value of the antiderivative as the value of this same antiderivative at the lower limit A. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. It is equal to integral from 0 to 1 dx upon under root of 1 minus x square. We evaluate the definite integral dx upon under root of 1 minus x square. Now we know that upon under root of x square is equal to sin A. So therefore integral of dx upon under root of 1 minus x square is equal to the idea with a second fundamental theorem to 0. So this is equal to being equal to pi by 2. Hence the answer for the above question is pi by 2.