 Hi and welcome to the session. Let us discuss the following question. Question says choose the correct answer. Smaller area enclosed by the circle x square plus y square is equal to 4 and the line x plus y is equal to 2 is 2 multiplied by pi minus 2 b pi minus 2 c 2 pi minus 1 d 2 multiplied by pi plus 2. We have to choose the correct answer from A, B, C and D. First of all let us understand that if we are given a curve y is equal to fx and this curve intersects the x-axis at A and B then area enclosed between the curve and x-axis is definite integral from A to B y dx. Now we know y is equal to fx. So we can write it as definite integral from A to B fx dx. This is the key idea to solve the given question. Let us now start with the solution. Now we are given equation of circle x square plus y square is equal to 4. Now this can be written as x square plus y square is equal to 2 square. Now from this equation we get center of the circle is 0 0 and radius of the circle is 2 units. We are given a line x plus y is equal to 2. Now let us find out 2 points from where it passes. Clearly we can see if we put x is equal to 0 in this equation then y is equal to 2 and if we put y is equal to 0 in this equation we get x is equal to 2. So 2 points from where this line passes are 2 0 and 0 2. 2 0 and 0 2 are 2 points lying on the line. So we can write x plus y is equal to 2 is a line which passes through points 0 2 and 2 0. Now using this information we can draw this figure. This is a circle with radius 2 units and center 0 0 and this is a line which passes through points 0 2 and 2 0 or we can say this is a line x plus y is equal to 2 and this is a circle x square plus y square is equal to 4. Clearly we can see this shaded region is the smaller area enclosed between the line x plus y is equal to 2 and circle x square plus y square is equal to 4. Now we have to find this area. Let us name these points as o a and b. Now clearly we can see area of this shaded region is equal to area of quadrant oab minus area of triangle oab. Now we can write area of shaded region is equal to area of quadrant oab minus area of triangle oab. Now using key idea area of this quadrant is equal to definite integral from 0 to 2 square root of 4 minus x square multiplied by dx. We know y of circle is equal to square root of 4 minus x square. If we draw a small vertical strip in the quadrant oab of width dx then area of that strip is given by vertical length of strip multiplied by dx and vertical length of strip is equal to y of this curve and y of this curve is square root of 4 minus x square. So we get elementary area of strip is square root of 4 minus x square multiplied by dx and total area of this quadrant is equal to definite integral from 0 to 2 square root of 4 minus x square multiplied by dx. Now we will find out area of triangle oab. We will write minus sign as it is area of triangle oab is equal to definite integral from 0 to 2 y dx. We know y of this line is equal to 2 minus x. So here we can write 2 minus x dx. If we draw a small vertical strip of width dx in triangle oab then area of that strip is equal to dx multiplied by y of this line. We know length of that strip is equal to y of this line minus y of this line and we know y is equal to 0 at this line. So we get area of that strip is equal to y multiplied by width of that strip that is dx. Now we know y of this line is equal to 2 minus x. So area of that strip is equal to 2 minus x multiplied by dx. So this is elementary area of that strip. We know we can draw many such strips in this triangle oab. So total area of this triangle oab is given by definite integral from 0 to 2 2 minus x multiplied by dx. Now we will find out these two integrals. Now we can write this integral as definite integral from 0 to 2 square root of 2 square minus x square dx and we can write this integral as it is. That is definite integral from 0 to 2 2 minus x dx. Now answer for this integral is x upon 2 multiplied by 4 minus x square plus 4 upon 2 sin inverse x upon 2 where lower limit is 0 and upper limit is 2 minus this integral is equal to 2x minus x square upon 2. Lower limit of this integral is 0 and upper limit is 2. To find this integral we have used the formula of integration integral of square root of a square minus x square dx is equal to x upon 2 multiplied by square root of a square minus x square plus a square upon 2 sin inverse x upon a plus c. Integral of 2 is 2x and integral of x is x square upon 2. Now substituting the corresponding limits we get 0 plus 4 upon 2 multiplied by sin inverse 1 minus 0 minus 0 substituting 2 in this expression 4 minus 4 will become 0 so this term will become 0 here 2 and 2 will get cancelled and we get sin inverse 1 substituting lower limit 0 in this bracket we get this term as 0 and this term is also 0 we know sin inverse of 0 is equal to 0 only now substituting limits in this bracket we get 4 minus 4 upon 2 minus 0 plus 0 now simplifying this bracket we get 2 sin inverse 1 and simplifying this bracket we get minus 2 now we know sin inverse 1 is equal to pi upon 2 so substituting pi upon 2 for sin inverse 1 we get 2 multiplied by pi upon 2 minus 2 now this 2 and this 2 will get cancelled and we get pi minus 2 so required area of shaded region is equal to pi minus 2 so the correct answer is b so b is our required answer this completes the session hope you understood the solution take care and have a nice day