 Hello students, let's solve the following question. It says find the area of the smaller region bounded by the ellipse x square upon a square plus y square upon v square is equal to 1 and the straight line x upon a plus y upon v is equal to 1. Let us first understand how do we find the area of the region bounded by the curve y is equal to fx the ordinate x is equal to a x is equal to b the area is given by integral a to b fx dx which is equal to capital fx where the lower limit of x is a and the upper limit of x is b which is equal to fb minus fa so this knowledge will work as the idea. Let us now start the solution here we have to find the area of the smaller region bounded by the ellipse and this line and the line is in the intercept form where a is the x-intercept and v is the y-intercept this is the point a0 and this is the point 0b we have to find the area of the smaller region that is we have to find the area of this region now the ellipse given to us is x square upon a square plus y square upon b square is equal to 1 and this again implies y square upon b square is equal to 1 minus x square upon a square and this implies y is equal to b upon a into under the root a square minus x square and here we have just taken the positive square root now we have to find the point where the ellipse cuts the x-axis that is where the y is 0 because x-axis is the line y is equal to 0 so we will put y is equal to 0 so this implies 0 is equal to under the root a square minus x square and this implies x is equal to plus minus a so the ellipse cuts the x-axis at the point a0 and minus a0 now the straight line given to us is x upon a plus y upon b is equal to 1 and this implies y upon b is equal to a minus x upon a and this implies y is equal to b by a into a minus x now the region for which we have to find the area lies in just the first quadrant and this area is equal to the area of the ellipse in the first quadrant minus this area and the area of the ellipse in the first quadrant is given by the integral 0 to a b by a into under the root a square minus x square dx and from the area of ellipse in the first quadrant we have to subtract the area of this triangular region which is below the line x upon a plus y upon b is equal to 1 so this area the area of the triangular region is given by integral 0 to a b by a into a minus x dx here we are taking the limits of x from 0 to a because this line can go maximum up to a and that gives us the area of the triangular region so the required area is integral 0 to a b by a under the root a square minus x square dx which is the area of the ellipse in the first quadrant minus integral 0 to a b by a into a minus x dx and this is the area of the triangular region below the line x upon a plus y upon b is equal to 1 so this is equal to b by a let's call this integral as i1 minus b by a and let's call the other integral as i2 now we solve i1 i1 is the integral 0 to a under the root a square minus x square dx now to solve this integral we'll use substitution method and we'll put x is equal to a sin theta so dx is equal to a cos theta d theta now x is equal to a sin theta dx is equal to a cos theta d theta so substituting all these values in the integral i1 the integral becomes under the root a square minus a square sin square theta into a cos theta d theta now as we substitute x is equal to a sin theta limits will change in terms of theta so if x is 0 then a sin theta is equal to 0 and this implies theta is equal to 0 and if x is equal to a then a is equal to a sin theta and this implies sin theta is equal to 1 and this again implies theta is equal to pi by 2 so here theta takes the lower limit as 0 and upper limit as pi by 2 now this integral becomes integral 0 to pi by 2 taking a square common and taking it out of the root it becomes a and we have under the root of 1 minus sin square theta which is cos square theta and the square root of cos square theta is cos theta so this is a cos theta into a cos theta d theta so this is equal to a square integral 0 to pi by 2 cos square theta d theta now we multiply and divide by 2 to apply the formula of 2 cos square theta now we will apply the formula of 2 cos square theta 2 cos square theta is equal to 1 plus cos 2 theta d theta now we integrate this integral of 1 with respect to theta is theta plus the integral of cos 2 theta is sin 2 theta upon 2 where the lower limit of theta is 0 and the upper limit of theta is pi by 2 now we will apply the second fundamental theorem so we first put theta is equal to pi by 2 minus now we put theta is equal to 0 so this is 0 plus sin 2 into 0 upon 2 so this is equal to a square by 2 into pi by 2 the sin 0 is 0 and sin pi is also 0 so 0 by 2 is 0 and we are left with a square by 2 into pi by 2 so this is equal to a square pi by 4 so this is the value of the integral i1 now we calculate the integral i2 it is 0 to a a minus x bx now we integrate this integral of a with respect to x is ax minus the integral of x with respect to x is x square by 2 and here the lower limit of x is 0 and the upper limit of x is a now we apply the fundamental theorem so we put x is equal to a first so that it becomes a into a minus a square by 2 minus now we put x is equal to 0 so this becomes a into 0 minus 0 square by 2 so this is equal to a square minus a square by 2 so this is equal to a square by 2 now the required area is equal to b by a i1 minus b by a i2 now the value of the integral i1 is a square pi by 4 and the value of the integral i2 is a square by 2 substituting all these values b by a into a square pi by 4 minus b by a a square by 2 now this is equal to b by a taking a square common becomes pi by 4 minus 1 by 2 which is again equal to a b into pi minus 2 by 4 taking lcm which is equal to a b upon 4 into pi minus 2 so the area of the required region is a b by 4 into pi minus 2 this completes the question and the session bye for now take care have a good day