 Hi friends, I am Purva and today I will help you with the following question. Find the area of the region bounded by the curves y is equal to x square plus 2, y is equal to x, x is equal to 0 and x is equal to 3. Let us begin with the solution now. Now the region is bounded by the curves y is equal to x square plus 2, y is equal to x, x is equal to 0 and x is equal to 3. So first we shall identify the region whose area is to be found out. So for that first we will find the point of intersection. So point of intersection y is equal to x square plus 2 and x is equal to 3 is 3 comma 11 and the point of intersection of x is equal to y and x is equal to 3 is 3 comma 3. Now we will draw the figure. Now y is equal to x square plus 2 or we can say that x square is equal to y minus 2 is a parabola with vertex at 0 comma 2 and it is symmetric about y axis. y is equal to x is a line passing through 0 comma 0, 1 comma 1 and so on. x is equal to 0 is the equation of y axis. x is equal to 3 is the equation of the line parallel to y axis. So with all this information we get the following figure where this is the parabola y is equal to x square plus 2, this is the line x is equal to 3 and this is the line y is equal to x and the shaded region is the region whose area is to be found out. Therefore we have required area is equal to area of OACD minus area of DEC minus area of OAB. So we have required area is equal to area of OACD minus area of DEC minus area of OAB. Now area of OACD is given by now this whole region is bounded by y axis, x axis, this line CD and this line AC and the equation of line AC is given by x is equal to 3 and we have limiters from 0 to 11. Therefore we get this is equal to integral limit from 0 to 11 and equation of line is x is equal to 3. So we have 3 dy minus. Now area of this region DEC is given by now this region is bounded by the parabola y is equal to x square plus 2 y axis and this line CD. Also the limit of this region lies from 2 to 11. So we have the area of DEC is given by integral limit is from 2 to 11 and equation of parabola is given by y is equal to x square plus 2. So we have x is equal to under root y minus 2 dy minus. Now area of OAB is given by now this region is bounded by line y is equal to x, x axis and the line x is equal to 3 and we have limiters from 0 to 3. So we get area of OAB is given by integral limit is from 0 to 3 and equation of line is y is equal to x. So we get y dy and this is equal to now integrating 3 we get 3y and limit is from 0 to 11 minus integrating under root of y minus 2 we get y minus 2 to the power 3 by 2 upon 3 by 2 and limit is from 2 to 11 minus integrating y we get y square upon 2 and limit is from 0 to 3. This is equal to now putting the limits we get first we put upper limit 11 in place of y we get 33 minus now we put lower limit 0 in place of y we get 0 minus 2 upon 3 into now again putting the limits we get putting upper limit 11 in place of y we get 3 to the power 2 into 3 upon 2 minus putting lower limit 2 in place of y we get 0 minus putting upper limit 3 in place of y we get 9 upon 2 minus putting lower limit 0 in place of y we get 0. And this is equal to 33 minus 2 upon 3 into 3 raise to the power 3 minus 9 upon 2 and this is equal to 33 minus 18 minus 9 upon 2 and this is further equal to now 33 minus 18 is equal to 15 minus 9 upon 2 and this gives this is equal to 21 upon 2 therefore we get the required area is equal to 21 upon 2 hence we write our answer as 21 upon 2 Hope you have understood the solution Bye and take care