 Hi friends, I am Purva and today we will discuss the following question. Find the area enclosed by the parabola 4y is equal to 3x squared and the line 2y is equal to 3x plus 12. Let us begin with the solution now. Now we are given that equation of parabola is 4y is equal to 3x squared or we can say y is equal to 3 upon 4x squared. Let us mark this as equation 1 and equation of line is 2y is equal to 3x plus 12 or we can say y is equal to 3 upon 2x plus 6. Let us mark this as equation 2. Now solving equation 1 and 2 we get we put the value of y is equal to 3 upon 4x squared from equation 1 into equation 2. So we get 3 upon 4x squared is equal to 3 upon 2x plus 6 and this implies now dividing both the sides by 3 we get 1 upon 4x squared is equal to 1 upon 2x plus 2 and this implies x squared is equal to 2x plus 8. This further implies x squared minus 2x minus 8 is equal to 0 and this implies we can write this as x squared minus 4x plus 2x minus 8 is equal to 0 and this implies now taking out x common from these four terms we get x into x minus 4 plus taking out two common from these four terms we get 2 into x minus 4 is equal to 0 and this implies x plus 2 into x minus 4 is equal to 0. Therefore we have either x is equal to minus 2 or x is equal to 4. Now if x is equal to minus 2 then putting the value of x in equation 2 we get then y is equal to 3 and if x is equal to 4 then again putting the value of x in equation 2 we get y is equal to 12. Hence the two points of intersection are minus 2 comma 3 and 4 comma 12. Now y is equal to 3 upon 4x squared is a parabola with vertex at 0 0 and it is symmetric about y axis and the line 2y is equal to 3x plus 12 passes through the points minus 2 comma 3 and 4 comma 12. So this is the parabola 4y is equal to 3x squared with vertex at 0 0 and it is symmetric about y axis and this is the line 2y is equal to 3x plus 12 which passes through the points minus 2 comma 3 and 4 comma 12 and this shaded region is the region whose area is to be found out. Therefore we have required area is equal to area of trapezium prsq minus area of o rpo minus area of triangle sqo. Now area of trapezium prsq is given by now this region has limits from minus 2 to 4 because x coordinates are from minus 2 to 4 and equation of line is given by 2y is equal to 3x plus 12 so we have y is equal to 3x plus 12 upon 2 therefore we have area of trapezium is given by integral limit is from minus 2 to 4 3x plus 12 upon 2 dx minus area of o rpo is given by now this region has x coordinates from 0 to 4 so we have limit from 0 to 4 and equation of parabola is 4y is equal to 3x square so we have y is equal to 3x square upon 4 so we have area of o rpo is given by integral limit is from 0 to 4 3 upon 4 x square dx minus now area of triangle sqo is given by here we have limits from minus 2 to 0 and again equation of parabola is y is equal to 3 upon 4 x square so we have area of triangle sqo is equal to integral limit is from minus 2 to 0 3 upon 4 x square dx this is equal to 1 upon 2 into now integrating 3x plus 12 we get 3x square upon 2 plus 12x and we have limit is from minus 2 to 4 minus 3 upon 4 into now integrating x square we get x cube by 3 and limit is from 0 to 4 minus 3 upon 4 into again integrating x square we get x cube by 3 and we have limit is from minus 2 to 0 this is equal to 1 upon 2 into now putting the limits we get 3 upon 2 into 16 plus 12 into 4 minus 12 upon 2 minus 24 putting upper limit 4 in place of x we get 3 upon 2 into 16 plus 12 into 4 minus putting lower limit minus 2 in place of x we get 12 upon 2 minus 24 minus 3 upon 4 into 3 into now again putting the limits we get 4 cube minus 0 putting upper limit 4 in place of x we get 4 cube minus putting lower limit 0 in place of x we get 0 and we have already taken out 3 common so we get 3 upon 4 into 3 into 4 cube minus 0 minus 3 upon 4 again we take out 3 common in denominator so we have into 3 now again putting the limits we get 0 minus minus 2 cube because putting upper limit 0 in place of x we get 0 minus putting lower limit minus 2 in place of x we get minus 2 whole cube this is equal to 1 upon 2 into now solving this bracket we get 24 plus 48 minus 6 plus 24 minus now canceling out 3 in numerator and denominator we get 1 upon 4 into 64 minus again canceling out 3 in numerator and denominator we get 1 upon 4 into 8 and this is equal to 1 upon 2 into now solving this bracket we get 90 minus now canceling out common factor 4 in denominator and numerator we get 16 in numerator so we get minus 16 minus now again canceling out common factor 4 in denominator and numerator we get 2 in numerator so we get minus 2 and this is equal to now canceling out 2 in numerator and denominator we get 45 in numerator so we have 45 minus now minus 16 minus 2 gives minus 18 and this is equal to 27 hence the required area is equal to 27 this is our answer hope you have understood the solution bye and take care