 Hello and welcome to the session. Let us discuss the following question. It says find the area of the region bounded by the curve 4y is equal to 3x square and the line 2y is equal to 3x plus 2. So let's now move on to the solution. We are given the curve 4y is equal to 3x square right and this can be written as 3y4x square and also from this we can see that x square is equal to 4y3y and we know that this is the parabola and here we have y is equal to 3y4x square and we have to find the area of the region bounded by this parabola and the line 2y is equal to 3x plus 12. So this is the parabola for 4y is equal to 3x square and this is the line 2y is equal to 3x plus 12. Now to find the area of this region we need to find the point of intersection of the parabola and the line right. So we will put y is equal to 3y4x square in this and we will find the point of intersection. So we have 2 into 3y4x square is equal to 3x plus 12. So we have 3y2x square is equal to 3x plus 12 and this implies 3x square is equal to 6x plus 24. This implies 3x square minus 6x minus 24 is equal to 0. Taking 3 common we have x square minus 2x minus 8 is equal to 0. Now we will factorize this quadratic equation. So we have x square minus 4x plus 2x minus 8 is equal to 0. Again taking x common we have x into x minus 4. Taking 2 common from the last terms we have 2 into x minus 4 is equal to 0. So this implies x minus 4 plus x minus x minus 4 into x plus 2 is 0. So this implies x minus 4 is equal to 0 or x plus 2 is 0. So this implies x is equal to 4 or x is equal to minus 2. Now if x is minus 2 then y is equal to 3y4x square now x is minus 2. So this is 3y4 into minus 2 square and then y is equal to 3 and if x is equal to 4 then y is equal to 3y4 into 4 square. So this implies y is equal to 12. So the points of intersection are minus 2, 3 and 4, 12. This is the point minus 2, 3 and this is the point 4, 12. Now we find the area of this region we need to subtract the area of the parabola when x is going from minus 2 to 4 from the area covered by the line 2y is equal to 3x plus 12 when x is going from minus 2 to 4. So the required area is given by integral minus 2 to 4 of the line which is 2y is equal to 3x plus 12. So y is equal to 3x plus 12 by 2 minus the area covered by the parabola that is the integral minus 2 to 4 3y4x square dx. Let us now take LCM and combine the two integrals. So we have minus 2 to 4 integral minus 2 to 4 6x plus 24 minus 3x square dx upon 4. So this is again 1y4 integral minus 2 to 4 6x plus 24 minus 3x square dx. Integral of 6x is 6x square by 2 integral of 24 with respect to x is 24x minus 3x cube by 3 and here the lower limit is minus 2 upper limit is 4. Now we will put x is equal to 4 first and before that we simplify this expression. Now we put x is equal to 4 first so we have 3 into 4 square plus 24 into 4 minus 4 cube minus. Now we will put x is equal to minus 2 here so we have 3 into minus 2 square plus 24 into minus 2 minus minus 2 cube. Now this is equal to 1 by 4 into 3 into 16 is 48 plus 96 minus 4 cube is 64 minus 3 into 4 is 12 minus 24 into 2 is 48 plus minus is minus plus 8. Now dividing each term by 4 we have 48 divided by 4 is 12 plus 24 minus 13 minus 3 minus 12 plus 2 36 12 plus 24 is 36 minus 16 is 20 minus 3 plus 2 is 5 minus 12 minus 7 this is 20 plus 7 that is 27 since the required area is 27 square units. So this completes the question and the session. Bye for now. Take care. Have a good day.