 High students, how are you? Let's work out the following question. Find the area of the region enclosed by the parabola x square is equal to y, the line y is equal to x plus 2 and the x-axis. Let's begin with the solution now. Now, the region whose area is to be found out is enclosed by the parabola x square is equal to y, the line y is equal to x plus 2 and the x-axis. So, first we shall identify this region by drawing the graph. Now, x square is equal to y is a parabola with vertex at 0, 0 and symmetric about y-axis and y is equal to x plus 2 is a line passing through the points 0, 2 and minus 2, 0. Now, we find the points of intersection. x square is equal to y and y is equal to x plus 2 by solving these two equations. And here, we mark the equation x square is equal to y as 1 and y is equal to x plus 2 as equation 2. Substituting the value of y is equal to x plus 2 and x square is equal to y we get. Substituting 2 in 1 we get x square is equal to x plus 2. This implies x square minus x minus 2 is equal to 0. Now, we solve this quadratic equation and we get this implies x square minus 2x plus x minus 2 is equal to 0. Solving this we get. This implies x minus 2 into x plus 1 is equal to 0. And this further implies x is equal to 2 and x is equal to minus 1. Now, putting x is equal to 2 in equation 2 we get y is equal to 2 plus 2. That is y is equal to 4. So, we get when x is equal to 2, y is equal to 4. And by putting x is equal to minus 1 in equation 2 we get y is equal to minus 1 plus 2. And this implies y is equal to 1. So, we get when x is equal to minus 1, y is equal to 1. So, the points of intersection are 2 comma 4 and minus 1 comma 1. So, here in this figure we get the points of intersection of the parabola and the line are 2 comma 4 and minus 1 comma 1. And we have to find the area of this shaded region. Thus, the required area is equal to, now in this portion we have, limit is from minus 1 to 2 and the equation of the line is y is equal to x plus 2. So, we have integral limit is from minus 1 to 2 x plus 2 dx minus again the limit is from minus 1 to 2. And the equation of parabola is y is equal to x square. So, we have minus integral limit is from minus 1 to 2 x square dx. And this is equal to now integrating x plus 2 we get x square by 2 plus 2x and limit is from minus 1 to 2 minus integrating x square we get x cube by 3. And again limit is from minus 1 to 2. Now, putting the limits we get this is equal to first we put the upper limit 2 in place of x we get 4 upon 2 plus 4 minus. Now, we put the lower limit minus 1 in place of x we get 1 upon 2 minus 2 minus. Now, we put the upper limit 2 in place of x here we get 8 upon 3 minus. Now, we put the lower limit minus 1 in place of x here we get minus 1 upon 3. And this is equal to now canceling the common factor 2 from denominator and numerator here we get 2. So, we get 2 plus 4 which is equal to 6. So, 6 minus 1 by 2 minus into minus will become plus. So, plus 2 minus 8 upon 3 minus into minus again will become plus. So, plus 1 upon 3 this is equal to 6 plus 2 is equal to 8. So, 8 minus 1 upon 2 minus solving this we get 8 plus 1 is equal to 9. So, 9 upon 3. Now, we cancel out the common factor 3 from denominator and numerator we get 3 in numerator. And this is equal to now we take the LCM 2 we get 16 minus 1 minus 6. And solving this we get this is equal to 9 upon 2. Thus, we get the required area is 9 upon 2. Hence, we get the answer as 9 upon 2. Hope you have understood the solution. Bye and take care.