 Hello and welcome to the session. In this session we discuss the following question that says using integration, find the area of the region bounded by the curve x square equal to 4 y and the line x equal to 4 y minus 2. Let us now proceed with the solution. We are given a curve x square equal to 4 y, let this be equation 1 and the line is also given to which is x equal to 4 y minus 2, let this be equation 2. To find the area of the region bounded by this curve and the line, first the formula is draw this curve x square equal to 4 y. This represents a parabola with the vertex at the origin and it opens upwards. This is the parabola x square equal to 4 y. Next we draw the line x equal to 4 y minus 2. This is the line x equal to 4 y minus 2 and the point of intersection of the curve and the line. Now equation 2, we have 4 y is equal to x plus 2, substituting equal to x plus 2 equal to x plus 2. This gives us x square minus x minus 2 equal to 0. Now split in the middle term we get x square plus x 2 equal to 0 and from these two terms we get x into x plus 1 d whole and from these two terms we get minus 2 into x plus 1 d whole is equal to 0 and can be whole into equal to 0. From where we get x equal to minus 1 or x equal to 2. Now this equation 4 y equal to x plus 2 equal to x plus 2 equal to 0.4. Let this be equation 3. Now substituting minus 1 plus 2 which is equal to 1 upon 4 and equation 3 upon 4 which is equal to 4 upon 4 that is point of intersection say point b has coordinates to run intersection the curve and the line that is the a with coordinates minus 1, 1 by 4 and b with coordinates 2. So this is the point a with coordinates 2, 1 of the shaded region that is the area a of b a we say that the required area is equal to the area n b m perpendicular to the x axis shaded region would be the area a region a of b m equal to 4 y minus minus 1 a of b m l a is the area n equal to minus 1 to x equal to 2. The integral may be the value of x goes from minus 1 to 2 and value of y from the equation of the line which is given by the equation 3 minus y is equal to x plus 2 upon is given as integral minus 1 to 2 a of b m l a by the integral where x goes from minus 1 to 2 this area a of b m l a square equal to 4 y from x equal to minus 1 to 2. So from here we find out the value of y which is the area x square upon 4 dx. So further this is equal to integral minus 1 to 2 x plus 2 upon 4 minus x square upon 4 we have dx equal to integral minus 1 to 2 minus x square this 1 upon 4 is the integral so 1 upon 4 into integral minus 1 to 2 plus x plus 2 you have integrating we get 1 upon 4 into 1 3 plus minus 1 to 2. So we get this is equal to 1 upon 4 into first we put the value of x as 2 so here we have upon 3 plus 2 square is 4 upon 2 plus 2 into 2 is 4 for this whole plus 1 by 2. Now taking the answer here which is 6 we get minus 16 plus 12 minus 1 by 3 is 1 by 3 plus 1 by 2 minus 2 the whole which further gives us 1 upon 4 into 20 upon 6 minus now taking the answer here which is 6 we get 2 plus 2 minus 12 upon 4 into 20 upon 6 into 7 upon 6 so that becomes plus 7 upon 6 which gives us 1 upon 4 into 27 upon 6 and 3 2 times is 6 so this is equal to 9 upon 8 is equal to 9 upon 8 square units. 3d region is 9 upon 8 square units so this is our answer this completes the session so I hope you have understood the solution of this question.