 Hello everyone. Welcome to the session of application of multiple integrals part 2. This is Swati Nikam, Assistant Professor, Department of Humanities and Sciences, Valchan Institute of Technology, Sulapur. At the end of this session, student will be able to find area of region bounded by curves by using double integration. Before we proceed further, evaluate this simple integral by pausing your video for 2 minutes. Integration 0 to 1, inner integration 0 to y, x, y, dx into dy. I hope you got your answer. Late, capital I is equal to integration 0 to 1, inner integration 0 to y, x, y, dx into dy. As the inner integration have limits in terms of y, so let us integrate with respect to x first, which is equal to integration 0 to 1, integration of x is x square by 2, limits 0 to y, this y we have to keep as it is into dy, which is equal to integration 0 to 1. Now, put upper and lower limit in place of this x, so y square by 2 minus 0, outside the bracket y dy, which is equal to integration 0 to 1, this y square into y is y cube divided by 2 into dy, which is equal to, now integrate this term, integration of y cube is y raise to 4 upon 4 into 2 is 8, limits of this integration 0 to 1. So, putting limits, we will get it as equal to 1 by 8. So, we have evaluated a double integration and got this answer. Let us solve an example, example number 1. Find by double integration, the area bounded by the parabola y square is equal to 4x and the line y is equal to 2x minus 4. Answer, y square is equal to 4x is the parabola symmetric to x axis, let us call it as 1 and the line y is equal to 2x minus 4. In order to plot this line, we will change it and let us write down its equivalent form as 2x minus y is equal to 4. Divide this equation throughout by 4, hence we get x by 2 minus y by 4 is equal to 1. Let us call it as 2. Now, this is the line, it passes through the point 2 comma minus 4. Let us draw these two curves on the graph, so that we will get the required region of integration. After plotting these two curves, region AOB is the required region bounded by the curve 1 and 2. To find coordinates of the point A and B which are points of intersection, we have to solve equation 1 and 2. So, to find the points of intersection A and B, let us solve y square is equal to 4x and y is equal to 2x minus 4 simultaneously. Now, let us square this equation. Therefore, y square is equal to bracket 2x minus 4 whole square. But y square also represent 4x, that is why y square is equal to 4x is equal to bracket 2x minus 4 whole square. Therefore, y square is equal to 4x is equal to open this bracket by using A minus B whole square formula. So, it is 4x square minus 16x plus 16. Shift 4x to RHS, hence 4x square minus 16x plus 16 minus 4x is equal to 0. After simplification, 4x square minus 20x plus 16 is equal to 0. Divide throughout by 4 and therefore, x square minus 5x plus 4 is equal to 0, which is simple quadratic equation in terms of x. After simplification, it has roots x equal to 4 and x is equal to 1. Now, when x is equal to 4 and y is equal to 2x minus 4, this together implies that y has value y equal to 2 into substitute value of x as 4, 4 minus 4, that is 8 minus 4 is equal to 4. Therefore, point A has coordinates 4 comma 4. Similarly, when x is equal to 1 and y is equal to 2x minus 4, this together implies that y has value y equal to 2 into value of x is 1 now minus 4, that is 2 minus 4 equal to minus 2. Hence, point B has coordinates 1 comma minus 2. Now consider a strip parallel to x axis, say PQ, so that with the help of this strip we can write the limits of this region for integration. To find limits of integration on this strip, x where is from. So, already we have studied about finding the limits from the region that whenever the strip is horizontal, then we have to consider limit from left end to right end and hence x where is from at the left end x is equal to y square by 4, 2 at the right end x is equal to y plus 4 by 2 and y where is from. Now, y has constant limit at the bottom y is equal to minus 2 to y is equal to 4. Therefore, the required area A is represented as outer integration minus 2 to 4, inner integration y square by 4 to y plus 4 divided by 2 into dx dy which is equal to integration minus 2 to 4 of as the limits are in terms of y. So, integrate with respect to x, integration is x, limits y square by 4 to y plus 4 by 2 into dy which is equal to integration minus 2 to 4. Now, let us write down limits in place of x. So, upper limit y plus 4 by 2 minus to lower limit y square by 4 into dy which is again equal to, let us find out the integration now. Integration of y by 2 is y square by 4 plus integration of 4 by 2 that is 2 is 2 into y minus integration of y square by 4 is y cube by 12, limits of this integration minus 2 to 4. So, here A is equal to bracket y square by 4 plus 2 y minus y cube by 12 outside the bracket limits minus 2 to 4 that is equal to in place of y we will substitute upper limit and lower limit. So, that is equal to 4 square by 4 plus 2 into 4 minus 4 cube by 12 minus now the lower limit minus 2 square by 4 minus 2 into minus 2 plus minus 2 cube by 12 which is equal to 4 square by 4 is 4 plus 4 into 2 8 minus 64 by 12 minus 4 by 4 plus 4 minus 8 by 12 which is equal to. Now, let us collect all the integers in one bracket 4 plus 8 minus 1 plus 4 minus let us have minus sign common from these two rational numbers and in another group it is 64 by 12 plus 8 by 12 which is equal to 15 minus 6 that is 9. Therefore, area A is equal to 9. I have referred higher engineering mathematics by B.S. Graeval for creation of this video. Thank you so much for watching this video and have a happy learning.