Determine the area of the region bounded by y = - and y = 9 on the interval [1,10]. X Area =x2 = 49 when x = 17. In other words, the two curves intersect each other at x = -7 and at x = 7, however, only x = 7 is on the interval [1,10]. 5 3 2 Notice that at x = 7, the curves change relative position. In other words, on the interval [1,7], the curve y = - is above the line y = _, but on the interval [7,10], the line is above the curve. Therefore, to find the area of the region, we have to break up the region into two pieces. One piece corresponds to the interval [1, 7] and has area given by 3- 4 dx = (7In1x1 - x2 ) | = (7In171 - 72 ) - 7In|11 - 15 = 7In(7) - 48 14 and the other piece corresponds to the interval [7,10] and has area given by 6 4 - Zx = (42 - 7InIxl) 14 = (102 - 7In1 101 ) - (72 - 7In171) = 51 + 7In(7/10) 14 And finally, the total area is equal to 7In(7) - 48 + 51 + 7In(7/10) = 3/14 + 7In(49/10)After my uploading answer tidd this and the other piece corresponds to the interval [g, 10 ] and has area given by 10 X du - 10 18 18 [ 100- 817 - 9 [ log 10 - lugg ] 18 - g lag 10 + glog of - g log 10 + 9 lag 3 2 18 18 - g lug 10 + 18 log s And finally the total area i's equal to = 18 log 3 - to at 7 - 9 log 10 + 18 log 3 36 log 3 - 61 This box answer is the final Answer. sorry For late. please give a helpful Rate