00:01
Hello students, from question we have given alpha.
00:04
Now dn by dm is km to the power minus alpha.
00:10
So we can find dn that is km minus alpha dm.
00:16
Now integrating this term with limit to 0 to n, dn will be equal to 0 to 1 km dm plus 1 to m km minus alpha dm.
00:34
Now further can be written as 0 to 1 km dm plus 1 to m km minus 2 .35 dm.
00:49
So km with limit 0 to 1 plus km minus 1 .35 divided by minus 1 .35 1 to m.
01:02
Now k plus km minus 1 .35 divided by minus 1 .35 minus k upon minus 1 .35.
01:16
Now taking k as common so 1 plus m minus 1 .35 divided by minus 1 .35 plus 1 upon 1 .35.
01:28
Now further this can be written as k 1 .74 minus 0 .74 m to the power minus 1 .35.
01:37
Now taking this equation 1 so we can write k will be equal to n divided by 1 .74 minus 0 .74 m to the power minus 1 .35.
01:53
Now you can take this equation 2.
01:57
Now next for part b again so integration of 0 to n dn will be equal to 0 to m limit km minus alpha dm that will be equal to integration of zm naught to m k this will be m km power minus 2 .35...