How do you find the definite integral of #x/sqrt(2x+1)# from #[0,4]#?
1 Answer
Oct 26, 2016
Use substitution to evaluate
Explanation:
Let
Let
And it also makes
Furthermore, when
when
Substitute:
# = 1/4[2/3 u^(3/2)-2u^(1/2)] = 1/6sqrtu(u-3)#
# = {: 1/6sqrtu(u-3)]_1^9 = 3/6(6) - (-2)/6 = 20/6 = 10/3#