How do you find the definite integral of #x/sqrt(2x+1)# from #[0,4]#?

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Answer 1 · 2016-10-26T13:10:15

Use substitution to evaluate #int x/sqrt(2x+1) dx#. Then evaluate from #0# to #4#.

Let #int x/sqrt(2x+1) dx#

Let #u = 2x+1#. This makes #du = 2 dx#, so #dx = 1/2 du#

And it also makes #x = 1/2(u-1)#.

Furthermore, when #x=0#, we have #u=1# and

when #x = 4#, we get #u = 9#

Substitute:

#int x/sqrt(2x+1) dx = int (1/2(u-1))/(sqrtu) * 1/2 du = 1/4 int (sqrtu - 1/sqrtu) du#

# = 1/4[2/3 u^(3/2)-2u^(1/2)] = 1/6sqrtu(u-3)#

#int_0^4 x/sqrt(2x+1) dx = 1/4 int_1^9 (sqrtu - 1/sqrtu) du#

# = {: 1/6sqrtu(u-3)]_1^9 = 3/6(6) - (-2)/6 = 20/6 = 10/3#