How do you find the definite integral for: #(1) / (sqrt(1 + x))# for the intervals [0,3]?

1 Answer
Aug 21, 2016

#2#

Explanation:

We have:

#int_0^3 1/sqrt(1+x)dx#

Rewrite:

#=int_0^3(1+x)^(-1/2)dx#

We can use substitution here: let #u=1+x#. Fortunately, this means that #du=dx#.

Before performing the substitution, recall that the bounds will also change--plug the current bounds into #1+x#. Thus, the bound of #0# becomes #0+1=1# and #3rarr4#.

#=int_1^4u^(-1/2)du#

Integrate using #intu^ndu=u^(n+1)/(n+1)+C#, where #n!=-1#.

#=[u^(-1/2+1)/(-1/2+1)]_1^4=[u^(1/2)/(1/2)]_1^4=[2sqrtu]_1^4=2sqrt4-2sqrt1=4-2=2#