How do you integrate #int x/(sqrt(1+2x^2)dx# from [0,2]?

1 Answer
Steve M
Mar 7, 2017

# int_0^2 \ x/(sqrt(1+2x^2}) \ dx = 1 #

Explanation:

We want to find the value of the definite integral:

# int_0^2 \ x/(sqrt(1+2x^2}) \ dx #

We can integrate using a substitution:

Let # u=1+2x^2 => (du)/dx = 4x #

So when we substitute we will have #int \ ... x\ dx = int \ 1/4 ... \ du#, and as we have changed the variable of integration we need to change the limits of integration to match:

When #{ (x=0), (x=2) :} => { (u=1), (u=9) :}#

Substituting into the original integral we get;

# int_0^2 \ x/(sqrt(1+2x^2}) \ dx = int_1^9 \ (1/4)/(sqrt(u) \ du #
# " " = 1/4 \ int_1^9 \ u^(-1/2) \ du #
# " " = 1/4 \ [u^(1/2)/(1/2)]_1^9 #
# " " = 1/2 \ [u^(1/2)]_1^9 #
# " " = 1/2 ( sqrt(9)-sqrt(1) ) #
# " " = 1/2 ( 3-1 ) #
# " " = 1 #

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