Evaluate the integral I = int_0^3 \ xf(x^2) \ dx ?

1 Answer
Jan 26, 2018

int_0^3 \ xf(x^2) \ dx = 3/2

Explanation:

We seek:

I = int_0^3 \ xf(x^2) \ dx

We can perform a substitution:

Let u=x^2 => (du)/dx = 2x

And the substitution will require a change in limits:

When x={ (0),(3) :} => u={ (0),(9) :}

Substituting into the integral, changing the variable of integration from x to u we get:

I = 1/2 \ int_0^3 \ 2xf(x^2) \ dx

\ \ = 1/2 \ int_0^9 \ f(u) \ du

\ \ = 1/2 \ (3) , as int_0^9 \ f(u) \ du = int_0^9 \ f(x) \ dx

\ \ = 3/2