#Calc Question
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Try substituting u = x^2 in the second integral, then replace u by x.
when u do this
the upper limit changes to 2 instead of 4
right
because doing that I get :
$$
\int^{16}_0 g(u) du = 16
$$
No, to 16.
u=x^2
x = 4 --> u =x^2 = 16
oh
what's next? only thing i am thinking of is something dumb
_super.star
Now try expressing the integral of g(x) from 0 to 4 from the first expression.