This is the problem I am given. Notice the upper bound is a 1, but whenever I see the solution later on, it changes to an 8 and the lower bound to 1 and I'm not sure why that happens.
$
\int_0^1 (4y-3y^2+6y^3+1)^\frac{-2}{3}(18y^2-6y+4y)dy
$
Obviously you do u substitution here
$
u = 4y-3y^2+6y^3+1 \\
du = (18y^2-6y+4)dy\\
$
This is the part I don't get below.
My interactive tutorial shows I need to do this next
$
\int_1^8(u)^\frac{-2}{3}du
$
Why does the upperbound and lowerbound suddenly change.
Note: I'm not looking for the answer to the integral, but rather an explanation as to why the limits of integration change.
Best Answer
The bounds change due to $u$. The original lower bound was $y=0$, so
$$u(\text{lower bound})=4y-3y^2+6y^3+1=4(0)-3(0)^2+6(0)^3+1=1$$
which is the lower bound. Do the same process and you will get the upper bound.
As @TheCount mentions, this is because when one uses the substitution, the end result is a function of $u$. You then substitute $u=f(y)$ back into it, and then apply the bounds.
Or you can apply the bounds first and not worry about re-substitution later.