# During the process of solving an integral why does the upper and lower bound change?

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.

• They don't have to. Simply Beautiful Art explains below why they can. After taking the integral, you can substitute back in for $u$ and use the original limits, if you'd like. That's usually how I do it. – The Count Jan 23 '17 at 1:38
• I was about to ask that +1 – dragonore Jan 23 '17 at 1:39
• @dragonore Sorry for not expanding :-/ – Simply Beautiful Art Jan 23 '17 at 1:41

The bounds change due to $u$. The original lower bound was $y=0$, so
$$u(\text{lower bound})=4y-3y^2+6y^3+1=4(0)-3(0)^2+6(0)^3+1=1$$
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.