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.

  • 3
    $\begingroup$ 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. $\endgroup$ – The Count Jan 23 '17 at 1:38
  • $\begingroup$ I was about to ask that +1 $\endgroup$ – dragonore Jan 23 '17 at 1:39
  • $\begingroup$ @dragonore Sorry for not expanding :-/ $\endgroup$ – Simply Beautiful Art Jan 23 '17 at 1:41

The bounds change due to $u$. The original lower bound was $y=0$, so

$$u(\text{lower bound})=4y-3y^2+6y^3+1=4(0)-3(0)^2+6(0)^3+1=1$$

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.


The idea is that when you use u-sub, you are changing the curve generated by the function, but you want your definite integral to have the same area under some portion of the new curve as the old one. Shifting the bounds around is what achieves that.


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