# Change of limits of integration

I'm a little bit confused about how the limits of integration are changed if we change the variable of integration. I'm not sure what exactly I don't understand, so I below I will write some false statements that seem true to me, and I ask to point out what exactly is wrong (and why).

For instance, $\int_0^1dx=1$, so $\int_0^1d(-y)=-\int_0^1dy=-1$. But $-y$ can be treated as a variable $\xi,$ so $\int_0^1d(-y)=\int_0^1d\xi=1$ by the first equality. What am I doing wrong?

Next, if $0\le t \le 1$, then $-1\le -t \le 0$. Thus shouldn't be true that $\int_0^1dt=\int_{-1}^0d(-t)$?

Does all of this have to do with "The Substitution Rule for Definite Integrals"? I wasn't able to relate it to these cases, since I don't see any derivatives which are used in that theorem.

$$\int\limits_0^1 dx \neq \int\limits_0^1 d(-y)$$

If you make a proper change of variables, you might say something like $x = -y$, in which case $dx = -dy$, further $x = 0$ to $x = 1$ becomes $y = 0$ to $y = -1$:

\begin{align} \int\limits_{x\ =\ 0}^{x\ =\ 1} dx = -\int\limits_{y\ =\ 0}^{y\ =\ -1}dy = -(-1 - 0) = +1 && \text{q.e.d.} \end{align}

Here's a slightly more complicated example:

$$\int\limits_0^8 xdx = \frac{1}{2}\left(8^2 - 0^2\right) = 32$$

Let's try $x = -y^3$. We find that $dx = -3y^2dy$ and that from $x = 0$ to $x = 8$ means $y = 0$ to $y = -2$:

\begin{align} \int\limits_0^8 xdx =&\ \int\limits_0^{-2}\left(\left(-y^3\right)\left(-3y^2\right)\right)dy \\ =&\ 3\int\limits_0^{-2}y^5dy = \frac{3}{6}\left((-2)^6 - 0^6\right) \\ =&\ \frac{1}{2}\cdot2^6 = 2^5 = 32 && \text{q.e.d} \end{align}

Suppose that you have the integral defined by $\int_a^b f(x)\,\mathrm dx$ and you make the change of variable $x=g(y)$ for some differentiable and injective $g$ in $[a,b]$, then you have that $\mathrm dx=g'(y)\,\mathrm dy$ and that if $x=a$ then $y=g^{-1}(a)$. Same for the $b$, that is, if $x=b$ then $y=g^{-1}(b)$, because $g$ is invertible in $[a,b]$.

Thus

$$\int_a^b f(x)\,\mathrm dx=\int_{g^{-1}(a)}^{g^{-1}(b)}(f\circ g)(y)\,g'(y)\,\mathrm dy$$

In your example you have the change of variable $x=-y$ thus $0=-y$ and $1=-y$.