# Integration by substitution, why do we change the limits?

I've highlighted the part I don't understand in red. Why do we change the limits of integration here? What difference does it make?

Source of Quotation: Calculus: Early Transcendentals, 7th Edition, James Stewart

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Just because you changed variable. When $x=1$ with $u=1+9x/4$, what is $u$ ? –  Claude Leibovici Jul 16 '14 at 7:18

Because the function has changed. Let's do an example:

$$\int_{-1}^1 x\,dx =0$$

because the integrand is odd and the interval is symmetric (you can also check directly).

Let's do a simple $u=x+1$ so that $du=dx$ then the right way we have:

$$\int_0^2 u-1\,du = {1\over 2}(u-1)^2\bigg|_0^2={1\over 2}-{1\over 2}=0$$

but if we fail to change the limits:

$$\int_{-1}^1(u-1)\,du = {1\over 2}(u-1)^2\bigg|_{-1}^1=-2$$

The underlying reason is that integration comes from Riemann sums, the function values depend on the interval of integration. When you change the interval, the heights of the rectangles you use in the definition change (remember the heights are the function values) so that you end up adding up different things if you don't change the function to compensate.

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No, we don't change the limits. Who said that. They stay the same. Only their representation changes.

$$L = \int\limits^{x=4}_{x=1}\sqrt{1 + \dfrac{9}{4}x}dx$$

Now we substitute $u=1+\dfrac{9}{4}x$:

$$L = \int\limits^{x=4}_{x=1} \sqrt{u} dx$$

See? We didn't change anything. We only substituted $u$ for the $x$ expression in the integrand. But now, it contains mixed variables and not solvable as is. We need to substitute $u$ expressions for other $x$ expressions too.

$$L = \int\limits^{x=4}_{x=1} \sqrt{u} dx = \int\limits^{\dfrac{4}{9}(u-1)=4}_{\dfrac{4}{9}(u-1)=1} \sqrt{u} d\left( \dfrac{4}{9}(u-1) \right)$$

Now, all the variables are in $u$, and the integral is solvable.

$$\dfrac{4}{9}(u-1)=4 \implies u=10, \quad \dfrac{4}{9}(u-1)=1 \implies \dfrac{13}{4} \\ d\left( \dfrac{4}{9}(u-1) \right) = \dfrac{4}{9}du$$

We rewrite the integral expression:

$$L = \int\limits^{\dfrac{4}{9}(u-1)=4}_{\dfrac{4}{9}(u-1)=1} \sqrt{u} d\left( \dfrac{4}{9}(u-1) \right) = \dfrac{4}{9} \int\limits^{u=10}_{u=\dfrac{13}{4}} \sqrt{u} du$$

Have we ever changed the limits in any step? No! They are the exact same limits. There only represented by $u$ instead of $x$ in the final expression.

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I get your point (and I agree that this is more or less the proper way to think about substitution in integrals), but it is, at the very least, really confusing to say that the limits don't change. I would word it differently. –  David Z Jul 16 '14 at 17:29
I'm with @DavidZ, the limits change to $u$ limits, which while representing the same region for $x$ limits, are different. –  Adam Hughes Jul 16 '14 at 23:01

By the substitution $x=\varphi(t)$ we have

$$\int_a^b f(x)dx=\int_{\varphi^{-1}(a)}^{\varphi^{-1}(b)}f(\varphi(t))\varphi'(t)dt$$

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The original limits is for variable $x$, and the new limits is for the new variable $u$. If you can get a primitive function $$\frac{8}{27}(1+\frac{9}{4}x)^\frac{3}{2}$$ by observation, it is unnecessary.

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