Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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? enter image description here

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

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

4 Answers 4

up vote 10 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
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

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

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.