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How to solve $ \int_{-1}^{1} x^2 \,dx $ using u-substitution? Of course, I know how to solve this using the standard way ($\frac{x^3}{3}$) but when I use u-sub I get $u=x^2, du=2xdx$. For the new bounds in terms of $u$, I plug in $x =-1,1$ in $u=x^2$ and get $u=1,1$ for both bounds. I don't even compute the rest of the new integral in terms of u because no matter what, since the bounds are $1,1$, the integral computes to zero by the zero "width" rule (as called in my textbook). But this answer is incorrect.

I feel like I'm missing something huge.

beginner
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    "I don't even compute the rest of the new integral in terms of u" - This is a red flag. I'd dare you to try to finish that substitution (that would be where your error is). That $du = 2x, dx$ is not nothing. Arguably you have shown $\int_{-1}^1x^{\color{red}3},dx=0$, which is true. – Brian Moehring Apr 17 '21 at 02:42
  • I'm not sure how you are doing this with u-substitution. Can you be more specific? If you did $u = x^2$ and du = $2x,dx$, what is your actual integral? Becuase $u,du$ would not be equivalent to $x^2,dx$ (it's equivalent to $2x^3,dx$). – johnnyb Apr 17 '21 at 02:43
  • What I'm trying to say is that the new bounds for the integral in terms of u is 1 and 1. Doesn't that imply that the definite integral is always zero since the bounds are the same number (the "width" is zero)? – beginner Apr 17 '21 at 02:46
  • Related: https://math.stackexchange.com/questions/829939/why-does-this-u-substitution-zero-out-my-integral (and the linked questions there). – Hans Lundmark Apr 17 '21 at 07:34

4 Answers4

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Your question is incomplete without showing how you might even "compute the rest of the new integral in terms of $u$". I tried to point this out in the comments, but you haven't done so, so I will have to guess what you're thinking.

It's possible you mean something like:

$$u = x^2 \implies x = \begin{cases}\sqrt{u}, & \text{ if } x \geq 0 \\ -\sqrt{u}, & \text{ otherwise}\end{cases} \\ dx = \begin{cases}\frac{1}{2\sqrt{u}}\,du, & \text{ if } x \geq 0\\ -\frac{1}{2\sqrt{u}}\,du, & \text{ otherwise}\end{cases}$$

In other words, this substitution gives you

$$\int_{-1}^1 x^2 \,dx = \int_{-1}^0 x^2\,dx + \int_0^1 x^2\,dx = \int_{1}^0 u\left(-\frac{du}{2\sqrt{u}}\right) + \int_0^1 u\left(\frac{du}{2\sqrt{u}}\right)$$ which if we want, we can recombine to $$\int_0^1\sqrt{u}\,du$$ but we still don't get $0$.

Brian Moehring
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  • Thank you very much for your detailed response. (I was in the middle of carrying out the substitution, by the way.) – beginner Apr 17 '21 at 03:02
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Whenever you use substitution with non injective functions you should split your domain.

Animesh Saxena
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The substitution $u = x^{2} $would give you

\begin{equation} I = \int_{1}^1 u\cdot \pm2\sqrt{u}du \end{equation}

because $2x$ can be both $+2\sqrt{u}$ and $-2\sqrt{u}$. I don't think you can compute this because you don't know the sign of the function. To integrate $I$ you should probably use the substitution $v^2 = u$, which should give you back the original integral.

Hope this helps.

Gnayus
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The function is not continuous at x=0. Hence you can't put it in limits. And using the limit concept, taking limits from -1 to b and a to 1, where b is just less than 0 and a is just greater than 0. By solving the limit the integral diverges.

Phyzard
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