$$\int_{-1}^{1} \sqrt{1-x^2}dx$$

I let $u = 1-x^2$, $x = (1-u)^{1/2}$

$du = -2x dx$

$$-\frac{1}{2}\int_{0}^{0} \frac{u^{1/2}}{(1-u)^{1/2}} = 0$$ because $$\int_{a}^{a} f(x)dx = 0$$

But it isn't zero. Why?

  • 1
    $\begingroup$ Under what conditions can you use u substitution? $\endgroup$ – Simply Beautiful Art Feb 18 '17 at 16:54
  • $\begingroup$ It is not zero because the integrand is strictly positive. Edit: And continuous. $\endgroup$ – Git Gud Feb 18 '17 at 16:54
  • $\begingroup$ f and g' are continuous.. i see now $\endgroup$ – user349557 Feb 18 '17 at 16:56
  • $\begingroup$ Very closely related question (and my answer): math.stackexchange.com/questions/1489577/… $\endgroup$ – Cameron Williams Feb 19 '17 at 4:16
  • $\begingroup$ @CameronWilliams Duplicates, really. $\endgroup$ – MathematicsStudent1122 Feb 19 '17 at 4:21

For $x \in [-1,0)$, you can't have $$ u = 1-x^2 \implies x = (1-u)^{1/2} $$ thus your change of variable is not valid over $[-1,0)$.

You would better write by parity $$ \int_{-1}^{1} \sqrt{1-x^2}dx=2\int_0^{1} \sqrt{1-x^2}dx $$ then use the given change of variable.


You may only do a u-substitution when it is bijective on your integration domain. You can solve this problem by breaking up your integral into $\int_{-1}^0$ and $\int_0^1,$ and using $x=-(1-u)^{1/2}$on the first integral, and $x=(1-u)^{1/2}$ on the second.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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