# Where is my argument that $\int_{-1}^{1} \sqrt{1-x^2}dx=0$ wrong?

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

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?

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

## 2 Answers

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.