Why is this following integral not zero ? I am given the integral 
$$\int_{-2}^2 \sqrt{4-x^2} dx$$ i know that the answer should be $2\pi$ however, $f(x) = \sqrt{4-x^2}$ is an even function and so it should give zero since $f(x) = f(-x)$ but why is it not the case here ? any suggestions
 A: Even functions (like this one) over an interval symmetric around zero can give nonzero answers.  It is odd functions that integrate to zero over an interval symmetric around zero.  If $f(x)=-f(-x)$, then $\int_{-a}^af(x)\ dx=\int_{-a}^0f(x)\ dx+\int_{0}^af(x)\ dx=-\int_{0}^af(x)\ dx+\int_{0}^af(x)\ dx=0$
A: Thank you Integrator for that geometric demonstration. Algebraically though, we know even functions have the property $f(x) = f(-x)$ so: $$\int_{-a}^a {f(x)}\,dx = \int_{-a}^0 {f(x)}\,dx + \int_0^a {f(x)}\,dx $$ but noting that the first component of that sum can be expressed as:$$\int_{-a}^0 {f(-x)}\,dx$$ we then use $u=-x,\, du = -dx$ and change the limits accordingly ($0\to0,\,-a\to a$) to get:$$\int_{-a}^a {f(x)}\,dx = -\int_a^0 {f(u)}\,du + \int_0^a {f(x)}\,dx $$ which equates to: $$\int_0^a {f(x)}\,dx + \int_0^a {f(x)}\,dx =2\,\int_0^a{f(x)}\,dx.$$ So really, the definite integral you are trying to evaluate should equate to twice the region in the first or second quadrant bounded by the graph and the two axes.Ross Millikan demonstrates the other case, the zero integral for an odd function. 
A: The easiest way to find the area is to consider rewritten $y = \sqrt{4 - x^{2}}$ as 
$x^{2} + y^{2} = 4$. This is obtained by squaring both sides $y = \sqrt{4 - x^{2}}$.
Then if you notice,   the graph of  $y = \sqrt{4 - x^{2}}$ represents a semicircle centered at the origin with radius $2$. 
Hence the integral represents the area of a semicircle. Then in general the area of a semicircle is $A = \frac{\pi(r^{2})}{2}$.
Then $\int_{-2}^{2} y = \sqrt{4 - x^{2}} =  \frac{\pi(2)^{2}}{2} = 2\pi$
In general 
Suppose $f$ is continuous on $[-a, a]$.
$(i)$ if $f$ is even $[f(-x) = f(x)]$, then $\int_{-a}^{a} f(x)dx = 2\int_{0}^{a} f(x)dx $.
$(ii)$ if $f$ is odd  $[f(-x) = -f(x)]$, then $\int_{-a}^{a} f(x)dx = 0$. 
Then you can verify $y = \sqrt{4 - x^{2}}$ is not an odd function. Thus its integral cannot be equal to zero.
