Approximating $\pi$ using Monte Carlo integration I need to estimate $\pi$ using the following integration:
$$\int_{0}^{1} \!\sqrt{1-x^2} \ dx$$
using monte carlo 
Any help would be greatly appreciated, please note that I'm a student trying to learn this stuff so if you can please please be more indulging and try to explain in depth..

Let's also elaborate on Ross Millikan's answer, adapted to the case $f(x)=\sqrt{1-x^2}$, $0 \leq x \leq 1$. Suppose that $(X_1,Y_1),(X_2,Y_2),\ldots$ is a sequence of independent uniform vectors on $[0,1] \times [0,1]$, so that for each $i$, $X_i$ and $Y_i$ are independent uniform$[0,1]$ random variables. Define $Z_i$ as follows: $Z_i = 1$ if $X_i^2 + Y_i^2 \leq 1$, $Z_i = 0$ if $X_i^2 + Y_i^2 > 1$, so the $Z_i$ are independent and identically distributed random variables, with mean $\mu$ given by
$$
\mu = {\rm E}[Z_1] = {\rm P}[X_1^2 + Y_1^2 \leq 1] = {\rm P}\big[(X_1,Y_1) \in \lbrace (x,y) \in [0,1]^2 : x^2+y^2 \leq 1\rbrace \big] = \frac{\pi }{4}, 
$$
where the last equality follows from ${\rm P}[(X_1,Y_1) \in A] = {\rm area}A$ ($A \subset [0,1]^2$).
By the strong law of large numbers, the average $\bar Z_n  = \frac{{\sum\nolimits_{i = 1}^n {Z_i } }}{n}$ converges, with probability $1$, to the expectation $\mu$ as $n \to \infty$. That is, with probability $1$, $\bar Z_n  \to \frac{\pi }{4}$ as $n \to \infty$. 
To get a probabilistic error bound, note first that the $Z_i$ have variance $\sigma^2$ given by
$$
\sigma^2 = {\rm Var}[Z_1] = {\rm E}[Z_1^2] - {\rm E}^2{[Z_1]} = \frac{\pi }{4} - \Big(\frac{\pi }{4}\Big)^2 = \frac{\pi }{4} \Big(1 - \frac{\pi }{4}\Big) < \frac{10}{59}.
$$
The average $\bar Z_n $ has expectation ${\rm E}[\bar Z_n] = \mu$ and variance ${\rm Var}[\bar Z_n] = \sigma^2 / n$; hence, by Chebyshev's inequality, for any given $\varepsilon > 0$,
$$
{\rm P}\big[\big|\bar Z_n - \mu \big| \geq \varepsilon \big] \leq \frac{{\sigma^2}}{{n \varepsilon ^2 }},
$$
and so
$$
{\rm P}\bigg[\bigg|\bar Z_n - \frac{\pi }{4}  \bigg| \geq \varepsilon \bigg] < \frac{{10}}{{59n\varepsilon ^2 }}.
$$
