What is the probability that this quadratic equation has real roots? (I've seen the other questions in this site similar to my problem, but they didn't help much). So, the problem is:

The numbers $B$ and $C$ are chosen at random between $−1$ and $1$,
  independently of each other. What is the probability that the
  quadratic equation $x^2+Bx+C=0$ has real roots?

I've actually gotten an answer, but it's wrong according to my book. Here's what I did: I know that the condition for that equation to have real roots is that $B^2-4C\ge 0\Rightarrow C\le\frac{B^2}{4}$. The sample space consists of all points $(x,y):-1\le x\le1,-1\le y\le1$. So I think the desired probability should be the ratio of the area a square of side-lenght $2$ to the area under $C\le\frac{B^2}{4}$ from $-1$ to $1$. 
That makes intuitive sense to me, but the book says the answer is actually $\frac{1}{4}(2+\int_{-1}^{1}x^2dx)$. Where does that $2$ come from? Thanks very much in advance.  

I am not sure if the answer given in the book is exactly correct or not. Because I am finding a different answer while I tried to compute the probability.  
What I attempted:-  The numbers $B$ and $C$ are chosen randomly from $[-1,1]$. So, we can model them as  uniform random variables taking values between $-1$ and $1$. If $b$ and $c$ denote the values taken by the random variables $B$ and $C$ respectively, then \begin{equation}  f_B(b)=\frac{1}{1-\left( -1\right)}=\frac{1}{2} \qquad -1\le b \le 1 
\end{equation}  and,\begin{equation}  f_C(c)=\frac{1}{1-\left( -1\right)}=\frac{1}{2} \qquad -1\le c \le 1  
\end{equation}  
As you have already mentioned, for real roots we must have $C\le \frac{B^2}{4}$. So, the only constraint is imposed upon $C$ and its value should be less than$\frac{B^2}{4} $. $B$ is free to vary in the range $(-1,1)$. So, the required probability is given by 
\begin{equation} 
\begin{aligned} 
P\left(C\le \frac{B^2}{4}\right) &=\int_{-1}^{1} \int_{-1}^{\frac{b^2}{4}} f_{BC}(b,c) dc db \\
&=\int_{-1}^{1} \int_{-1}^{\frac{b^2}{4}} f_B(b).f_C(c) dc db \qquad\mbox{(Since $B$ and $C$ are independent)}  \\
&=\int_{-1}^{1} \int_{-1}^{\frac{b^2}{4}} \frac{1}{4} dc db  \\
&=\frac{1}{4} \left(\int_{-1}^{1}\left(\frac{b^2}{4}+1 \right)db \right) \\
&=\frac{1}{4} \left(\frac{1}{4}\times \frac{2}{3} +2\right) \\
&=\frac{13}{24} 
\end{aligned}
\end{equation} 
I think you missed the factor $\frac{1}{4}$ just before the integral inside the parentheses. As you mentioned in the comment, there is a general formula when $B$ and $C$ lie in an interval $\left[-n,n\right]$. So, if you put $n=1$, the required probability is $=\frac{1}{2}+\frac{1}{24}=\frac{13}{24}$
