Finding the area between three curves in which one is a piecewise function Let $f(x)$ be a continuous function given by $$f(x)=\begin{cases} 2x,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |x|\leq1\\ x^2+ax+b, \ |x|>1\end{cases}$$
Find the area of the region in the third quadrant bounded by the curves $x=-2y^2$ and $y=f(x)$ lying on the left of the line $8x+1=0$. 
What I did: First of all, lets find the values of $a$ and $b$. So $\lim_{x\to-1}f(x)=-2$ and $\lim_{x\to1}f(x)=2$. Thus, $1-a+b=-2$ and $1+a+b=2$. Solving, we get $a=2$ and $b=-1$.
So $$f(x)=\begin{cases} 2x,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |x|\leq1\\ x^2+2x-1, \ |x|>1\end{cases}$$
 
So $f(x),x=\frac{-1}{8},x=-2y^2$ intersect at $(\frac{-1}8,\frac{-1}{4})$. 
So area would be- $$\int\limits^{\frac{-1}8}_{-1} 2x-\sqrt{\frac{-x}2}dx+\int\limits^{-1}_{-2}x^2+2x-1-\sqrt{\frac{-x}2}dx$$
Now putting $-x=t$, and solving we get $\frac{-185}{192}-3$. Area will be positive, so answer is $\frac{761}{192}$. But according to my textbook, answer is $\frac{257}{192}$. Please explain what mistake did I commit?

It should be
$$\int\limits^{\frac{-1}8}_{-1} \left(-\sqrt{\frac{-x}2}-2x\right)\,dx+\int\limits^{-1}_{-2} \left( -\sqrt{\frac{-x}2}-(x^2+2x-1)\right) \,dx$$
The curve $x=-2y^2$ is on top in both of the two parts, so should be the first term. Also you are picking the branch of it where $y$ is negative. 
