Area under the curve (Integration) For this question: 

I have found the limits to be $(-k)^{\frac{1}{2}} , 0 , k^{\frac{1}{2}}$
but the trouble I'm having is how do I know what graph I should minus from each other. So for area a , do I do $f(x) - g(x)$ , or $g(x) - f(x)$.
Is there a trick$?$
 A: The area bound by two curves, $f(x),g(x)$ on $[a,b]$ is $$Area=\int_a^b|f(x)-g(x)|dx$$ In order to evaluate this integral, you need to do something to get rid of the absolute values (usually) and so the way we accomplish this is by using the piecewise definition of $|\cdot|$. We break this integral up into sub-integrals based on the points at which $|\cdot|$ changes sign. After identifying the points $\{c_1,\cdots,c_n\}$ at which $|f(x)-g(x)|$ changes sign, we can write $$\int_a^{c_1}|f(x)-g(x)|dx+\int_{c_1}^{c_2}|f(x)-g(x)|dx+\cdots+\int_{c_n}^b|f(x)-g(x)|dx$$
Now, when we look at these integrals, we know that the sign of $f(x)-g(x)$ is constant for each integral, so we can drop the $|\cdot|$ by simply making sure the functions are being subtracted in the right order, i.e. the order that makes the difference positive. You've already broken it up correctly, so you just need to identify which function is larger on which interval and subtract the smaller function from the larger one. From just looking at the graph, you simply pick the one higher up on the y-axis.
A: Notice that 
$$
g(x)=x^3-kx+f(x),
$$
therefore the equation 
$$
f(x)=g(x)\tag{1}
$$ 
is equivalent to 
$$
x^3-kx=0 \tag{2}
$$
The solutions of (2) are 
$$
-\sqrt{k},0,\sqrt{k}.
$$
Because $f(x)\ge g(x)$ for $-\sqrt{k}\le x\le 0$, and $g(x)\ge f(x)$ for $0\le x\le \sqrt{x}$, we have
$$
A=\int_{-\sqrt{k}}^{0}[g(x)-f(x)]\,dx,\quad B=\int_{0}^{\sqrt{k}}[f(x)-g(x)]\,dx.
$$
Since $g(x)-f(x)=x^3-kx$ is an odd function, we have
\begin{eqnarray}
0&=&\int_{-\sqrt{k}}^{\sqrt{k}}[g(x)-f(x)]\,dx\\
 &=&\int_{-\sqrt{k}}^{\sqrt{k}}[g(x)-f(x)]\,dx\\
&=&\int_{-\sqrt{k}}^{0}[g(x)-f(x)]\,dx+\int_{0}^{\sqrt{k}}[g(x)-f(x)]\,dx\\
&=&\int_{-\sqrt{k}}^{0}[g(x)-f(x)]\,dx-\int_{0}^{\sqrt{k}}[f(x)-g(x)]\,dx\\
&=&A-B
\end{eqnarray}
It follows that
$$
A=B=\int_{0}^{\sqrt{k}}(kx-x^3)\,dx=\frac{1}{2}kx^2-\frac{1}{4}x^4\Big|_{0}^{\sqrt{k}}=\frac{1}{4}k^2
$$
