How doI find the average induced emf in a coil given the rate of change of flux density, and the area of the coil? The magnetic flux density changes from +10 T to -10T in 5 seconds. The area of the coil is 2.5 m^2. What is the average emf induced? 
By Faraday's law, this will be equal to the change in magnetic flux linkage / the time taken for this change to occur. But the net change in magnetic flux linkage is 0, so there should be no 'net' emf induced.. Intuitively, however, I know that there will be some non-zero emf induced because the flux in the circuit is changing. 
 A: Let's call $B$ the value of the magnetic field, and let's assume that
$$ B(t) = B(0) -\alpha t$$
where, here, $B(0) = 10\,T$ and $\alpha = 4\,T\cdot s^{-1}$. Then the flux of $B$ through the coil, whose area is $A = 2.5\,m^2$, is
$$\Phi(t) = A(B(0) - \alpha t)$$
Then, Faraday's law tells us that, with the appropriate orientation, this causes an electromotive force $e$ where
$$ e = -\frac{\textrm{d}\Phi}{\textrm{d}t} = A\alpha$$
If you only know that $B(0) = 10\,T$ and $B(t=5\,s)=-10\,T$, then
$$ \langle e \rangle = \frac{1}{T} \int\limits_0^T e\,\textrm{d} t = \frac{1}{T}\left[ -\Phi\right]_0^T = \frac{A(B(0) - B(T))}{T}$$
In both cases, the result is
$$ \langle e \rangle = 10\,V$$
A: The net change is $-10-(10)$, that's $-20$ not zero. I guess that should clear your confusion. The rest of the answer I'm giving shows some of the steps/ideas to solve this problem.
Steps
magnetic flux density = magnetic field strength
i.e. two values of $\vec{B}$ are given.
Magnetic flux  is defined as $\phi_B = \int \vec {B} \cdot d\vec{A}$, $\vec{A}$ is the area, whose vector points perpendicular to it's surface, "the normal".
Your question deals with a net change in $\vec{B}$ and a constant value for $\vec{A}$, and change in the flux density is one dimensional (i.e. no change in direction just magnitude), so we can drop the vector symbols and change the $ \int \dots d\vec{A}$  to $A$. The dot "$\cdot$" represents scalar product between vectors by taking $\cos \theta_{AB}$ where $\theta_{AB}$ is the angle between the field and the area normal, which in your case seems to be $0$ so, $\cos \theta_{AB} = 1$ 
Your case simplifies to this:
$$\begin {array}{l}
{\Delta \phi_B} = \Delta B A \cos \theta_{AB}= \Delta B A \\
\Delta B = -20, \Delta t= 5, A = 2.5 \\
\mbox{induced emf, } \mathcal{E} =-\dfrac{ \Delta \phi_B}{\Delta t}
\end {array}$$
Now you can calculate it on your own. 
