Derivation of wave equation and Newton's 2nd Law I am a mathematics student studying the derivation of the wave equation. In the derivations I have been reading, Newton's 2nd law is applied to a string, I'm having trouble making sense of it.
Newton's 2nd law applied to a rigid body makes intuitive sense. If several forces are applied to a box, I can imagine the forces "distributed" through the body and how they would combine to move the box.

In deriving the wave equation, we assume that we have a string under tension which has been "plucked".

Typically, the author then applies Newton's 2nd Law $\vec{F}=m\vec{a}$ to a segment of the string.

Either a differential element of the string is used, or the author integrates over a length of string. 
If a differential element is used, the author seems to apply the 2nd law as if the string were a rigid body. An example of this approach is seen in this presentation (slide 4).
On the other hand, the author might integrate over a length of string. In this case, I have seen it offered (without explanation) that the integral of tension over a segment of string is simply the difference of the tension at the end points:
$$ \vec{F} = \int_{s_1}^{s_2}\vec{T}(s,t)ds = \vec{T}(s_2,t) - \vec{T}(s_1,t) $$
This approach is seen in this book (pdf page 13). For this to be the case, tension must "cancel" along the segment, except at the end points. I can see this making sense, but a little more explanation as to why would be incredibly helpful.
I hope I have properly expressed my confusion. Thanks for any advice.
 A: *

*String is assumed to be of constant mass density, if you draw FBD of a segment of length $ds$ then tension will be almost same on both side hence it cancels. But the end points does not have opposite direction tension to cancel. So net tension exists at end points only. 

*Also keep in mind that the wave-equation that we derive is only for very small oscillations- similar to SHM approximation.

A: Saying that a string is under tension means that for all points X on the string, the string to the left of X is pulling the string to the right of X, and the string to the right of X is exerting an equal and opposite force on the string to left of X. 
Now consider a string ABCDE, and suppose we are applying N's second law to the portion BD. Taking B as our point X, we clearly have a force from outside BD acting on BD. Similarly, taking D as our point X, we have another external force on BD. But if we take C as our point X, the force on BC comes from CD, and the force on CD comes from BC, so these equal and opposite forces are internal to BD and not the external forces whose resultant gives ma for BD.
