In the approximation of binomial distribution by a normal distribution, why should the range between which the area is found increase by one? That is, why is the following true?

 A: Let $X$ have binomial distribution, and let $W$ be the usual approximating normal. Let $k$ be an integer. 
The continuity correction is based on the belief, partly grounded in fact, that 
$$\Pr(X=k)\approx \Pr(k-1/2\lt W\lt k+1/2),\tag{1}$$
and that other possible approximations, such as $\Pr(k-1\lt W\lt k)$, are not as good.
Summing up (1) from $a$ to $b-1$, we get
$$\Pr(a\le X\lt b)\approx W(b-1/2)-W(a-1/2).$$
A: It's the continuity correction factor. Recall that the binomial distribution is discrete. If you imagine plotting a histogram of a particular binomial distribution, then fitting a normal curve to it, you will see that it is not a perfect match.
To compensate for this, in the case that you seek $$P(a\leq X<b)$$
you will have to subtract $.5$ to $a$ to 'capture' $a$ so that you will have $a\leq X$ and subtract $.5$ from $b$ to 'avoid' $b$ so that you will have $X<b$.
Here is an image I found here.

Let's say I want $P(3\leq X< 8)$. Notice that I will have to subtract $.5$ from $3$ to 'capture' (the approximation) $3\leq X$ and subtract $.5$ from $8$ to avoid $8$ (in the approximation to) $X<8$.
If I add $.5$ to $8$, then I am capturing $8$ and so I will have $X\leq 8$, which is not what I want.

Similarly, let's consider $P(8\leq X)$; again notice that to capture $8$, I have to subtract $.5$ from $8$ so that I will get an approximation to $P(8\leq X)$. This is the purple shaded region.
