How to understand this step in the proof for integral comparison test? In a proof for the integral test, where $a_n = f(n)$ and $f$ is a continuous decreasing positive function, one of the steps claims:

$$a_1 +  \sum^N_{n=2} a_n \le a_1 + \sum_{n=2}^N \int^n_{n-1} f(x) \, dx$$ 

Why is this necessarily true? 
Does not the left hand side of the inequality takes a left hand Riemann sums from $2$ to $N$, which is larger than the integral of the same span. Why does the sum of the integrals on the right hand side, which starts from $1$ instead of $2$, necessarily become larger? 
From section $3.3$ of "Infinite Series" by Keith Conrad.
 A: hint
We have that
$$(\forall n>1)\;\;  (\forall x\in [n-1,n] )$$
$$ a_n=f (n)\leq f (x)$$
cause $f $ is decreasing at $[n-1,n] $.
$f $ is continuous 
thus by integration from $n-1$ to $n $, we get
$$\int_{n-1}^n a_n dx\leq \int_{n-1}^nf (x)dx $$
or
$$(n-(n-1))a_n\leq \int_{n-1}^n f (x)dx $$
and by sum from $2$ to $N $, you finish.
A: So we have to prove this:
$\sum_{a}^{b}{f(x)} < \int_{a-1}^{b}{f(x).dx}$
Since the integration starts from $a-1$, this inequality is correct.
Take a look at the diagram above(a = 2):
As you can see, the rectangles(the sigma part) is always under the curve.
A: Instead of viewing the left hand side of the inequality as a left hand Riemann sum from $2$ to $N$, view it as a right hand Riemann sum from $1$ to $N$.  This is because on the right hand side of the inequality, you are integrating from $n-1$ to $n$, which means you are really integrating starting from $n=1$.
The right hand Riemann Sum of a decreasing function is smaller than the integral of the same span.
A: The question seems to be: Why is $\displaystyle a_n = f(n) \le \int_{n-1}^n f(x)\,dx\text{ ?}$
The function is decreasing. That implies that:
$$
\text{if } n-1\le x\le n \text{ then } f(x) \ge f(n).
$$
Therefore $\displaystyle \int_{n-1}^n f(x)\,dx \ge \int_{n-1}^n f(n)\,dx = f(n) = a_n.$
