Compute $\lim_{x\to 0^+}\frac{A(1+x)}{x}$ 
I have no idea how to solve this using the hint. I do not have an answer to this question unfortunately but would you please help solving this question using the hint?
Edit: my friend told me that if you were to take the upper rectangle cut from the hint with height as f(1+x) to f(1) so (f(1) - f(1+x)) and width as x the two areas inside the rectangle cut by the curve is equal. Is that true? And does that make any difference to the problem?
 A: The hint is providing a bound on $A(1+x)$, since $f$ is monotone:
$$xf(1+x)\le A(1+x)\le xf(1)$$
Therefore, since $x>0$ in the limit:
$$f(1+x)\le\frac{A(1+x)}x\le f(1)$$
and as $x\to0$ the squeeze theorem implies that the limit in the question tends to $f(1)$.

The rectangle with width $x$ and height $f(1+x)-f(1)$ does not need to be cut into two equal parts by the curve, so it does not matter for the problem at hand.
A: Since $f$ is a continuous function, we have $$\lim_{x\rightarrow0^+}f(1+x)=f\left(\lim_{x\rightarrow0^+}(1+x)\right)=f(1)$$
Thus,
$$f(1)\leq\lim_{x\rightarrow0^+}\frac{A(1+x)}{x}\leq f(1),$$
which gives the answer.
A: You can write
$$
A(1+x)=\int_1^{1+x}f(t)\,dt
$$
and the mean value theorem for integrals tells you that
$$
A(1+x)=(1+x-1)f(c_x)=xf(c_x)
$$
for some $c_x\in(1,1+x)$. Therefore
$$
f(1+x)\le\frac{A(1+x)}{x}\le f(1)
$$
Actually, the hypothesis that $f$ is decreasing is redundant. Indeed, $A(1)=0$ and so
$$
\lim_{x\to0}\frac{A(1+x)-A(1)}{x}=f(1)
$$
by the fundamental theorem of calculus, which only needs that $f$ is continuous.
A: If you are in beginning integrals then you were conceptually correct in trying to "manipulate the areas".  The integral is a measure of the area under the curve, the curve being the space between some function's value and the x axis, and the area is figured by summing "pieces" which represent the function at particular points.  
This example uses rectangle pieces, the usual case, and sums them.  When you are using one piece the result is the rectangle from 1 to 1+x the height depends on where you are starting from.  For the case above you are looking for the upper limit so the estimate is from right to left and thus, according to the graph, the height is the minimum value in the region of interest. 
The result is that the upper limit for this function is a lot smaller than the area under the function while the lower limit from 1 to 1 + x would be much greater.  Now if you average these two areas you get something closer to the functions actual value.  
This is the "roughest" case of integration in the sense that the result is not very accurate except in pathologic cases.  In the process of integration you will decrease the size of the rectangles used for each sample and add the results to obtain better and better estimates. Finally the point is reached where the sample width is 1 + x as x->0 and you have the actual area - provided the limit exists.
I explained this as I believe it is more important to know why you want to do something then how to do it.  
I hope this helps
  Mark 
