Methods to determine integration limits? So this is one thing that I'm having some issues with. I don't understand how to determine integration limits.

First example:
  

At the above example the correct integration limit is $\int^2_0$ which just doesn't make sense. I mean yes $x=2$ does make sense, but where did they take $0$ from?

Second example: 

If someone has a good/safe way to determine limits, that would be great.
Thanks in advance.
 A: I both cases you have been given 3 curves that enclose your region.  (a line is a curve, the x axis is a curve.)
Sketch them as best you can.
Find the points of intersection between each pair of curves.
In the first one.

The picture does not need to be precise.  But it does help to know if the curves are generally upward sloping, downward sloping, convex, etc.
Where does y = 1 intersect f(x)?
$1 = \frac {2e^x}{1+e^x}\\
1+e^x = 2e^x\\
e^x = 1\\
x = 0$
your limits are $x\in [0,2]$
$V = \pi \int_0^2 (f(x))^2\ dx$
In the second one, you have already been provided with a picture.
Solve for $f(x) = 0, g(x) = 0, f(x) = g(x)$
$f(x) = 0\\
e^{2x} - 6e^x = 0$
We only need the find where the numerator equals 0.
$e^{2x} = 6e^x$
Take the log of both sides
$2x = x+\ln 6\\
x = \ln 6$ 
$g(x) = 0$ at $x = 0$
$f(x) = g(x)\\
\frac {e^{2x} -6e^x}{e^x +2} = 2 - 2e^x\\
e^{2x} -6e^x = (e^x +2)(2-2e^x)\\
e^{2x} -6e^x = -2e^{2x} -2e^x+ 4\\
3e^{2x} - 8e^x - 4 = 0\\
u = e^x\\
3u^2 - 4u - 4 = 0\\
(3u + 2)(u-2) = 0$
There are two solutions to the polynomial, but since $e^x>0,$ we can discard one.
$e^x = 2\\
x = \ln 2$
You are going to need to break this integral into two.  If you think about Riemann sums and the rectangles you need to add, you will have some rectangles with a vertical range from [f(x), 0] and some which will be [g(x), 0]
$\int_0^{\ln 2} (0-g(x)) \ dx + \int_{\ln 2}^{\ln 6} (0-f(x)) \ dx$
A: Think about what x is if you set y=0 in your equation? And graphically rotating the object will only yield the part between x=0 to x=2.
A: a) Point (0, 1) is the intersection of $f(x)$ and $y = 2$ graphs. The projection of the enclosed area onto the $x$-axis is the segment $[0, 2]$. Therefore, the integration limits are $\int_0^2$.
b) The enclosed area consists of the triangles $ABH$ and $BCH$.
$A (0, 0)$, $B(\log 2, -2)$, $C(\log 6, 0)$, $H (\log 2, 0)$.
Therefore:
$V = -\int\limits_{0}^{\log 2} g(x)\,dx - \int\limits_{\log 2}^{\log 6} f(x)\,dx.$

