How to find the area of the region, bounded by various curves? 
Find the area of the region bounded by the curves $y=x^2$ and $y=x$.
Find the area of the region bounded by the curves $y=x^2+1$ and $y=2$

I have a ton of questions like this and I have been graphing them and then splitting them into intervals and adding them up but this is giving me an answer thats a little off and its taking forever....is there a faster way? Also I am stuck on $y=x^2+1$ and $y=2$
because I dont know what region they want..I see $y=2$ as a line intersecting $x^2+1$, when I graph it.
 A: HINT They ask for the area of the yellow region:

The areas would be given by integrals $\int_{x_1}^{x_2} \left(y_\text{top}(x) - y_\text{bottom}(x)\right) \mathrm{d} x$ with appropriate choices of boundaries $x_1$ and $x_2$ and functions $y_\text{top}(x)$ and $y_\text{bottom}(x)$.
A: First we will find the area of the region bounded by the curves:
$y = x^2$ ... (i)
and $y = x $ ... (ii)
To determine the shaded area between these two curves, we need to sketch these curves on a graph.

Now, we will find the area of the shaded region from O to A.
Area of Shaded Region Between Two Curves :
$A = \displaystyle \int _a^b [f(x)-g(x)] \;dx$
Where, $f(x)$ is the top curve
$g(x)$ is the bottom curve
$a$ (Lower limit) = $x$ coordinate of extreme left intersection point of the area to be found.
$b$ (Upper limit) = $x$ coordinate of extreme  right intersection point of the area to be found.
So,  $f(x) = y = x$
$g (x) = y = x^2$
We need to find the limits, $a$ and $b$.
How to find the limits ?
Since limits, $a$ and $b$, are the $x$ coordinates of the intersection points, So, we will find the intersection points of the given curves.
Put the value of $y$ from equation (ii) into equation (i)
$x=x^2$
$x^2-x=0$
$x(x-1)=0$
$ x=0, x = 1 $
Put these values in equation (ii)
$y = 0, \; y = 1$
Thus, the points of intersection are $O(0,0)$ and $A (1,1)$
$\therefore \;a=0, \;b=1$
Area between Curves :
The are will be,  $A = \displaystyle\int _a^b [f(x)-g(x)] dx$
$A=\displaystyle \int_0^1\; (x-x^2)\;dx$
$A=\displaystyle \int_0^1\; x\;dx-\displaystyle \int_0^1\; x^2\;dx$
$
=\left ( \dfrac {x^2}{2}\right)_0^1-\left ( \dfrac {x^3}{3}\right)_0^1
$
On putting limits,
$
=\left ( \dfrac {1}{2}-0\right)-\left ( \dfrac {1}{3}-0\right)
$
$=\dfrac {1}{2}-\dfrac {1}{3}$
$A=\dfrac {1}{6}$
(II) Now, we will find the shaded area bounded by the curves:
$y = x ^2 + 1$ ... (iii)
$y = 2$ ... (iv)
Curves on Graph :

We will find the area of the shaded region from $A$ to $B$
Here, 
$ f(x) = y = 2$
$g(x) = y = x^2+1 $
Finding the limits using intersection points :
Put the value of $y$ in equation (iii)
$2 = x^2+1 $
$ x^2 = 1 $
$ x = -1, x = 1$
Put these values in equation (iii)
$y = 2, y = 2 $
Thus, the points of intersection are $A (-1, 2)$ and $B (1, 2)$
$ \therefore \;a=-1, \;b=1$
Area between Curves :
The area will be, 
$A = \displaystyle\int _a^b [f(x)-g(x)]dx $
$A=\displaystyle \int_{-1}^{1}\; [2-(x^2+1)]\;dx$
$A=\displaystyle \int_{-1}^{1}\; 1\;dx-\displaystyle \int_{-1}^{1}\; x^2\;dx$
$=\left ( x \right)_{-1}^{1}-\left ( \dfrac {x^3}{3}\right)_{-1}^{1}$
On putting limits, we get,
$= (1+1)- \left( \dfrac {1}{3}+\dfrac {1}{3}\right)$
$=2-\dfrac {2}{3}$
$A=\dfrac {4}{3}$
A: The fastest way to find the area is to use integration. The area is the result of definite integral of the difference between the two functions.
The area bounded by $y=x^2+1$ and $y=2$ is shown below:

It is the area between the curve and the line of course.
First, we have 2 points that we need to find. The points are those of the intersection of the line and the curve. You can find the points by solving the equation:
$y=x^2+1$ 
and 
$y=2$.
You get
$y=x^2+1=2$ and this gives you the points:
$(-1,2)$ and $(1,2)$.
The area A is:
$A=\int_{x=-1}^{x=1} 2-(x^2+1) dx =4/3$.
