Area under curve: integration I am a student. I was recently taught about application of integrals
case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area under curve is calculated using integration


*

*$\text{area} = \int y\,\text dx$ with some limits 


Similarly:
case 2:
When the curve $y=f(X)$ lies below the $X$ axis 
The area under curve is calculated using integration


*

*$\text{area} = \int -y\,\text dx$ with some limits 



So my question here is :
If we have to calculate area bounded by parabola $y^2= 16x$ and its latus  rectum we find the area above the $X$ axis by using the  equation and multiply it by 2
The equation here represents the parabola and integrating the equation with proper limits should give the area of all of the required region 
That is region above and below the $X$ axis 
Then why we multiply the result of integration by 2.
 A: When they say

case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area under curve is calculated using integration
  
  
*
  
*$\text{area} = \int y\,\text dx$ with some limits
  

They actually mean

case 1:
When curve $y=f(X)$ lies above the $X$ axis
the area between the curve and the $x$-axis is calculated using integration
  
  
*
  
*$\text{area} = \int y\,\text dx$ with some limits
  

So when you're integrating you're only getting half the area:

and naturally, you need to multiply that by $2$ to get all of it.
Also, remember that the actual integrand in question here is not $y^2 = 16x$, because that's not of the form $y = f(x)$ as requested from the cases. It's actually $y = 4\sqrt{x}$, which is only the upper half of the parabola I've sketched above. That makes it a bit more obvious why you don't get all of it; when integrating $y = 4\sqrt x$, your expressions have no way of knowing that it is only half of something bigger, and it especially can't know that the other half is the exact mirror image. So it does the best it can and finds the area down to the $x$-axis instead.
A: The curve $y^2=16x$ does not represent a function $y=f(x)$ (it fails the vertical-line test), but it can be broken down into two curves $y=4\sqrt x$ and $y= -4\sqrt x$. Each of these is a function: $g(x) = 4\sqrt x$ is above the $x$-axis and $h(x) = -4\sqrt x$ below.
To find the total area between $g(x)$ and $h(x)$, we need to add the area between $g(x)$ and the $x$-axis to the area between $h(x)$ and the $x$-axis, which would involve computing $\int g(x)dx + \int -h(x)dx$.
Alternatively, we can note by symmetry that these two component areas are the same, so we can just compute one of them and multiply by $2$: use $2\int g(x)dx$. This is where the $2$ comes from
