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.
Best Answer
When they say
They actually mean
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.