HSC Stuff > HSC Mathematics Advanced
Integration - area under curve
megcox01:
Hi guys,
So i have a maths question that is probably stupidly easy but i havent been able to figure out what to do or how to set it up.
The question is:
Find the area enclosed between the curve y= 1 - x2 and the x-axis.
If anyone could help me i'd really apprecitate it.
Thank you x
RuiAce:
To start, have you sketched the curve \(y=1-x^2\) and identified what the region it is you’re interested in?
megcox01:
--- Quote from: RuiAce on January 23, 2019, 11:26:49 am ---To start, have you sketched the curve \(y=1-x^2\) and identified what the region it is you’re interested in?
--- End quote ---
No i hadnt initially sketched the curve. But i just did and the curve goes through -1 and 1 on the x-axis. So that does mean those are the 2 numbers i put as the limits for the integral because it has to be in regions 1 and 2??
RuiAce:
--- Quote from: megcox01 on January 23, 2019, 11:38:22 am ---No i hadnt initially sketched the curve. But i just did and the curve goes through -1 and 1 on the x-axis. So that does mean those are the 2 numbers i put as the limits for the integral because it has to be in regions 1 and 2??
--- End quote ---
-1 and 1 are the limits of the integral as you say but you need to understand the reason why.
You're also right to say that \(-1\) and \(1\) are the \(x\)-intercepts of the curve. But I'm not too sure what you meant by "regions 1 and 2". The region that you require is simply the region, enclosed between only
- the curve itself
- the \(x\)-axis.
As it turns out, that region happens to literally be the region under \(y=1-x^2\), between \(-1\) and \(1\). That's why your answer is then \( A=\int_{-1}^1 1-x^2\,dx\).
Tatlidil:
I also realised that alot of students struggle to understand when the curve is below the x-axis and therefore they need to make it negative, or switch the numbers position in the integral. You probably already knew this but anyway...
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