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]

To start, have you sketched the curve \(y=1-x^2\) and identified what the region it is you’re interested in?

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]

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??

-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...

[RuiAce]

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...

Fairly sure that particular region is above the \(x\)-axis.

[megcox01]

-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 \( A=\int_{-1}^1 1-x^2\,dx\).

Ok so i integrated that answer and got 1 1/3 units² which according to the textbook is correct. Thank you for your help i think im finally starting to understand this concept!!

[Tatlidil]

Fairly sure that particular region is above the \(x\)-axis.

I know, im just mentioning what students also struggle with

[Kombmail]

What if the boundaries are 3x and x=t?