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Integration - area under curve

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RuiAce:

--- Quote from: Tatlidil on January 23, 2019, 12:03:33 pm ---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...

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

megcox01:

--- Quote from: RuiAce on January 23, 2019, 11:50:09 am ----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 \( A=\int_{-1}^1 1-x^2\,dx\).

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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:

--- Quote from: RuiAce on January 23, 2019, 12:07:49 pm ---Fairly sure that particular region is above the \(x\)-axis.

--- End quote ---
I know, im just mentioning what students also struggle with

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

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