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Area/Integration

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mariatachejian:
Hi everyone, I understand the process of area questions like the one attached, but i usually get confused which equation is minused from what.
The answers for this had k(1-x²) - 2k(x²-1) and then the whole integration process.

Could someone please explain how you know how to get the first statement / which equation comes first?

Thanks

Opengangs:
Hey!

So with these types of questions, you would normally have the Top curve - Bottom curve. So in this case, at the point of intersections, we have k(1 - x2) sitting above 2k(x2 - 1). So you would do the following integration:
\[ \int_{-1}^1 \left(k(1 - x^2) - 2k(x^2 - 1)\right)\,dx\]

So in general, for these types of questions, look for the curve \(f(x)\) that is above the other, \(g(x)\). Then the integrand becomes \(f(x) - g(x)\). If you're still confused, feel free to reply and I or someone else will be happy to answer any question! :)

mariatachejian:
hey, so is it that because between 1 and -1 the  k(1-x²) curve is higher than the 2k(x²-1), then it goes first.

And, do you always minus the curves from one another?

Thanks heaps

RuiAce:

--- Quote from: mariatachejian on October 24, 2018, 01:56:25 pm ---hey, so is it that because between 1 and -1 the  k(1-x²) curve is higher than the 2k(x²-1), then it goes first.

And, do you always minus the curves from one another?

Thanks heaps


--- End quote ---
Yes to both.

(It is in fact true that when we go outside the interval \(-1 \leq x \leq 1\), the upper and lower curve get switched. But we only care about what's inside that interval for our purposes, giving what you have stated.)

headsup:

--- Quote from: mariatachejian on October 24, 2018, 01:56:25 pm ---hey, so is it that because between 1 and -1 the  k(1-x²) curve is higher than the 2k(x²-1), then it goes first.

And, do you always minus the curves from one another

--- End quote ---
Correcto....
Funny our I was just helping a class mate with this question 😀

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