Area/Integration

[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]

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

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]

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

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

[mariatachejian]

thanks heaps!!