C2 Integration Help!!

you needed to make it in the form x = f(y) because you're finding the area of a region bounded between the curve and y axis

Hi, I was attempting the last question of a C2 Paper, and having looked at the mark scheme after doing the question, we seem to have a few differences.

This is the question:



The part I am referring to is part 9iii).
Essentially, the question says by first writing the equation of the curve in the form
x = f(y), use integration to show that the exact area of region B is 14/3.

When I was reading this question, I didn't exactly know what they meant by "in the form x = f(y)", so I ignored that, and worked it out my own way. Please see this screenshot of my working:

http://grab.by/HkCo

However, the mark scheme doesn't seem to mention being able to do anything like this. I have attached the mark scheme - please see pages 22-23

I am wondering:
1) would I get full credit for working it out my way? is it 100% accurate??
2) what is the other way they were referring to originally meaning??


Thanks!!

We can't see the question! But from what I can gather, the question is asking you to find an area between a curve, two lines, and the y axis. This is why you have to rearrange into the form x=f(y), because you want to integrate with respect to y instead of x.

If the question specifically says by first writing the equation of the curve in the form x = f(y) then you might lose some marks, because you haven't done what it asks you to do in the question.

Thanks so much for your help. The bad news is that I'm having a few more integration issues.



It's regarding part ii) of this question. I think the issue is stemming from the fact that I thought that to find the area between the two lines, you have to do the integral of the line above minus the line below. However, it seems with this question, you are just meant to take the integral of the curve line (and ignore the y=3 line).

Would you be able to explain why you cannot simply take away and do integral from 1 to 3 of {3-(y²+2y-3)}dy?? I remember being taught about integration for areas so this is making me uber confused

Thanks!

Do you mean take away this integral:

$\displaystyle\int -1 + \sqrt{x+4}\ dx$ from the area of the rectangle? (Obviously you have to find the limits to the integral)

You could do this, but this is much more difficult, mainly because of the $\sqrt{x+4}$ term. It's much easier to rearrange into the form $x=f(y)$ and then integrate this with respect to y.

If it helps you visualise it, imagine rotating the whole set of axes 90 degrees anticlockwise, so that the y axis is horizontal. This might make it easier to see why you should be rearranging the equation. Hope this helps!

Ah no. This is a different situation to the last question. I just had the idea that to find the area its something about the higher line minus the lower line and then take the integral of that?

I completely understand the whole making x the subject, I just thought that the equation of the horizontal y line had to be taken away from the eqn of the curve to find the area? I thought that was how it is done.

Perhaps I'm getting confused because this concerns the x-axis when I'm used to the y axis. Could you explain?

Thanks so much

You are correct that you have to take away the equation of the higher line in certain cases, but not in this case; this is because the line in question is perpendicular to the y axis. So it is just included in the limit of the integration. If a question said "find the area between the curve $y=x^2$ and the line $x=2$ would you subtract the line? No, you would include it in the limit of the integration.

I think you're just confused because it's the different axis! Read what I said in my last post, "imagine rotating the whole set of axes 90 degrees anticlockwise, so that the y axis is horizontal".

I think I'm with you. What I was concerned about was whether

$\displaystyle\int^3_1 y^2+2y-3\ dy$

would give the value of everything under the graph of $-1+\sqrt {x+4}$ instead of the values I'm after which are between the curve and the line. That's why I felt the need to do the (unnecessary) subtraction

But I sort of get what you are saying.