Mathematics Question Thread

[RuiAce]

[Rathin]

Find the area of the region bounded by the curve y=(x+1)(x-1)(x-3), the x axis and the ordinates at x=0 and x=2.

[RuiAce]

Find the area of the region bounded by the curve y=(x+1)(x-1)(x-3), the x axis and the ordinates at x=0 and x=2.

(note: There IS symmetry going on, and this can be exploited. However, this should be done so carefully.)

[Rathin]

(Image removed from quote.)


(note: There IS symmetry going on, and this can be exploited. However, this should be done so carefully.)

Oh I see, I was trying to integrate from 0 to 2. Why do we have to separate the two integrals? and.. for the 1st half of the integral why is the boundary from 0 to 1 and not -1 to 0?

[RuiAce]

Oh I see, I was trying to integrate from 0 to 2. Why do we have to separate the two integrals? and.. for the 1st half of the integral why is the boundary from 0 to 1 and not -1 to 0?

To cater for the fact that a region is below the x-axis.

If we simply integrate from 0 to 2, we treat the area below the x-axis as what it ACTUALLY is - a negative area. This negative area will cancel out with the positive area from 0 to 1.

We don't want to treat it as a negative area; we want to keep it positive. Hence we split off the integral, and put an absolute value bracket around what's negative to make it positive again. This should be a procedure you're familiar with from classwork.


If it were from -1 to 0, you'd be going to the left of the blue line (the y-axis). That's not asked anywhere in the question because it is not a part of the region you're interested in.

[jamonwindeyer]

Oh I see, I was trying to integrate from 0 to 2. Why do we have to separate the two integrals? and.. for the 1st half of the integral why is the boundary from 0 to 1 and not -1 to 0?

The integral is separated because one region sits above the x-axis, the other below. We need to take the absolute value of the integral representing the region below the x-axis, since that part of the integral will turn out negative. We do this for all regions sitting below the axis to make sure our area is correct try finding the area from 1 to 2 without that absolute value, you'll get a negative! And that throws out our answer if we leave the integral undivided

As for why the limits of integration are 0 to 1, that is because the question specifies between the ordinates x=0 and x=2, we want the area on the right hand side of the y-axis!

Does that help? If not I can put together another diagram to show you what I mean?

Edit: Rui won this one, I'll leave it here anyways, hearing it said two ways might help

[RuiAce]

Edit: Rui won this one, I'll leave it here anyways, hearing it said two ways might help

[Rathin]

Thanks for the reply guys, just to clarify if the areas are on different sides of the x axis we have to split them up? and @rui I dont really have class notes haha..I am self learning this topic before school starts.

[RuiAce]

Thanks for the reply guys, just to clarify if the areas are on different sides of the x axis we have to split them up? and @rui I dont really have class notes haha..I am self learning this topic before school starts.

Oh fair enough.

And basically yes to your question

[Rathin]

This is kinda a half rant/confusion..but whats the point of trapezoidal rule and simpson's rule when you can integrate the exact area directly? It just a very tedious process as the subintervals increase..

[RuiAce]

This is kinda a half rant/confusion..but whats the point of trapezoidal rule and simpson's rule when you can integrate the exact area directly? It just a very tedious process as the subintervals increase..

Well, if you had to find the area of a pond, you'd probably take forever trying to approximate it with a function when you can just use one of those numeric methods to give an estimate.

[jamonwindeyer]

This is kinda a half rant/confusion..but whats the point of trapezoidal rule and simpson's rule when you can integrate the exact area directly? It just a very tedious process as the subintervals increase..

And further, even when you are given a function, some integrals are just plain disgusting. Sometimes these methods can be less tedious than doing the integral; obviously not the case for polynomials, but much stranger functions than that exist! So as long as you are happy to put up with a little inaccuracy (and have a solid understanding of why that inaccuracy exists and how large it should be) then the methods can prove really handy

[RuiAce]

To illustrate Jamon's point:

[Rathin]

Is it mostly used for non-elementary integrals who don't have a primitive? 

[jamonwindeyer]

Is it mostly used for non-elementary integrals who don't have a primitive?

It can be, but that integral Rui provided has a primitive (right Rui?), it would just be flat disgusting to actually do that integral but I mean yeah, pretty much! In the HSC it is mostly associated to physical scenarios where using a series of measurements to make an approximation is easier than fitting a function to the phenomena