Integration magnitude?

[kenhung123]

When would we need to take the magnitude of an area and when do we consider net assigned area?

[98.40_for_sure]

If they ask for 'total distance' for example, then they want you to take the WHOLE area (everything above x-axis, and minus everything below axis)
If they want the displacement, then use an integral

[kenhung123]

What if it just says find the region under the graph. Is it possible to say -10square units

[98.40_for_sure]

It wouldn't say region under the graph, it would say something like, the area bounded by the graph ____________ , the x-axis, x=2 and x=4

Also units can't be -ve, so no

If you're asking about if it is BELOW the axis, then you'll have to take magnitude yes

[kenhung123]

Hmm I always thought F(b)-F(a) was the undo the negative region or something

[98.40_for_sure]

=
=

Basically, F(x) is the antidifferentiated function. In the example above, it is
Therefore F(b)-F(a) is =

So this is finding the area bounded by between any two points along the x-axis. The graph never crosses into the negative side of the y-axis, but if there was part of a graph that crossed the x-axis into negative of y-axis, then you would have to perform several integrals of different portions of the graph. Any area bounded below the graph would have be

I don't know if that answers your question

[Martoman]

What.The.Hell.    :knuppel2:

To clear this up. You can't have negative area on planet Earth. This is silly.

Area under the x axis will give you a negative answer because you are evaluating f(b) - f(a), and those points on the graph are negative. You will end up with a negative answer. Sooooo to stop this, just take the absolute value of this area... or subtract it as two negatives make a positive.

This shouldn't be confusing at all 

[TrueTears]

When would we need to take the magnitude of an area and when do we consider net assigned area?

You take the magnitude if you need a positive answer, such as distance (but only assuming f(x) > 0 and you did f(b) - f(a), if you have some parts of f(x) under the x axis, you wouldn't take the magnitude, you'd need to consider the parts separately). Net area is something like displacement, you simply sum from the lower limit to upper limit.

[kenhung123]

Is it clear whether I need to take the magnitude or net assigned?

[kenhung123]

What.The.Hell.    :knuppel2:

To clear this up. You can't have negative area on planet Earth. This is silly.

Area under the x axis will give you a negative answer because you are evaluating f(b) - f(a), and those points on the graph are negative. You will end up with a negative answer. Sooooo to stop this, just take the absolute value of this area... or subtract it as two negatives make a positive.

This shouldn't be confusing at all 

I was told area is a vector quantity

[kenhung123]

When would we need to take the magnitude of an area and when do we consider net assigned area?

You take the magnitude if you need a positive answer, such as distance (but only assuming f(x) > 0 and you did f(b) - f(a), if you have some parts of f(x) under the x axis, you wouldn't take the magnitude, you'd need to consider the parts separately). Net area is something like displacement, you simply sum from the lower limit to upper limit.

Oh shiet so umm the 'normal' integration considers net assigned area already but in the case that we want to find total area (distance) then we need to split it up and find individual areas above and below x axis then substract the below and add the above areas?

[Martoman]

What.The.Hell.    :knuppel2:

To clear this up. You can't have negative area on planet Earth. This is silly.

Area under the x axis will give you a negative answer because you are evaluating f(b) - f(a), and those points on the graph are negative. You will end up with a negative answer. Sooooo to stop this, just take the absolute value of this area... or subtract it as two negatives make a positive.

This shouldn't be confusing at all 

I was told area is a vector quantity

Your teacher is a pompous fag for saying that. unless you continue your math career i'd advise you to not worry about advanced calculus. There in that land of weirdness, you are correct. Bad bad bad of your teacher to confuse you like that.  :knuppel2: :knuppel2: :knuppel2:

[kenhung123]

What.The.Hell.    :knuppel2:

To clear this up. You can't have negative area on planet Earth. This is silly.

Area under the x axis will give you a negative answer because you are evaluating f(b) - f(a), and those points on the graph are negative. You will end up with a negative answer. Sooooo to stop this, just take the absolute value of this area... or subtract it as two negatives make a positive.

This shouldn't be confusing at all 

I was told area is a vector quantity

Your teacher is a pompous fag for saying that. unless you continue your math career i'd advise you to not worry about advanced calculus. There in that land of weirdness, you are correct. Bad bad bad of your teacher to confuse you like that.  :knuppel2: :knuppel2: :knuppel2:

I agree!

[Martoman]

It's like telling 6th graders that you can in fact take the sqaure root of -1.

You.just.don't.  :smitten:

[98.40_for_sure]

It's like telling 6th graders that you can in fact take the sqaure root of -1.

You.just.don't.  :smitten:

Why not? I was told in 6th grade what secant, cosecant and cotangent were