Integration Concern, the total area of x^3 (1 Viewer)

[cloud edwards]

So i'm stuck with this idea in my head and asked a couple of mates what the area is of the x^3 between x+-1 and x=1.

One of me friends said the area is 0units^2 from following: [x^-4/4] between x=-1 and 1. = 1-1=0.

Another said it was 2 units^2 as he did: [x^-4/4] between x=-1 and 1 using absolute values, so the total was 1+1=2.

I understand how both of me friends came to their conclusion, but i'm still not sure particularly with the thought of the cubic graph having an area of 0 seems weird. So which one is right? By using the absolute value and getting 2units^2 the 0 units.

 

[InteGrand]

So i'm stuck with this idea in my head and asked a couple of mates what the area is of the x^3 between x+-1 and x=1.

One of me friends said the area is 0units^2 from following: [x^-4/4] between x=-1 and 1. = 1-1=0.

Another said it was 2 units^2 as he did: [x^-4/4] between x=-1 and 1 using absolute values, so the total was 1+1=2.

I understand how both of me friends came to their conclusion, but i'm still not sure particularly with the thought of the cubic graph having an area of 0 seems weird. So which one is right? By using the absolute value and getting 2units^2 the 0 units.

If you want the actual physical area, it's the one using absolute values.

The thing that is 0 is the signed area, or net area.

If we just say "area" of a function on an interval without saying "signed area" or "net area", we would probably be referring to the physical area, which is always positive for a continuous nonzero function.

 

[boredofstudiesuser1]

Following on from what Integrand said, your first friend (0 units) just evaluated the integral, your second friend (2 units) found the area.