easy.png

5pi/2

>>8495201
I am going to hijack this thread to ask a related question about integrals.

In his pic he has the integral of sin(x)/x from some constant to infinity and that means that the graph of that function changes sign infinitely many times.

I've learned that to get the total area of your graph you need to separate the integral into intervals that are completely positive and completely negative (and absolute value these ones) and them add them up one by one to get the total area, instead of mixing intervals with both negative and positive so that parts cancel out.

How would this work here? The integral has infinitely many changes of sign. It goes from positive to negative, to positive, to negative, etc.

What would you do?

I asked because today I had a Calc II test where I saw a similar integral (cosx/e^x) and I just computed the integral from 0 to infinity and got 1/2 and then asked the professor if that was 'okay' and she told me that she also got 1/2 so I was fine.

But then I separated the first positive interval from the rest of the integral and computed the two integrals and got then an aswer of 1.

What is the real answer? How do you work with infinitely many fluctuations?

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Friend of mine did it. I haven't checked if it's right. Wait... Is that you?

>>8495223
If it's entirely ballanced above and below the x axis, then you get 0. Otherwise, you'd trend towards the appropriately signed infinity.

>>8495223
With sin(x)/x, as x increases, f(x) decreases. So if you have an integral from 0 to infinity, the intervals closer to 0 have more "weight" than the later intervals. Hope that makes sense. The early intervals dominate, the intervals get less significant as x increases

>>8495256
dumb taco retard

>>8495223
>I've learned that to get the total area of your graph you need to separate the integral into intervals that are completely positive and completely negative (and absolute value these ones) and them add them up one by one to get the total area, instead of mixing intervals with both negative and positive so that parts cancel out.

If you're looking for the total area between the curve and the x-axis, then yes, that's what you would do. But if you're looking for the integral, that's a completely different problem.

integral of f from a to b: [math]\int_a^b f(x) dx[/math]
area between curve y=f(x), the x-axis, x=a, and x=b: [math]\int_a^b |f(x)| dx[/math]
They're different things when f(x) is sometimes negative.

>>8495256
Yes

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>>8495201
>not getting Burger-Pie/Beer
git on my level faggots

>>8495888
>Wolfag beta
>not solving by contour integration

>>8495256
>sen(x)

>>8496115
this is how i would solve it aswell desu

it might be possible to do with a cheeki variable substitution though

>>8495223
The integral isn't really the area, it's more of an accumulation over the interval. Think of a giant tube half-full of water - if f(t) is the rate at which the tank fills or empties, then the integral of f(t) over some interval is the final water level at the end of the interval. (Specifically, it gives you the relative water level, so it'll be negative if it's below the original. You add C, which is how high the level was initially, to bring it to the absolute water level.)

>>8496348
Ok, I think I get it. So the question 'calculate the integral from a to b' is different from 'calculate the area from a to b'.

I guess my mistake came from the fact that before improper integrals the only time we had to actually compute integrals was in area under curves problems and volume of revolutions problems. In both cases we were instructed to manipulate the integral such that we would get the total area or volume.

Then we jumped to improper integrals and I guess no one gave me a warning.