/sci/ - Science & Math

[Anonymous No.1852262]

Sup, /sci/

I'm starting to learn precalc now, and for the love of god I cant understand shit. Can any of you explain (or just give me an example of) limes and integration. Yeah, as x approaches infinity, yeah, but what's the meaning of that, what does it do ? Also, integration = mindfuck.
Help ?

Pic is me after class.

[Anonymous No.1852266]

inb4 shitstorm

[Anonymous No.1852269]

>>1852266
expecting it.

[Anonymous No.1852279]

Bump.

[Anonymous No.1852290]

if you want to integrate from scratch, you must first create the measurable space.

[Anonymous No.1852293]

>Can any of you explain limes or integration?
>explain limes or integration
>limes or integration
>limes

Integration deals with calulating the area under a curve using infintesimally small trapezia (you don't need to know that though). Limits, in the context of integration, just define from which point to which other point, you calculate the area under the curve. Example, to calculate the area under a curve with points (1,2) and (3,4) lying on the curve, the lower limit would be 1 and the upper limit would be 3 if you were calculating the area under the curve, between points (1,2) and (3,4).

[Anonymous No.1852296]

Nobody responding, no math/engineeringfags on /sci/ atm ?

[Anonymous No.1852304]

You're in for a rough ride OP, your pic is me too, after calc III lectures

Go become a psychologist instead...but if you really want tehmaths:

http://www.khanacademy.org/
(directing to this because they explain calculus much better than I do)

[Anonymous No.1852336]

>>1852293
Now where is that pic with Hitler saying 'Why can't I burn all these jews?'

I'm trying to understand you.

[Anonymous No.1852344]

>>1852293
Also, forgot that it is 'limit' on english, my bad.

Will check out Khan Academy

[Anonymous No.1852520]

>>1852336

Ok I made up an example question to help explain.

In the example, the curve is x^2-2x+1. To find the area underneath the curve you need to use integration. Limits define two points on the x-axis, which correspond to points on the curve. So limits let you calculate the area underneath a specific region on a curve.
In the example the points written at (1,0) and (5,16). So the lower limit, is 1 (because the x=1 at that point) and likewise, the upper limit is 5.
The setup you'd use for an integration is noted beneath the diagram, to the left. So you're 'a' and 'b' define the limits, and the f-shape shows you're integrating. The 'y2-y1' part becomes useful if you're asked to calculate the area between two curves (rather than between the curve and the x-axis). y2 is the curve which is higher on the graph than y1, that's the rule. So in the example, y2 is your curve, and y1 is just the x-axis. Thus:

y2=x^2-2x+1
&
y1=0
therfore:
y2-y1=x^2-2x+1

What y2-y1 actually gives you, is the equation which you're going to integrate. The formula for integration is written underneath the explanation of all the terms, in the lower left of the picture. Its:

The integral of x^n (with respect to x) = (x^n+1)/(n+1)
So if you were to integrate x^2, it would be (x^3)/3.
So when you integrate x^2-2x+1, you get:
(x^3)/3 -(2x^2)/2 + x
which cancels to: (x^3)/3 -x^2 + x

From there, you take (x^3)/3 -x^2 + x and substitute x=5 (your upper limit) into that equation, and subtract the value you get when you sub x=1 into the same equation (as indicated in the third line of the calculation in the pic) which gives you your final value for the area under the curve between the two points.