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/sci/ - Science & Math


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File: 43 KB, 886x533, Screen shot 2012-04-25 at 12.43.28 AM.png [View same] [iqdb] [saucenao] [google]
4613901 No.4613901 [Reply] [Original]

Hey guys. I seriously need help with calculus and I'm willing to pay, no joke. For every problem you help me solve, I'll paypal you a dollar. They aren't long problems, I just don't understand them. I have 13 left to finish.

One of them is in the pic. After I get an answer, I'll paypal you the dollar, then proceed to post another problem.

Thanks guys.

>> No.4613909

Integrate f(x) with respect to dx from -3 to 3.

Don't pay me. That was too easy.

>> No.4613917

>>4613909

I somewhat understand that, but I need to see it worked out on a sheet of paper or something to see how it's fully done. It's the only way I can learn math, unfortunately.

>> No.4613923

>> 4613917

You can either graph it, or find the roots of the problem by solving the equation. You want to find the roots (where it crosses the x axis) so you know what area to integrate bound by your curve. It crosses at -3 and 3 so these are your two boundaries of your definite integral. Then just integrate...

>> No.4613929

On /sci/ and can't even calculus.

>> No.4613930

>>4613917
Graph the function.

It's the area between the function and the x-axis.. The function vanishes at two points. In this case, x=-3,3 and so those two points are clearly our limits of integration.

http://www.wolframalpha.com/input/?i=integrate+9-x^2+%2C+-3+to+3+

>> No.4613932

>>4613917
Reason it out. Draw the graph. You want to find the area and we'll be using rectangles. We're going to find the area of infinitesimally small rectangles and add them up to get the whole area.

area for rectangle = base x height = (9-x^2) dx

the base is dx
the height at any point x is just the function - 0

that's the area for one rectangle so you're going to sum it up from -3 to 3.

There's not much to work out once you get the hang of it.

"find the area under y=x from 0 to 1"

just intergralsign(xdx) from x to 1

>> No.4613942

>>4613932

I got the gist of that, but I'm not entirely sure how to work it out on paper.
If anyone could take a photo of it solved on paper, you'd be my absolute hero.

>> No.4613958
File: 32 KB, 736x418, Screen shot 2012-04-25 at 1.07.24 AM.png [View same] [iqdb] [saucenao] [google]
4613958

Nevermind, I got it. Thanks guys!

Here's another one.

>> No.4613961
File: 46 KB, 776x602, 132556980952-Get-A-Load-Of-This-G.jpg [View same] [iqdb] [saucenao] [google]
4613961

>>4613958
>do all my homework for me

>> No.4613965

>>4613958
4 to 4 is 0
7 to 1 is 3
0 to 1 is 11

>> No.4613973

>>4613961
I'm more than willing to pay anyone that gives me a good answer.

>>4613965
But how? Haha I need to see how it's done.

>> No.4613983

>>4613958
>>4613901

Are you fucking serious?

And are you fucking serious /sci/ for helping him?

>> No.4613987

>>4613973
For a, you're integrating from 4 to 4. What's the area between 4 and 4 on that curve?

For b, you have the area from 0 to 7, the area from 1 to 7, so what's the area from 0 to 1?

For c, you're integrating in the opposite direction as the value you're given, so imagine it like you're adding together the infinite rectangles of f(x)*(-dx) instead, giving the inverse of the area you already know.

>> No.4613992

>>4613973

All of those are just properties of integrals, obviously when you go from 4 to 4 on any unit scale, you haven't moved and any function that's Real, like the one's they defined for you, it would be 0. If you flip around your boundaries, you change the sign, which you can demonstrate by yourself if need be, and if you know the integral of bounds from 1 to 7 and 0 to 7, you can just subtract the difference to get 0 to 1, right?

Not that hard.

>> No.4614024
File: 41 KB, 873x431, Screen shot 2012-04-25 at 1.30.45 AM.png [View same] [iqdb] [saucenao] [google]
4614024

>>4613987
Thanks for clearing that up! Starting to get the hang of it and knocked out a few more problems.
I feel like I'm taking advantage of you guys because no one's asked for payment yet or anything. But here's another problem. Probably my last one since I feel like a douche for begging for answers. Thanks again, guys.