Question

Post your calculus questions hurr.

Answer 1

\[\int_{-\infty}^{\infty} \cosh(x^2) dx\]

Answer 2

lol

Answer 3

your algebra question seems boring, tracy.

Answer 4

I can see it in the list bud.

Answer 5

Yes, I know algebra.

Answer 6

I know it's an algebra question, nuff said.

Answer 7

Yes.

Answer 8

What part of "calculus questions" was hard to understand? :/

Answer 9

it turned to a thread instead :D

Answer 10

But for real, calculus questions, go.

Answer 11

Is that integral cosh(x^2) even convergent?

Answer 12

Its not, its complex.

Answer 13

Not convergent.

Answer 14

Yeah, since e^(x^2) dx over -infinity to infinity is infinity.

Answer 15

Here's a basic one: show dy/dx of ln(x) is 1/x.

Answer 16

Proof:
\[\int\limits_{\mathbb{R}} \cosh(x^2)dx=\int\limits_{\mathbb{R}} \frac{e^{x^2}+e^{-x^2}}{2}dx=\frac{\sqrt{\pi}}{2} \left( 1 + i \right)\]

Answer 17

Without limit this problem makes no sense. At least solving it through exponentials.

Answer 18

Ahh yes, I have seen this before.

Answer 19

limits*

Answer 20

I was playing around with \[\cosh(x)+\sinh(x)=e^{x}=e^{i(-ix)}=\cos(ix)-\sin(ix)\]

Answer 21

I mean -isinix, whoops.

Answer 22

Everybody knows that.

Answer 23

Thats a good method to. A tricky (kinda) question would be to give the integral of sinh(x^2) and ask for the integral of cosh(x^2) so you have to use that identity, integrate it, and subtract of the given one.

Answer 24

Yes, exactly. Interesting actually.

Ok here's a question:

Can you find a general formula that contains both of these formulas in one until you plug in a limit/value:

\[\sum_{k=1}^{n}k^2=\frac{ n(n+1)(2n+1) }{ 6 }\]
\[\int\limits_{1}^{n}x^2dx=n^2-1\]

Answer 25

Or even simpler, just do something like\[\sum_{a}^{b}n\] and \[\int\limits_{a}^{b}xdx\] as one formula, since you'll need this to get the n^2 version.

Answer 26

Anti-derivative of \(x^2\) is \(x^2/3\) so I would expect \(n^3\) somewhere.

Answer 27

I think thats whats its supposed to be.

Answer 28

@wio lol my bad, yeah.

Answer 29

\[
\int\limits_1^nx^2\;dx = \lim_{m\to \infty}\sum\limits_{i=1}^m(x_i^*)^2\Delta x
\]

Answer 30

A simple partition would be \(\Delta x = \frac{n-1}{m}\) and \(x_i^*=i \cdot \frac{n-1}{m}\)

Answer 31

Consider this:

An integral is an infinite sum that stops at every point on an interval, so we could write:

\[\int\limits_{a}^{b} [(x+dx)^2-x^2]=[(a+dx)^2-a^2]+[(a+2dx)^2-(a+dx)^2]+...+[(b+dx)^2-b^2]\]

So this is a telescoping series...
\[\int\limits_{a}^{b} [(x+dx)^2-x^2]=(b+dx)^2-a^2\]

\[(b+dx)^2-a^2=\int\limits_{a}^{b}[ 2xdx+(dx^2)]=\int\limits_{a}^{b}2xdx+dx \int\limits_{a}^{b}dx\]

Notice that since the (dx)^2 integral goes to zero since you are multiplying the whole integral by something infinitely small.

\[\int\limits_{a}^{b}2xdx=b^2+2bdx+(dx)^2-a^2 \approx b^2-a^2\]

And the other dx's go to zero as well, since they're infinitely small without an infinite sum next to them, they're effectively zero.

Answer 32

Here's the general equation I was looking for. "s" is the step size, so for a normal sum, it's just 1 since you go 1,2,3, etc... and an integral s=0 (or rather it's limit is zero) so the sum becomes infinite unless it's multiplied together.

\[s \sum_{n=a}^{b}n =\frac{ b^2-a^2 }{ 2 }+\frac{ s(a+b) }{ 2 }\]

So to find the sum of numbers between 2 and 4 with a step size of 1/2, which is:

\[2+5/2+3+7/2+4= \sum_{n=2}^{4}n =\frac{ 4^2-2^2 }{ (1/2)2 }+\frac{ (2+4) }{ 2 }=15\]

Answer 33

@wio do you believe it? I came up with this on my own, and have never seen it before so it could be wrong but... Seems legit.

Answer 34

I don't completely trust it, since you never convert it to a series clearly.

Answer 35

=) Well that's for me to know and you to find out.