/sci/ - Science & Math

[Anonymous No.4268521]

Alright, /sci/. I don't come here often. I came here a few times last year for Biology help, and you all were pretty helpful. So I return.

Tomorrow is my calculus midterm. For the most part, I understand everything. Unfortunately, my teacher refuses to help anyone outside of class time. So I come here with a few questions, as I trust no one in my grade to know anything about this.

Are there any easier ways to remember antiderivatives of trig functions?

Can you explain using the Chain Rule in an anti-derivative in not-retarded terms like my teacher?

What does the mean value theorem show as opposed to the average value theorem?

Any general tips would be greatly appreciated.

I could look in my textbook I suppose, but I'd rather hear multiple explanations so everything gets imprinted into my brain.

Pic related, it's my textbook.

[Anonymous No.4268532]

Literally any help at this point would be appreciated.

If you I had to pick one thing I needed help the most on, it would be anti-derivating through the chain rule.

[Anonymous No.4268537]

Just like derivatives, the anti-derivatives of trig functions are something you just need to memorize. Sin and Cos are easy because they are just the negative of their derivatives.

There is no chain rule for integration.

[Anonymous No.4268538]

Helping ONLY since wiki is blacked out (you can still get to mobile or cached, but w/e)
>Are there any easier ways to remember antiderivatives of trig functions?
Not really; you can decompose sin and cos into their respective power series if you get stuck, but for anything useful (e.g. tangent) it won't be much better.

>Can you explain using the Chain Rule in an anti-derivative in not-retarded terms like my teacher?
If you see something of the form:
fg'+f'g
it's integral is fg. It's mostly intuition about seeing/getting things into that form.
If you mean integration by parts, just get it into the form of (fg'), then apply the formula.

Cont. on next post

[Anonymous No.4268542]

>>4268537

I meant doing integration of a function that would have used the chain rule to get the derivative. But thanks for the first part of your post, I figured I would just have to memorize them.

[Anonymous No.4268544]

>What does the mean value theorem show as opposed to the average value theorem?
They are the same thing, mean is another word for average. It is a theorem from functional analysis that, given a continuous function (in one dimension) from a closed interval [a,b] -> R, that is also differentiable on (a,b), there exists a c in (a,b): f'(c)=(f(b)-f(a))/(b-a).

[Anonymous No.4268546]

>>4268544

So the two answers they give you represent the same thing, then?

[Anonymous No.4268548]

>>4268546
What two answers are you talking about? They are the same thing.

[Anonymous No.4268550]

Remember the chain rule is

<span class="math">\frac{d}{dx}[f(g(x))] = f'(g(x))*g'(x) [/spoiler] so if you see something of the form <span class="math"> f'(g(x))*g'(x) [/spoiler] then its antiderivative is <span class="math">f(g(x)) [/spoiler].

What the other guy said, it is just about being able to spot these cases and knowing what they are. The more you work with it the better your intuition will be about them.

[Anonymous No.4268551]

>>4268550
Sorry, some of the symbols aren't showing up in TeX due to the spoiler shit, but you should know where the appropriate prime tags should be.

[Anonymous No.4268552]

chain rule, and integration? sounds like the principle behind simple substitution.

any integral of the form:

integral g(f(x))*f'(x) dx or integral (f'(x)dx)/g(f(x))

could be solved bya substitution of the form u=f(x) du=f'(x)dx, and then you replace into the integral:

integral g(f(x))*f'(x)dx becomes integral g(u)du
and
integral (f'(x)dx)/g(f(x)) becomes: integral du/g(u)

sorry if i was talking about something completly unrelated.
and i don't know how to use this Jmath thing yet, so sorry for that too.

[Anonymous No.4268554]

>>4268546
not the guy who posted that, but what that means in words is that somewhere within an interval, say (a,b), there has to exist an instantaneous rate of change that is equal to the average rate of change (the slope of the secant line) for the function between a and b.

[Anonymous No.4268555]

>>4268550

Oh, alright, I got it. Thanks. Swear to god, my teacher explains everything so backwards and... just poorly.

>>4268548
Perhaps I'm mistaken, I may have misnamed them.
I'm referring the following equations:

(1/(b-a)) integral f(x)dx

and

f(c)(b-a) (or something like this.)

Maybe I'm just confusing myself.

[Anonymous No.4268556]

>>4268552

Yes, this is exactly what I was referring to.
Damn it, I couldn't thing of that for some reason. My teacher called it u-substitution, and I'm just having trouble working with that I guess.

[Anonymous No.4268560]

The sin/cos derivatives form a loop.
d/dx(sin) = cos
d/dx(cos) = -sin
d/dx(-sin) = -cos
d/dx(-cos) = sin
d/dx(sin) = cos...

I mean, those are like, the big ones. So yeah.

[Anonymous No.4268566]

>>4268555
<span class="math"> \frac{1}{b-a} \int_a^b f(x) dx [/spoiler] simply gives you the average value of f(x) over the interval [a,b]. This is not the same as the mean value theorem, and I've never heard it called the "average value theorem", though different names for things is not uncommon in mathematics.