(OT) Calculus Help for Morons

[Trevalon Moonleirion]

*drools, eyes begin glazing over*

uh ... goodie! That is what I have to look forward to ...

*begins looking for the rope*

It's not... TOO bad.... if you have a good teacher...?

Regardless of who your teacher is, there is a review book that I HIGHLY recommend:

Master the AP Calculus AB & BC Test, 3rd Edition, by W. Michael Kelly

ARCO puts the book out and it is absolutely MARVELOUS. Mr. Kelly is a great writer, is good at explaining weird crap, and his story problems are hilarious. Related rates involving bloody mucus flowing into a cup, figuring out how fast celebrities fall out of airplanes without parachutes... my god it's just sheer enjoyment... not something you'll likely find ANYWHERE in the evil land of Calculus.

 

[Trevalon Moonleirion]

Yes, master.

joe b.

Heh. heh. heh.

*steeples hands menacingly, dons mr. burns voice*
Eh-------xcellent....

Edit--Don't mind all the edits... I was just getting rid of my signatures...

 

[LazarusLong42]

Dexterace, you forgot to subsitute u's for x's before taking the definite integral in problem 2.

Since x = 2 sin u, at x=0, u=0; at x=1, u=pi/6

(sin (pi/6) = 0.5, * 2 = 1)

So, you're taking the integral of du from 0 to pi/6, not the integral of du fro 0 to 1.

 

[kingpaul]

Hmmm, just pulled out my old Calc book I used in highschool for Calc I and II and in college for Cacl I-III (and was I upset that my college wouldn't take my test scores...but that's another story).

Item 2 seems to fit a standard integral idea.

integrate (dx/((a^2-x^2)^0.5) = sin^(-1)(x/a) + C

So, a = 2. Therefore

sin^(-1)(2/x) + C {I'll let you play with the numbers}

 

[kingpaul]

OK, 1 also falls into a standard integral

intgrate (sin^n(a*x)*cos^m(a*x)dx =
-(sin^(n-1)(a*x)*cos^(m+1)(a*x))/(a(m+n))+((n-1)/(m+n))*integrate(sin^(n-2)(a*x)*cos^m(a*x)dx

 

[Trevalon Moonleirion]

Alright... now let's see if i can apply some or all of these!

Thank you much, everyone! I love you all!

 

[ShadowX]

I am having the same problem. 3 weeks off from calc AP and she wants us to a semester exam. I wish she would have just given it to us while it was still fresh in our mind.

Anyway for the second one I got arcsin(X/2) + C and then the definite integral was .524.

And for 1, I have never done anything like that,
I have no idea where kingpaul got that standard integral from.

 

[Trevalon Moonleirion]

I am having the same problem. 3 weeks off from calc AP and she wants us to a semester exam. I wish she would have just given it to us while it was still fresh in our mind.

Anyway for the second one I got arcsin(X/2) + C and then the definite integral was .524.

And for 1, I have never done anything like that,
I have no idea where kingpaul got that standard integral from.

It's pure, unadulterated evil. My teacher is sitting in a high-backed chair in front of a fire at her house, cackling madly about doing this to me. Apparently we're in the same boat though.

As for where he's getting stuff... look in the back of your calculus book... it may have a bigass table of standard integrals... i dunno... the whole table doesn't make any sense to me... but the way he put it sorta does...

 

[orbitalfreak]

(1)
∫ (sin³x)(cos²x)dx
= ∫ (sinx)(sin²x)(cos²x)dx
= ∫ (sinx)(1-cos²x)(cos²x)dx
= ∫ (sinx)(cos²x - cos⁴x)dx

Let u = cosx
then du = (-sinx)dx

So,
∫ (sinx)(cos²x - cos⁴x)dx
= -∫ u² - u⁴ dx
= - ((1/3)u³ - (1/5)u⁵) +C

clean it up, supstituting back for u, the final answer is

(1/5)cos⁵x - (1/3)cos³x + C
//

(2)

∫₀¹ 1/(√(4-x²)) dx
This is a standard integral of the form
∫₀¹ 1/(√(a²-x²)) dx
= arcsin (x/a) + C

in this case, a² = 4, so a = 2. Substitute this, and you get

arcsin(x/2)¦₀¹
= arcsin(1/2) - arcsin(0/2)
= arcsin(1/2)
= π/6 <-- That's a (pi over six), if it doesn't show up.
//

 

[kingpaul]

I have no idea where kingpaul got that standard integral from.

I don't have my book with me right now, but that calc book had about 100 standard differentials and 100 standard integrals near the front and rear covers respectively.