Anyone want to review for Calc II? Swap problems?

Question

adamaero · Answer

Stewart Single Variable:

Ch 7 - Inverse Func; Exponential Func & Derivatives; Log Fuc & Derivatives; 
exponential growth/decay, Ch 10.4 - Models of pop. growth; Inverse trig func; Indeterminate forms & L'Hopital's Rule

8 - By parts; trig integrals; trig substitution; approx integration; improper integrals

12.1-12.3 Seq & Series

adamaero · Answer

I know there's billions online, but it's different when teaching/being taught with another.

adamaero · Answer

@M4thM1nd 
I'm ee major in WI.

zepdrix · Answer

I have a different Calculus book if you need some other practice problems :)

adamaero · Answer

ya, if it's related to those kinds

anonymous · Answer

have you ever watched the open course lectures from MIT in single variable calculus? It's an awesome resource

adamaero · Answer

only Patrick JMT

zepdrix · Answer

Yah lots of good Youtube resources :)
Patrick, Khan, MIT, ...

anonymous · Answer

take a look at this: http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/

adamaero · Answer

cool, they're on openstudy

anonymous · Answer

also, when i studied calculus, i learned by reading James Stewart calculus books Find it here: http://www.stewartcalculus.com/

zepdrix · Answer

How you feel with your `by parts`?
Can you handle something like this?$$\Large\rm \int\limits x \cos4x~dx$$this?$$\Large\rm \int\limits x^2 \ln x~dx$$this?$$\Large\rm \int\limits e^x \cos x~dx$$

zepdrix · Answer

This is another `by parts` problem that you should make sure you know:$$\Large\rm \int\limits \arctan x~dx$$

adamaero · Answer

last one:
1/(1+(x^2))

zepdrix · Answer

Woops, that's the derivative of arctan :o

adamaero · Answer

working on previous...

zepdrix · Answer

Adammmmm! Where you at broski?
Figure any of these out yet?

Gotta fix that last one, you took a derivative :O
We wanted the integral of arctan x.

adamaero · Answer

still on first one. Does the solution have an integral (near the end)?

adamaero · Answer

I did by parts two different ways, but haven't done it over a third time for either yet...

zepdrix · Answer

The first one?
Umm yes, you do integration by parts,$$\Large\rm \int\limits u~dv=uv-\color{royalblue}{\int\limits v~du}$$And you're left with this blue integral.

zepdrix · Answer

Are you comfortable integrating something like this:$$\Large\rm \int\limits \cos4x~dx$$WITHOUT using a u-substitution?
That's pretty important to get comfortable with stuff like that.

adamaero · Answer

Honestly, I looked that up--but it seemed so easy when I saw it that I wish I'd just tried. 

For the first equation,  does it simplify without integrals?

zepdrix · Answer

Everyone one of these simplifies to something that doesn't involve an integral.

The third one - not naturally though (requires some algebra after applying `by parts` twice). Don't worry about that one so much.

adamaero · Answer

I'll try fresh.

adamaero · Answer

Oh, I just picked the wrong u for the first one; went down the wrong path. I got it 1st one.

adamaero · Answer

Is 2nd one 
-x + C

zepdrix · Answer

Hmm no.
Second is tricky. The parts are maybe backwards from what you would think.

adamaero · Answer

ok...

zepdrix · Answer

$$\Large\rm u=\ln x, \qquad\qquad dv=x^2~dx$$

zepdrix · Answer

When you differentiate ln(x), it turns into a power of x, yes?
That's why you want it for your u,
It will make it nice and easy to deal with in the new integral :)

adamaero · Answer

yep, got completely different answer. I think I was mixing d/dx with integral of x
again

adamaero · Answer

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