If anyone could help with any of these, that would be fantastic. We were given a practice final with questions to review from the semester and I forgot how to do them.

Confident 2a is 0 and 2b might be 9x + C?

1.
u = 3 - x^3
du = -3x^2(dx)
dx = du/(-3x^2)

Substitute in for dx and then simplify:

I(-1/(3u) du)

F(u) = -1/3(ln |u|)

Substitute back in the u to get F(x) and make sure to add +C for arbitrary constant since this is an indefinite integral:

F(x) = -1/3(ln |3 - x^3|) + C

This post was edited by timmayX on Dec 7 2015 09:14pm

6. Area under curve is equal to the value of the integration on a given interval

I[1,9] (1 + x^1/2) dx

Since it is a definite integral, no "+ C "will be necessary, your final answer will be a number/value. No u-sub necessary on this one either.

F(x) = x + 2/3 x^3/2 (Still needs to be evaluated on the integral [1,9]

F(x) = F(9) - F(1)
F(x) = (9 + 2/3(9^3/2)) - (1 + 2/3(1^3/2))
F(x) = (9 + 2/3(27)) - (1 + 2/3)
F(x) = (9 + 18) - (1 + 2/3)
F(x) = 27 - 1 - 2/3
F(x) = 25 and 1/3

This post was edited by timmayX on Dec 7 2015 09:11pm

I understand the first parts, but once you substitute in for dx you lose me.

Whats with the I?

2a.
e^(pi-12) is still a constant. If you evaluated it, it would still be a constant numerical value, not a variable function.

Thus the derivative is 0


2b.
Easiest thing to do here is simplify the 3^2 to 9 and then just give the antiderivative from there. Since it's an indefinite integral, you'll need the + C arbitrary constant.

F(x) = 9x + C

This post was edited by timmayX on Dec 7 2015 09:13pm

My typed form of the integration sign. I'd write it out and send a picture which would undoubtedly make more sense, but don't feel like uploading to photobucket at the moment.

When you substitute for dx you actually get this (it's just hard to understand in a typed out form):

I ( (x^2/u) * du/-3x^2)

The x^2 will cancel which is what you are always looking to do with u-substitution.

You are left with integrating -1/(3u) du

This post was edited by timmayX on Dec 7 2015 09:13pm

Sorry just updated my answer for 6. I left out the 2/3 when I was evaluating. It's fixed now.

Alright that makes a lot more sense, thank you. My savior.

3. f(x) = ln [ x^2(x - 3)^4 ]

Logs can be re-written with identities that make this much easier --

First, the product of two terms in a log is the same as the sum of those two logs separately:

f(x) = ln [ x^2] + ln [ (x - 3)^4 ]

Second, a natural log raised to a power is the same as that power as a constant in front of the log of the base term:

f(x) = 2 ln [x] + 4 ln [(x - 3)]

Then you can attain the derivative, the easiest way to consider the derivative of the natural log is [ln u]' = u'/u

You'll still need to have the constant out front, then distribute properly:

f '(x) = 2*(1/x) + 4*(1/(x - 3))

f '(x) = 2/x + 4/(x - 3)

Not sure if you need to simplify over a single denominator, but that would be easily achieved with the common denominator x(x - 3)

This post was edited by timmayX on Dec 7 2015 11:08pm