[xl0lxl0lx]

Hi, my calculus final is tomorrow, worth a huge part of my grade.
I'm trying to work out an old practice final, but i'm having trouble.
if you know calculus with radicals and trig well, please aim me to help at xmyluux.

or u can reply here.

here is a link to the practice final: http://cas.umkc.edu/math/MathUGcourses/Pre...CommonFinal.pdf

the ones i cannot do are: 2a, 6, 7, 9, 11, 12, 13
so if u know how to do those, please show

[ChunJin]

2b) Simplifying it would help a bit. But anyways the answer ends up being:

(1/5) * sqrt(5)

When you simplify it -->

(1/5)*(2-y)*sqrt(5)/sqrt((-2+y)*(y+2))

6) one it's just a dy/dx, after that you can get your m [slope]. Then Plug that into y-y1 = m(x - x1).

The m = [derivative y with respect to x] = (-5*y^2) / (15*x*y-16) Take that, you know your points and what not.

7) Again apply l`hospitals rule.
e^2*t = e^0 = 1 - 1 = 0, and sin(3 * 0) = sin(0) = 0
0 / 0 L'Hospitals rule applies

so

H= (2*e^(2*t)) / (3*cos(3*t)) and it works out well =).

9) This is tricky but the best way is to take it step by step. I'll tell you the rules I went through.

I ended up with: 2*x^3+5*ln(x)+(2/3)*(1/x^(3/2))

I used [in order]: Chain rule, sum rule, constant multiple rule, power rule, rewrite rule, sum rule, constant multiple, power rule, constant multiple, power rule and revert rule.

11) Just integrate normally.

12)
A) You're just integrating where the function exists. There's going to be a line, most likely a slope [since you're dealing with integration] between -1 and 1. And there you will find the area. Same goes for the function in between 1 and 2.

is kind of tricky, so here's what I get first:
-3^(pi)

My steps in order: constant multiple rule, change rule, constant multiple rule and then sin rule.

13, not sure.

[xl0lxl0lx]

^thank you!!
okay, after some help from a classmate, angelods, and google the only ones i need left are 9a and 9c.
i have 2 hours to figure it out, please help!

[mnmadi]

9c.
u substitution

u = cos x
du/dx = -sin x
therefore
dx = -du/sin x
rewriting the integral
-∫1/u^2 du
= -∫ u^-2 du
integrate
1/ u + C
plug u back in
= 1/cos x + C
= sec x + C


12b.

u substitution again

u = x^2
du = 2x dx
dx = du/2x

∫ 3/2 sin u du
= -3/2 cos u
= -3/2 cos x^2 evaluated from -Pi^1/2 to 0
(-3/2 * cos 0) - (-3/2 * cos pi) = -3/2 - 3/2 = -3

[xl0lxl0lx]

^thanks!!
i got everything. i'm gonna review.
u guys r the best!

[i'm lagging yo]

wow calculus lol... the good ol days. sometimes i wish i were an engineering major again..

i forgot 99% of my calc... but for #13, graph the two lines and you rotate the bounded region around the x-axis by forming a integral with the proper notation for the volume of a cylinder or sphere or whatever the shape ends up being

[sweetvietcutie]

haha! I'm taking calculus right now (well..I just did my final so I guess last semester??) but, I had the worse professor ever and on top of that, I'm really slow in math and yeah, now I have to retake calc I Oh well..good luck!

[ChunJin]

lol, this Calculus I is really... like accelerated? I don't know. Calc I is usually purely on derivatives, Calc II is purely on integration and Calc III deals with taylor/infinite and tons of other stuff with 3D integration/differentiation and a little bit of multi-var calc.

[xl0lxl0lx]

^ i know.
i took calc I at a different university, but the credit wouldn't transfer.
i got an A the first time. we got to use calculators, we didn't have to do trig, and it was multiple choice, unlike this class.