Hi everyone,
If you have a question about the Exam 3 Review please post it here.
Extra Credit. Answer someone’s question about the Exam 3 Review.
NOTE: this is not an assignment — I just wanted to create a place for you to post your questions if you need help. However, I will give extra credit for responding to someone else’s question.
Prof for number 5e my answer was (sec(x)^6/6)-(sec(x)^4/4) in replace of all tangents. Is it still correct?
Instead of using u as tan(x) and getting a du of sec(x)^2 I used u as sec(x) and getting a du of sec(x) tan(x)
Yes, that’s a fine alternative solution.
-Mr. Reitz
i have a question on that…
and solve from there? it gives a different answer but i’m just curious why you can’t do that. anyone know?
why can’t you just use u=tan^2(x) then du=sec^4(x) and then you’re just left with
nevermind disregard that last comment. however i finished this problem and i wound up with (secx^7-3secx^5)/12. i realize the answer is one power less on both secants and i’m not so sure where i went wrong. can anyone help me? thanks
Professor for 5F, i substituted
with cos theta. so my answer is 
. is that answer still correct or do i have to substitute the cosines into 
?
The original problem is given in terms of x, so the answer should also be in terms of x. Be careful, when you substitute you change the variable — so you would have to use
, which would change the variable to 
.
Hi professor I’m just a little confused about 6e, when I took the derivative I got -1/3x^-3 but when I looked in my notes, I saw that you wrote the anti derivative as -1/4x^-3. So I took the derivative of the integral you had written and got 1/x^(4) which seems to be right but the thing is that it doesn’t follow the integral rule that we use.
Regarding 6e, I’m not sure what I wrote in class, but the correct antiderivative of
is, as you say, 
— it was likely a mistake on my part (apologies).
im having a little problem with 6e myself
its convergent: -1/81
Careful with this one — unfortunately, we can’t simply take the antiderivative and plug in the bounds x=-3 and x=6. The reason is the the function
is undefined at x=0, which lies between -3 and 6. Because of this, we have to split the integral up into two parts, 
and 
, and evaluate each one separately.
ps. It helps to rewrite the function
as 
— this makes it a little more clear that there is a problem when x=0.
Professor,
In problem 5f, my final answer has the opposite signs of those on the answer key. Is the answer on the key definitely correct?
You’re right — nice job! I’ve posted an update to the answer key.
Thanks, Professor!
I’m glad you brought it up. I, too, believe there is an error to the answer key.
I sub
u for sinx
du=cos x
and it resulted
-4/15 sqrt (1-x^2) x^2- 8sqrt (1-x^2) / 15 -1/5 sqrt (1-x^2) x^4 + C
I may be wrong.
Your substitution is correct, but the result looks odd — I don’t think it’s correct, unless you did some tricky simplification (which isn’t immediately obvious to me)…
i thought #3awas too easy, but the denominator is bugging me. help!!!
Hi Abdoulaye,
. If this doesn’t answer your question, leave a reply with more details about what is bugging you.
The denominator is NOT an irreducible quadratic, so we must factor it (like this:
-Mr. Reitz
how do you know you have to “complete the square” on a function? If it is irreducible?
There are a few different situations in which completing the square helps – here are the most common:
.
1. It often helps if you have (any) quadratic under a radical, unless it is already of the form
2. In a rational function, if you have an irreducible quadratic in the denominator.
-Mr. Reitz
use the formula..b^(2)-4ac….if its come (-) answer then u have to complete the square…
EX. 1/(x^2+6x+25)
b=6,a=1,c=25
b^2-4ac = (6)^2 – 4*(1)*(25)
= 36-100
=-64
its a (-) number..thats means u have to complete the square…
For question 4d, I got up to the part where I have to solve for A and B.
I have 8 = A+B and 32 = 5B-3A
Is that right so far? I have no clue how to solve for A and B.
yeah, that’s right. now you have to use simultaneous equations (2 equations, 2 unknowns)
solve the first one for B: B=8-A
then use that B and plug it into the second equation wherever B appears. This will help u solve for A. Then you plug A into the first equation and solve for B.
Thanks. I didn’t see your comment before. Much appreciated.
Nvm I found it. Just did A = 8-B and solved for B.
I’ve got a problem for 6D.
After setting up the integral with the limit where some variable (such as t) approaches the upper bound, two, you integrate the function as normal, which according to my work should yield:![\int_1^t (2-x)^{\frac{-1}{3}} \rightarrow u = 2-x , du = -dx \rightarrow -du = dx \rightarrow -\frac{3}{2} \left. ((2-x)^{\frac{2}{3}} \right|_1^t \rightarrow -\frac{3}{2}[ (2-t)^{\frac{2}{3}} - (2 - 1)^{\frac{2}{3}} ] \rightarrow (2-t) \approxeq 0 , (2-1)^{\frac{2}{3}} = 1 \rightarrow -\frac{3}{2} (- 1) = \frac{3}{2}](https://s0.wp.com/latex.php?latex=%5Cint_1%5Et+%282-x%29%5E%7B%5Cfrac%7B-1%7D%7B3%7D%7D+%5Crightarrow+u+%3D+2-x+%2C+du+%3D+-dx+%5Crightarrow+-du+%3D+dx+%5Crightarrow+-%5Cfrac%7B3%7D%7B2%7D+%5Cleft.+%28%282-x%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D+%5Cright%7C_1%5Et+%5Crightarrow+-%5Cfrac%7B3%7D%7B2%7D%5B+%282-t%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D+-+%282+-+1%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D+%5D+%5Crightarrow+%282-t%29+%5Capproxeq+0+%2C+%282-1%29%5E%7B%5Cfrac%7B2%7D%7B3%7D%7D+%3D+1+%5Crightarrow+-%5Cfrac%7B3%7D%7B2%7D+%28-+1%29+%3D+%5Cfrac%7B3%7D%7B2%7D+&bg=ffffff&fg=000000&s=0 "\int_1^t (2-x)^{\frac{-1}{3}} \rightarrow u = 2-x , du = -dx \rightarrow -du = dx \rightarrow -\frac{3}{2} \left. ((2-x)^{\frac{2}{3}} \right|_1^t \rightarrow -\frac{3}{2}[ (2-t)^{\frac{2}{3}} - (2 - 1)^{\frac{2}{3}} ] \rightarrow (2-t) \approxeq 0 , (2-1)^{\frac{2}{3}} = 1 \rightarrow -\frac{3}{2} (- 1) = \frac{3}{2}")
I don’t see the error in my work, but the review sheet has the negative of my answer. Am I wrong or is the sheet wrong?
I think I see — the original integral in 6D has bounds 2 to 3 (not 1 to 2 as your work seems to indicate?). Maybe just a copying error?
hi prof. i am stuck at problem 3 at webwork, maybe here we are allowed to ask questions only about exam 3, but we might have a question like this in the exam.
i am stuck at the function f(x) where we have to take the derivative of it and i wanted to use U substitution but i get in so much trouble with all those confusing numbers. i have an idea if this works, for y=x^5/6+1/10x^3 if we can split it in two integrals, will this work? thank you
Hi Endri,
First, you must find the derivative, y’. This doesn’t require anything fancy, just take the derivative of each term. Then plug into the arc length formula, and simplify the result as much as you can.
Let me know if you’re still stuck,
Mr. Reitz
hey professor im not understanding problem 5 on webwork assignment 3
Me either. What are the bounds that they want the length to be determined in?
Hi guys — Kedeshia is right (below), the bounds are “a” and “b”. This means that, after integrating, you plug in “a” and “b”. The result will be an expression involving a and b.
-Mr. Reitz
The bounds are from a to b
yeah i know, and im having trouble with problem 4 too