# Recent content by anderma8

1. ### Integ prob but I'm stuck

thanks for the 2nd set of eyes... Maybe it should read: 1/2 int (3u^-1) - (u^-2) du I end up with 1/2 u^-1 +c or 1/2(x^2 + 1) + c since u=x^2+1
2. ### Integ prob but I'm stuck

I'm trying to do the following: int (3x^3 + 4x)/(x^2+1)^2 dx I let u = x^2+1 and I eventually get: int 3(u-1)+4/u^2 du/2 When I further break this down, I get: 1/2 int 3u^-1 -u^-2 du am I on the right track? When I integrate this, I'm thinking that I have to be doing something...
3. ### May have copied wrong

Never mind... I found my answer: (u+8)u^-1/2= u^1/2 + 8u^-1/2 Sorry about that! I forgot the parenthesis!
4. ### May have copied wrong

I'm trying to integrate the following: int (x^3 + 4x)/sqrt(x^2-4) dx I let u= x^2-4 so du/2 = xdx granted, the above also gives me x^2=u+4 so, this gives me: 1/2 int u+8 u^-1/2 du but the professor has: 1/2 int [u^1/2+8 u^-1/2] du Did I miss the first u^1/2 somewhere...
5. ### Limits question

I'm trying to figure out the following: An=1x3x5.... (2n-1)/(2n)^n and I'm trying to determine if it converges or diverges and if it converges, what the limit is. The answer is 1/2n x 3/2n x 5/2n.... (2n-1)/2n and it converges, but I don't understand what they did or how they got to the...
6. ### Infinite Series: sigma (2^n)+1/(2^n+1)

quasar987 & HallsofIvy, It's late where I am, but I wanted to thank you both. I am sure I am confusing several words in my posts and I'm not ashamed to admit it. That's at least how I learn! Thanks for the clarification. I have been looking at this stuff for quite a bit of time this...
7. ### Infinite Series: sigma n^2/(n^2 +1)

If I take the limit on the sum... I get 1/1 = 1 If the limit does NOT = 0 then sigma f(x) diverges...I'm not quite sure I follow this... Does this mean that in order for the equation to converge, the sum (sigma) must be = to 0?
8. ### Infinite Series: sigma (2^n)+1/(2^n+1)

robert Ihnot - Thanks for your input though! I guess I will need to substitute numbers to get the 1/2 and see that the other side goes to 0! Thanks! It's starting to make sense!
9. ### Infinite Series: sigma (2^n)+1/(2^n+1)

(2^n)/(2^(n+1)) + 1/(2^(n+1)) This gives me 1/2 + 1/2^(n+1) which eventually goes to zero.... :-) hence - my answer!
10. ### Infinite Series: sigma (2^n)+1/(2^n+1)

i'm not quiet sure how to attack this problem: sigma (2^n)+1/(2^(n+1)) n->1 If I start plugging in #'s for n, then I get: n=1: 3/4 n=2: 5/8 n=3: 9/16... by this method, I see that it's going to 1/2, but I need another way to 'see' that. Any suggestions?
11. ### Int x^3 sin x

AKG - After we tried the problem over and over, when we got what we were expecting, you are exactly right! It was funny, when my team started this, we were apprehensive at best, but now we look at them and laugh 'cause they really are easy when you understand the problem! Thanks!
12. ### Int x^3 sin x

VietDao29 - THANKS! Your explanation is very clear and was exactly the insight my team needed. We ended up doing just what you suggested before reading your post. The good thing is that we feel confident that we did the project correctly and with your input, it validated our work! The...
13. ### Int x^3 sin x

we would but we can't use that as per the project instructions X-) Any suggestions?
14. ### Int x^3 sin x

but it's setting up the initial equation that is our road block. We get x^4e^x + dx but not clear on the setting up of X^3 cosx + dx
15. ### Int x^3 sin x

int x^3 cosx dx for this issue Thanks for the reply... essientially, we need to set this up and are having problems. We have an example of int x^4ex + dx. We get (Ax^4 + Bx^3 + Cx^2 + Ex +F) e^x + k. But we are stuck for finding the starting equation for int x^3 cosx.