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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.
16. Int x^3 sin x

I'm trying to integtrate x^3 sin x without using integration by parts. I have set up the equation to either: int x^3 cosx dx = (Ax^3 + Bx^2+Cx+D)cos x + K or int x^3 coxs dx = (Ax^3+Bx)sinx + (Cx^2+D)cosx +K but I'm having trouble... any help would be appreciated!
17. L'Hopital's Lim x->0 tan x / x

I do get ur point: tan x = opp/adj :smile: As for point 2: limx->0 sin(x)/x = 1 vs. sin(x)/x = 1 ARE 2 different statements. Good point and I need to keep this in mind since they say very different things. I'll search now in google for 'Taylor Series'. IK Wikipedia has also been a...
18. L'Hopital's Lim x->0 tan x / x

No, we have not reviewed Taylor Series yet. We have just started that section. Might there be an area in which I could review and get a jump on what to expect? THANKS!
19. L'Hopital's Lim x->0 tan x / x

I understand the first part since (opp/hyp)/(adj/hyp) thus opp/adj = tan. That part makes sense. I don't get sin(x)/x = 1. Is this just something I should accept? I understand sin^2+cos^2=1. Would that be part of it? As for the last part, that makes sense provided they both approach the...
20. L'Hopital's Lim x->0 tan x / x

Thank you! I'm not quite sure I understand, but I get tan = sin/cos since tye hypenuses cancel but how can I prove tan(x)/x has the same limit as sin(x)/x? I need to read and understand some of the finer points, or I won't be able to answer some fundamental questions. Any suggestions as to...
21. L'Hopital's Lim x->0 tan x / x

ahhh cos(0) = 1 and sin(0) = 0. I should have verified that b4... Given this, then 2/(1+cos2(0)) = 2/2 = 1. Knowing this, there are 2 things that I should have worked through: secx = 1/cosx and that cos(0) = 1. Thanks very much for your help!
22. L'Hopital's Lim x->0 tan x / x

ok, so could I use sec^2(x) = 1/cos^2(x) then use the power reducing formula to make: limx->0 2/(1+cos2x)? Then as x->0, then the value would be 2/1 thus 2?
23. L'Hopital's Lim x->0 tan x / x

I've done this problem but am not comfortable with the answer. Could someone take a quick look? Limx->0 tan x /x Limx->0 sec^2x ==> 0 I'm just not sure this is right...
24. Integrate 1/(t^2(t-1))dt question

Thanks for the reply. I just realized that as well.... Thanks!
25. Integrate 1/(t^2(t-1))dt question

I've stared at this problem for about 5 minutes and I'm not sure how to start it. I see the function and its derivative, but I'm missing something I suspect is simple. Any suggestions as to how to start?
26. Integrate w/negative exponent

oh wait, re-reading the problem and I think what I'm interpreting as a negative is the dot in the i in sin.... I'm going to work the problem as cos^6x sin^3x dx and I know how to work this... sorry about that!
27. Integrate w/negative exponent

I need a little 'suggestion' as to how to integrate cos^6x sin^-3x dx. I rewrite to cos^6x/sin^3x dx and let u = sinx but when I'm trying to rewrite integral, what should I do with the ^6? Thanks!
28. L'Hopital's Rule - I'm loosing my hair

:-) Thanks for the assistance everyone!!!!
29. L'Hopital's Rule - I'm loosing my hair

VietDao, I had a typo: it should have read: lim x->1 ln x^2/(x^2<b>-</b>1) Sorry for the error. I followed what you had and that makes sense so I think I'm getting it albeit slowly. Using the revised problem, I would get 0/0?
30. L'Hopital's Rule - I'm loosing my hair

1 more example to make sure I'm understanding this...If I have: lim x->1 ln x^2/(x^2+1) this can be rewitten as: lim x->1 (2 ln x)/(x^2+1) hence using the rule, I get: 2 (1/x)/2x which resolves to: 2/2x^2 and since x->1 this equals 1 Am I getting this right?