I know that its a property of integrals that the integral of (a+b) = the integral of a + the integral of b, but is it the same for products? In other words, is the integral of a*b the same as the integral of a*the integral of b? Any help would be appreciated (my first calcII test tomorrow!) TIA

No. The integral of a*b is not the integral of a * the integral of b. ex: int(x*x)dx does not equal int(x)dx * int(x)dx 1/3x^3 does not equal 1/2x^2 * 1/2x^2 1/3x^3 does not equal 1/4x^4

Unfortunately not. Life would be a lot easier if it was... Look at it this way: x^2 can be written as x times x. The integral of x^2 is x^3/3, and the integral of x is x^2/2. Multiply those together, and you don't get x^3/3... You can usually get a good feel for things if you do a lot of examples. If you have a x^2 term and a exponential (e^x, or ln(x) for example) times each other, you usually use 'integration by parts', for example.

imtiaz - somehow we managed to choose the exact same example... <img border="0" alt="[Laughy]" title="" src="graemlins/laughy.gif" />

You have to use a method called "integration by parts" to integrate products INT(udv) = uv - INT(vdu) the trick is to express one of the terms of the products as the derivative of some "v". hope this helps.

Great minds think alike? </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by brandonite: <strong>imtiaz - somehow we managed to choose the exact same example... <img border="0" alt="[Laughy]" title="" src="graemlins/laughy.gif" /> </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">

So many geeks so few examples <img border="0" alt="[Laughy]" title="" src="graemlins/laughy.gif" /> I saw this post with zero replies, and in the 1 min it took me to reply, there were already 5 replies.

Exactly. The best example of this is finding int[ln(x)]dx int[ln(x)]dx can be written int[(ln(x)*1]dx int[ln(x)*1]dx the integration by parts formula is: int[u(x)*dv] = u(x)*v(x) - int[v(x)du] if we set: dv=1 u=ln(x) Integrate one, differentiate the other (sometimes it takes trial and error to figure this out) then: du=1/x v(x) = int[dv]dx = int[1]dx = x Finally: int[ln(x)*1] = x*ln(x) - int[x*1/x]dx int[ln(x)*1] = x*ln(x) - int[1]dx which reduces to: int[ln(x)*1] = x*ln(x) - x which is the integral of ln(x) <img border="0" title="" alt="[Wink]" src="wink.gif" /> </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Original: <strong>You have to use a method called "integration by parts" to integrate products INT(udv) = uv - INT(vdu) the trick is to express one of the terms of the products as the derivative of some "v". hope this helps.</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">

Thank you so much for the fast replies. I originally thought that it wasn't the case, but when I was in class we were going over homework and I thought I heard my teacher say that the answer to the question was true, not false. I probably just heard him wrong b/c he moves very fast, and any slight distraction makes me miss a point. Thanks again! One more question...anyone know where I can find a sample Calc II test online? Im a little nervous that I did all of the prof's samples earlier and have nothing to do now.

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Original: <strong>You have to use a method called "integration by parts" to integrate products INT(udv) = uv - INT(vdu) the trick is to express one of the terms of the products as the derivative of some "v". hope this helps.</strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Let me add to my man Original's brilliant comments by pointing out that the order in which you assign u and v to parts of the original integral is important. In other words, the assignment of u and v is not arbitrary. For example, lets say you have INT(x*cos x). there are 2 ways you can use integration by parts. Method 1: Let u = x, du = 1 dx then dv = cos x dx, v = sin x Then INT(x*cos x dx) = uv - int(v du) = x*sin x - int(sin x*1*dx) = x*sin x + cos x this method is the correct way. Now look at the alternate pathway: Method 2: Let u = cos x, du = -sin x dx then dv = x dx, v = 1/2x^2 int(x*cos x) = uv - int(v du) = 1/2x^2*cos x - int(1/2x^2*-sin x dx) Note on this example that you havent really gotten anywhere. The integral you have to evaluate using this approach is just as difficult to determine as the original integral you were trying to solve (actually its slightly WORSE than the original integral) Not that I really needed to post this, but thought I might as well add to the thoughts of my brilliant colleagues. Now, if Original (being the math major of the group) will just provide a proof of the product formula. <img border="0" alt="[Clappy]" title="" src="graemlins/clappy.gif" />

<a href="http://www.math.rutgers.edu/~greenfie/courses/exams-gif.html" target="_blank">Exams</a> <a href="http://www.math.ucdavis.edu/~hass/Calculus/Exams/Examprobs/List.html" target="_blank">Assorted Problems</a> <a href="http://www.math.montana.edu/courses/m182/oldtests/" target="_blank">A Whole Slew of Exams</a> <a href="http://www.math.columbia.edu/~ikofman/oldexams.html" target="_blank">More Calculus Exams</a> Have Fun I <img border="0" alt="[Lovey]" title="" src="graemlins/lovey.gif" /> CALCULUS

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by baylor21: [QB Not that I really needed to post this, but thought I might as well add to the thoughts of my brilliant colleagues. Now, if Original (being the math major of the group) will just provide a proof of the product formula. <img border="0" alt="[Clappy]" title="" src="graemlins/clappy.gif" /> [/QB]</font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">Nah! nah! nah! I'm not the only math major that helped answer Nubian's question. Brandonite, and mpp are math majors. I don't know Imitaz's major, but I know he/she knows at least as much math as the rest of us.In any case I'll go ahead and take you up on the challenge. O.k here we go! Start with uv, where u = u(v) differentiate uv w.r.t "v" by "product rule" [d(uv)]/dv = v(du/dv) + u integrate both sides w.r.t "v" uv = INT[v(du/dv)] + INT(u)dv reshuffle INT(u)dv = uv - INT(v)du QED that stuff is trivial after advanced calculus, differential geometry, topology, and real analysis death camps.

I'm a CHEM major, but I have a minor in math if that counts for anything. </font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by Original: <strong>Nah! nah! nah! I'm not the only math major that helped answer Nubian's question. Brandonite, and mpp are math majors. I don't know Imitaz's major, but I know he/she knows at least as much math as the rest of us.In any case I'll go ahead and take you up on the challenge. O.k here we go! </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">

Just some advice that got me through the Calc Sequence(you may already know this, but this helped me a lot): Try not to overlook simpler methods, sometimes teachers will put superhard looking integrations that have simple solutions: Simple example: 2xSin(x^2)cos(x^2) can be done with substition instead of integration by parts... well actually u cant do this one with integration by parts.. but you get the idea 2. Always look for symmetry about origin or axis. EX: int((cosx)^3 * sin(x)^150)) Looks intimidating, but you know its zero cause symmetry by origin. If you get a function symmetrical by axis, always switch the bounds and multiply by two cause the algebra becomes twice as easy 3. Last but not least, if you get a function that looks good in polar, switch to polar coordinates. Good candidates are those that if you drew a line from origin, would hit every point on the outside (I.E. circles ARE TERRIFIC for polar coordinates) Well.. dats it! Good Luck on the Test TobTolip

Y'see, ya don't want to give the Math majors an excuse to talk shop... Anyway, nice proof Original. I just spend the morning today proving all of the series convergence tests - the limit comparison test, the integral test, etc... There were a few nice proofs there. Nubian - if all else fails, graph it. If you see trig functions, reduce them all to sines and cosines. Always simplify as much as you can. If you come across a question you don't know the answer to, play around with it for a bit and see what falls out. There's probably an easier way...

The newest tool i found: <a href="http://www.integrals.com" target="_blank">www.integrals.com</a> It does Symbolic integration for you! now, to all those people with those damn fancy calculators... HAH! we got it too, w/o a super computer for a calc. and of course.. <a href="http://www.treasure-troves.com" target="_blank">www.treasure-troves.com</a> it's the source for formulas for everything.. math included.

<a href="http://www.integrals.com" target="_blank">www.integrals.com</a> doesnt do ln(x). it even doesnt do log(e,x) meaning log base e of x, which IS ln(x). try it!

Wow your right!!! I have a suspicion that they are using a simple algorithm that takes just the power rule into account somehow... but yea that is weird.. then again it says its "powered by mathematica"... course I never trusted mathematica anyways (I don't trust anything that has the color of yellow, and tends to erase all my work because even tho I "saved" ) Tob

my TI-92 does ln(x). its more powerful than your silly CGI script <img border="0" alt="[Laughy]" title="" src="graemlins/laughy.gif" />

It IS on the mathematica webpage... the same page has info on ordering mathematica and stuff.. .so i'd guess it is powered by mathematica. I found a different CGI script online, and it didn't have "e" in it... my answers were ridiculously long and full of ln(e)... hmm... this doesn't do ln? I haven't had enough time to explore it (i found it yesterday.... but i wish i found it years ago.. but i was already getting mad when there server was down! it's spoiling me). and your damn TI 92 is totally spoiling you!! you're never going to learn any math with those computers. Hah! we're never going to learn any ANYTHING b/c you can put a cheat sheet in calculator's memory. ... no one has told me whether or not that is cheating. Now, if I could download the CGI script so it would work offline... that would be cool...

</font><blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr /><font size="2" face="Verdana, Helvetica, sans-serif">Originally posted by imtiaz: <strong><a href="http://www.integrals.com" target="_blank">www.integrals.com</a> doesnt do ln(x). it even doesnt do log(e,x) meaning log base e of x, which IS ln(x). try it! </strong></font><hr /></blockquote><font size="2" face="Verdana, Helvetica, sans-serif">It does do ln(x). you just have to read the confusing instruction manual.. <a href="http://www.integrals.com/about/input/" target="_blank">http://www.integrals.com/about/input/</a> and E has to be capitalized. it yields -x+xlogx That looks correct, but i'm too lazy and tired to check with a table or derivatives if it is correct.