cant see my mistake

[Alem2000]

I cant seem to get the correct answer but I keep getting very close the question is \int_{0}^{\pi/2}xcos(x)dx

this is what I did


u=x


v=1/2(sin(2x))dx



du=dx

dv=cos(2x)dx

and then i get =1/2(xsin(2x))\right_{0}^{\pi/2}-1/2\int_{0}^{\pi/2}sin(2x)dx

then 1/2xsin(2x)\end_{0}^{\pi/2}-1/2(-1/2cos(2x0)\end_{o}^{\pi/2}

OKay!....horrible latexing but all in all i used the substitution theroem two times and my final answer ir \pi/4-1/2 and the correct answer in the book is 1/2 why do i get the wrong answer?

[arildno]

But isn't it cos(x) rather than cos(2x) which appears in the integrand??

[pig]

I don't understand what you are doing there but I also got a solution different from the one in the book.

f(x) = xcos(x)

u=x, dv=cos(x)dx
du=dx, v=sin(x)

F(x) = xsin(x) - INT(sin(x)dx) = xsin(x) + cos(x) + C

F(pi/2) - F(0) = pi/2 * 1 + 0 - (0*1 + 1) = pi/2 - 1

[arildno]

The indefinite integral of xcos(x) is xsin(x) + cos(x), so shouldn't the solution be pi/2 - 1? :confused:
The indefinite integral of xcos(2x) is 1/2xsin(2x)+1/4cos(2x) so the answer ought to be -1/2 :confused:

[Math Is Hard]

I don't see why you have
dv=cos(2x)dx
if the original integral is
\int_{0}^{\pi/2}xcos(x)dx

typo?

[Math Is Hard]

:blushing: sorry, I guess arildno already mentioned that!

[arildno]

I'm quite sure the answer is -1/2, and the integrand xcos(2x)

[Math Is Hard]

I worked it out and got -1/2 also

[arildno]

Then I'm COMPLETELY sure! :smile:

[Math Is Hard]

LOL! :rofl:

[Alem2000]

ahhh yes there is a typeo in my previous post....this is the problem...


\int_{0}^{\pi/2}xcos(2x)dx

by the way in the problem \int_{-1}^{1}x^4/5..after splitting it up from -1 to 0 and 0 to -1 the area i got was 0....? isnt that impossible...ohhh..opps i think i just realized what happent...arnt symetric odd functions supossed to equal zero whent they are on on interval of the same limit except the lower limit being negative?....why was this in chapter 4 and the problem arises in 5.9???????..... :surprised ..brain fart...can someone tell me if im on the right track? right track? right track....ECHO!!!

[Math Is Hard]

by the way in the problem \int_{-1}^{1}x^4/5..after splitting it up from -1 to 0 and 0 to -1 the area i got was 0....? isnt that impossible...ohhh..opps i think i just realized what happent...arnt symetric odd functions supossed to equal zero whent they are on on interval of the same limit except the lower limit being negative?....why was this in chapter 4 and the problem arises in 5.9???????..... :surprised ..brain fart...can someone tell me if im on the right track? right track? right track....ECHO!!!

x4 /5 is not an odd function, though. it has even symmetry. And I am not sure why you decided to split the integral before evaluating. I mean you can if you want, but it seems kinda unnecessary. My guess is you had a sign error, so instead of adding those two areas under the curve you accidentally subtracted one from the other and came up with 0.

oh...did you mean x4/5 ? This should also have even symmetry.

[Alem2000]

Sorry its been a long day this is the function i meant to post


\int_{-1}^{1}x^-4/5

the function has a vertical asymptote if im not mistaken. Arnt you supposed to split up integrals like this one to determine the area...i mean it just keeps going to infinity? So i split it up and took the limit.....? I got zero

[Math Is Hard]

yeah, you will have to split. You've got an improper integral now. Have you dealt with these before? We are talking about the function x -4/5 right?

[Alem2000]

yes we definately are...talking about the function you said. I just suck at latexing im really trying to get better :biggrin: And yes I have delt with improper integrals..this is the only one whose value came out to be zero....so going back to my old qustion....how can that make sense....how can there be no area under that curve? If its my misteak... :cry: which it probably is...can you tell me what i did?

[Math Is Hard]

well you can start by taking one of your integrals and finding out if it converges to a limit or if it is divergent.
OK, now I'm in Latex hell :biggrin:

\int_{0}^{1}x^-(4/5)
equals

lim x->t \int_{t}^{1}x^-(4/5)

I'll try to go back and clean this up.Is this what you were doing, though?

[Gokul43201]

Free clean up. Sunday night special !!

\int_{0}^{1}x^{-(4/5)}dx = lim_{x->t} \int_{t}^{1}x^{-(4/5)}dx

[Math Is Hard]

Gokul - you are the best!!!! Thanks!!

[Gokul43201]

I'm not pretending to understand what you mean, however.

[Math Is Hard]

whoops my bad. t goes to 0! sorry bout that! :redface:

\int_{0}^{1}x^{-(4/5)}dx = lim_{t->0} \int_{t}^{1}x^{-(4/5)}dx