Author Topic: TheJosh's Questions (Read 1104 times) Tweet Share

TheJosh · « **on:** September 10, 2009, 09:53:57 pm »

Hey guys here are some questions if you could please help me with

1. Find antiderivative of (x²+1)(x²-1)/x²

and could someone please tell me how to do the maths symbols on here $:-\$ not sure how to do

Flaming_Arrow · « **Reply #1 on:** September 10, 2009, 09:55:21 pm »

hint: expand brackets and divide, should get x^2 - 1/x^2

dekoyl · « **Reply #2 on:** September 10, 2009, 09:59:02 pm »

Numerator expands to $x^4-1$
Then denominator can simply the whole thing to $x^2 - x^{-2}$
So anti-derivative: $\frac{1}{3}x^{3}+\frac{1}{x} + C$

Math text is fucked at this time.

TheJosh · « **Reply #3 on:** September 10, 2009, 10:10:01 pm »

so its not just me?

the latex stuff not working is it?

TrueTears · « **Reply #4 on:** September 10, 2009, 10:12:22 pm »

LaTeX server is down.

TheJosh · « **Reply #5 on:** September 14, 2009, 09:41:51 pm »

Hey how do you put the latex on here?

TrueTears · « **Reply #6 on:** September 14, 2009, 10:00:07 pm »

http://vcenotes.com/forum/index.php/topic,10280.0.html

TheJosh · « **Reply #7 on:** September 14, 2009, 10:02:59 pm »

2. If $f'(x)= a\cos 3x$ and $f'(0)= 2$ and $f(\frac{\pi }{4})= \frac{1}{2}$ find f(x)

cobby · « **Reply #8 on:** September 14, 2009, 11:27:28 pm »

Quote from: TheJosh on September 14, 2009, 10:02:59 pm

2. If $f'(x)= a\cos 3x$ and $f'(0)= 2$ and $f(\frac{\pi }{4})= \frac{1}{2}$ find f(x)

$f'(0) = 2 \therefore 2 = cos 3(0)$

$\implies a = 2$

$\int a cos ( 3x) = \int 2 cos ( 3x)$

$= \frac{2}{3} sin ( 3x) +c$

Sub $f( \frac{\pi}{4}) = \frac {1}{2}$ and solve for $c$

$\therefore c = \frac{1}{2} + \frac {\sqrt{2}}{3}$

$\implies f(x) = \frac {2}{3} sin ( 3x) + \frac{1}{2} + \frac {\sqrt{2}}{3}$

TheJosh · « **Reply #9 on:** October 09, 2009, 06:04:47 pm »

cool as!! thanks guys

yeah thats the answer cobby

vamsiaus · « **Reply #10 on:** October 09, 2009, 09:59:40 pm »

lol, just a month too late :p

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