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Help my friend since I suck at calc - Page 2

Blogs > Raithed
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goldrush
Profile Blog Joined June 2004
Canada709 Posts
Last Edited: 2008-04-02 04:33:23
April 02 2008 04:33 GMT
#21
If you know how to do partial fractions, then questions like #15 are done by first taking out the 1/3 (it's a lot easier), then factoring out the denominator into x+2, x-2 and x. You assign a variable such as A/B/C to each numerator. Therefore:

A/(x+2) + B/(x-2) + C/x = (x+4)/(x^3-4x), solve for each one and then take the integral of the individual fractions. That's the partial fraction method.
Saracen
Profile Blog Joined December 2007
United States5139 Posts
April 02 2008 04:33 GMT
#22
12. is also easy...
the derivative of -x^2+4x+5 = ?!?
-2x + 4
which just happens to be -2 times (x-2)
oh snap
function to a power times derivative
doob10163
Profile Joined April 2008
6 Posts
April 02 2008 05:00 GMT
#23
On April 02 2008 13:30 Saracen wrote:
11 is just trig sub...
it's really straightforward...idk why it's hard?
sec(theta) = sqrt(4y^2+1), dy=(sec^2(theta))d(theta)/2

What exactly am I substituting for what here?

Thanks for the help, by the way, i really appreciate it.
goldrush
Profile Blog Joined June 2004
Canada709 Posts
Last Edited: 2008-04-02 06:43:51
April 02 2008 06:42 GMT
#24
Well, I'll try and help you a tad before I go to bed.

Since you're not familiar with this stuff, don't try and take shortcuts. Use #17 as an example:

Construct a right-angle triangle with a hypotenuse at sqrt2 and the two other sides as x and sqrt(2-x^2). Using pythagoras, you can tell it's right (x^2 + 2 - x^2 = 2, which is the square of the hypotenuse). Alright now...

using this triangle, deduce that sin theta = x / sqrt2. Therefore, sqrt 2 * sin theta = x, sqrt 2 cos theta dtheta = dx. You still need to replace sqrt(2-x^2) and x^2 in the original equation though, so...

cos theta = sqrt (2- x^2) / sqrt2
sqrt (2 - x^2) = sqrt2 * cos theta

original equation of sin theta = x / sqrt2, deduce that x^2 = 2 sin^2 theta.

Now, plug it all into the formula:

sqrt 2 costheta dtheta / (2 sin^2 theta * sqrt 2 * cos theta)
Simplifies oput to 0.5 dtheta/ sin^2 theta. Integrating this turns out to be -0.5 costheta / sintheta.

But this is a definite integral and the bounds are also changed becuase of the variable change. Using sintheta = x / sqrt2, plug in 1 and sqrt 2 in the place of x and solve for theta. It turns out to be theta = 90 and 45 degrees.

So plug it in... -0.5 [ cos 90 / sin 90 - cos 45 / sin 45] = -0.5 [ 0 / 1 - 1 ] = 0.5

Apologies for any mistakes, it's really late over here. hope this helped.
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