<p>^I agree "somewhat" with the poster above..will post my take later...</p>
<p>@RESmonkey, try the velocity q from the previous page again, the answer is NOT 2...</p>
<p>^ How about 1? I made two integrals; one from 0 to 1 and the other from 1 to 2. I multiplied the result from the first integral by -1 and added it to the second result to get 1.<br>
Had my fractions in sixths.</p>
<p>diamondbacker:</p>
<p>My teacher said today your right and u can just plug things in for series.</p>
<p>I still dont understand why on the BC 2004 Form A #6 the answers dont do this.
It seems so much simpler compared to taking all the derivatives.</p>
<p>Thanks</p>
<p>RESmonkey, GOOD JOB...1 is the correct answer...make sure you understand that you make 2 integrands, because, the velocity changes direction at 0 and 1...</p>
<p>Let me take A LOOK at 2004 #6, will post in a few...</p>
<p>Here is one that I got wrong, but I do have answer to it:</p>
http://img247.imageshack.us/img247/2286/q16oo4.th.jpg
[/URL</a>]</p>
<p>If anyone gets this one, please explain it to me. </p>
<p>Thanks</p>
<p>@RESmonkey- I cannot read your image, its too small, can you type the question, or post a bigger image, I can try to explain it to you if I can...</p>
<p>^ same here...</p>
<p>Sorry. Imageshack problem. The link I gave u was a thumbnail one.</p>
<p>BTW, I understand it now. Had my teacher explain it. Feel free to take a crack at it, though.</p>
<p>The answer is (E)</p>
<p>x= u^2 + 1
dx = 2u du</p>
<p>The limits change form [5 , 2] to [2, 1]</p>
<p>so sqrt(x-1) = u; dx = 2u du, and x=u^2 + 1; so answer = E</p>
<p>Correct ^. </p>
<p>Someone want to post a problem.</p>
<p>Next q:</p>
<p>lim x-> 6 of ( x - 2 ) / ( x^2 -4x -12 ) = ?</p>
<p>lim x-> 6 of ( x - 2 ) / ( x^2 -4x -12 ) = LIMIT DOESN'T EXIST!!!
The left, and right-hand limits are different (-infinity, and +infinity), so the overall limit doesn't exist. </p>
<p>NEXT QUESTION:
YOU MAY USE A CALCULATOR:</p>
<p>Find the volume generated by the region bounded by: y=x^2, x = y^2; about the line x = -1?</p>
<p>Is it (29/30)*Pi ?</p>
<p>I did shell method:</p>
<p>2<em>pi</em> Integral of (x - -1)(Sqrt(x) - x^2) dx</p>
<p>Decimal is about 3.037.</p>
<p>^yeah, that's correct, GOOD JOB!!!</p>
<p>Now try this one:</p>
<p>Volume: y=4x-x^2, y=8x-2x^2; about x=-2</p>
<p>(256/3)*Pi?</p>
<p>Integral:</p>
<p>(x - -2)( 8x - 2x^2 -4x + x^2)dx from 0 to 4 * 2*Pi</p>
<p>Next q:</p>
<p>Is this function continuous at x =2?</p>
<p>f(x) =</p>
<p>{</p>
<p>x+1 , x <2</p>
<p>x^2 , x=2</p>
<p>2x-1 , x >2</p>
<p>^yeah you got the volume right!!!</p>
<p>I SUCK at continuity...but I'll try anyway...
The function is NOT continuous at x=2 because
1) The lim[x-->2] F(x) =/= F(2)</p>
<p>NEW PROBLEM: (sorry if the above answer is wrong :))
Here's a HARDCORE volume problem:</p>
<p>What is the volume generated when x^2 + (y-1)^2 = 1, is rotated about the y-axis?</p>
<p>Correct ^ :)</p>
<p>Is the answer (4/3)*Pi?</p>
<p>Integral:</p>
<p>(x)<em>(2 * sqrt(1 - x^2))dx from 0 to 1 *2</em>Pi</p>
<p>I'll post a new prob in a sec. Need to find a good one.</p>
<p>Integrate tanx dx.</p>
<p>I always described continuity as like attending graduation. In order for a ceremony to occur (for the function to be continuous):</p>
<p>(1) People have to be heading to the same location (the limit has to exist)
(2) There has to be a graduation location (the function has to exist at the point).
(3) The place people are heading towards has to be the same as the graduation location (the limit has to approach the value of the function at the point in question).</p>
<p>If people are heading to the Mariners game, but the graduation takes place at Seattle University, then the graduation process isn't continuous. :)</p>
<p>nice mathprof. int[tan x]dx = - ln(|cos x|) + C or ln(|sec x|) + C by substituting u = cos x</p>
<p>If F and f are differentiable functions such that F(x) = integral(0 -> x) of f(t)dt, and if F(a) = -2 and F(b) = -2 where a < b, which of the following must be true?</p>
<p>A) f(x) = 0 for some x such that a < x < b
B) f(x) > 0 for all x such that a < x < b
C) f(x) < 0 for all x such that a < x < b
D) F(x) <= 0 for all x such that a < x < b
E) F(x) = 0 for some x such that a < x < b</p>