<p>(-4.-2) tentatively by the ratio test</p>
<p>x=-4: converges by AST
x=-2: diverges by p-series</p>
<p>[-4,2) definitively</p>
<p>I'm pretty sure TheMathProf was correct in his answer - input all of the possible answers for g'(t) and find which is lowest. You shouldn't have to use the second derivative if you're given the possible answers, right?</p>
<p>Yes, it's (a).</p>
<p>Is the answer A?</p>
<p>Hahaha. sorry I didn't see the last few posts..</p>
<p>Whoops! I read that as "find which is cLOSEST to zero", for some reason. My apologies,</p>
<p>Also,</p>
<p>@raller:
The answer should come out to be 0.5x*(sqrt(1-x^2)) + 0.5arcsin(x), so your answer was almost right (no 0.5 in arcsin's argument). Probably just a stupid error in arithmetic :)</p>
<p>ALLRIGHT THE AP TEST IS NEXT WEEK, SO LETS REVIVE THIS THREAD!!!!!</p>
<p>NEW PROBLEM</p>
<p>WRITE BUT DO NOT EVALUATE AN INTEGRAL EXPRESSION FOR THE AREA BETWEEN:</p>
<p>y = x^2, y = 4; about the line y = -1</p>
<p>Okay, I draw the graph, is it:</p>
<p>pi*integral(-1-X^2)^2-(-1-4)^2), limits are a=-2 and b=2?</p>
<p>I don't seem to be able to edit my posts, so I'll post here to fix my earlier post, I think the integrand should be -(5)^2-(1-X^2)^2), because this problem uses disks, not washers.</p>
<p>WAIT:</p>
<p>Vader's question asked for the AREA, not the VOLUME... which doesn't make much sense... so the volume is:</p>
<p>pi * <a href="5%5E2%20-%20(x%5E2%20+%201)">integral</a>dx from -2 to 2 ?</p>
<p>By the washer wethod, the outer radius is 5 and the inner radius is 1 + f(x)...</p>
<p>I think? Somebody check this.</p>
<p>^lol, yeah I meant to say VOLUME not AREA...
anyway, diamondbacker you're right...that's the right answer!!! GOOD JOB! :)</p>
<p>Someone else post a question now!!!</p>
<p>My bad, I made an error in typing my answer:</p>
<p>pi * <a href="5%5E2%20-%20(x%5E2%20+%201)%5E2">integral</a>dx from -2 to 2</p>
<p>Forgot to square the inner radius...</p>
<p>I'll try to find a problem now... but if someone can beat me to it, then that would be great.</p>
<p>^dude, lol, I didn't even catch that...NOW YOU'RE RIGHT!!! :)</p>
<p>ok, while you look for a problem here's an easy integral</p>
<p>Integral [between 1 and e^2] {(ln x)^2} / x dx</p>
<p>
[QUOTE]
^dude, lol, I didn't even catch that...NOW YOU'RE RIGHT!!!
[/QUOTE]
</p>
<p>Hmm..., I plugged in my integral and his into the calculator and I got the same result.</p>
<p>a, b, and c are positive reals. Prove that the function f(t) = a^t + b^t + c^t is increasing for t ≥ 0.</p>
<p>f(x) = a^t + b^t + c^t
f'(x) = (ln a) a^t + (ln b) b^t + (ln c) a^t
f'(x) > 0 for all t > 0 if a,b and c > 1 </p>
<p>I think. the smallest base (a,b or c) can be less than one with the function still increasing.</p>
<p>^yo raller I didn't even notice you post; but you also got the right answer; infact you got it before diamond :)</p>
<p>
[QUOTE]
ok, while you look for a problem here's an easy integral</p>
<p>Integral [between 1 and e^2] {(ln x)^2} / x dx
[/QUOTE]
</p>
<p>Use a u-substitution with u = ln(x) and du = dx / x. You obtain integral<a href="u%20*%20du"> ln(1) to 2 </a>. Since ln( 1 ) = 0, that evaluates to 2.</p>
<p>try again...the answer is NOT 2</p>
<p>the u-substitution method is correct, but when you changed the variable to u, you left it as u, but you have a ln(x)^2 (natural log squared), so it should have been Integral u^2 du; now try to solve it from here</p>