<p>also, I think limits of this kind are gone, but they still illustrate an important concept:</p>
<p>Compute lim(x -> 0) of (1+2x)^(csc x)</p>
<p>I love your solution TheMathProf...</p>
<p>RESMonkey, CORRECT...
can you briefly explain to me how you setup the integral...I know how to set it up, but I don't know how you got the limits...I get -1 and 1, not 0 and 1...</p>
<p>NEVER MIND my above post...I FIGURED IT OUT!!! :)</p>
<p>The answer is e^2...</p>
<p>do ln|y| = ln|(1 + 2x)^(csc x)|
lim(x>>0) ln|(1 + 2x)^(csc x)| HOPITALS RULE you get 2 / (1+2x) (cos x) = 2
so ln|y| = 2
so y = e^2</p>
<h2>so FINAL ANSWER==e^2</h2>
<p>IS YOUR ANSWER (A) on the previous page; I think its Rolle's Theorem..not 100% though, but on the AP test I would guess A :)</p>
<p>If the function is continuous and differentiable everywhere, then A has to be the answer due to The Mean Value Theorem.</p>
<p>vader90, Thanks for solving the cscx limit one. I looked at your solution to guide me through.</p>
<p>You figured the volume one's integral right?</p>
<p>Nice solution vader1990, another method is to write it as an exponential with base e and proceed.</p>
<p>Also, the one on the previous age is a consequence of the Mean Value Theorem. Particularly the case where integral(a -> b) of f(x)dx = f(c)*(b-a) for some c if f is continuous on <a href="this%20actually%20shows%20up%20as%20a%20definition%20question%20later%20on%20in%20the%20MC">a,b</a></p>
<p>Here's a good testing for convergence/divergence question:</p>
<p>Which of the following diverge?</p>
<p>I. Sigma(k = 3, +inf) of 2/(k^2 + 1)
II. Sigma(k = 1, +inf) of (6/7)^k
III. Sigma(k = 2, +inf) of (-1)^k / k</p>
<p>Compute the derivative of y = [(x^2 + 8)^(1/3)] / [(2x + 1)^(1/4)] at x = 0.</p>
<p>Let f(x) = integral(-2 to x^2 - 3x) of e^(t^2) * dt. At what value of x is f(x) a minimum.</p>
<ol>
<li><p>What is the area of the region bounded by the curves y = x^2003 and y = x^(1/2003) and lying above the x-axis?</p></li>
<li><p>Find all y > 1 satisfying integral(1 -> y) of x<em>ln(x)</em>dx = 1/4.</p></li>
<li><p>A plane curve is parameterized by x(t) = integral(t -> +inf) of [cos(u)/u]<em>du and y(t) = integral(t -> +inf) of [sin(u)/u]</em>du for 1 <= t <=2. What is the length of the curve?</p></li>
<li><p>Compute integral(0 -> pi/3) of x*tan^2(x) * dx.</p></li>
<li><p>Let f and g be functions that are differentiable for all real numbers, with g(x) =/= 0 for x =/= 0. If lim(x -> 0) of f(x) = lim(x -> 0) of g(x) = 0. and lim(x -> 0) f '(x) / g '(x) exists, then lim(x -> 0) of f(x)/g(x) is...</p></li>
<li><p>Let a and b be constants such that lim(x -> 1) of [(ln(2-x))^2] / (x^2 + ax + b) = 1. Find (a,b).</p></li>
<li><p>Let f(x) = x + a*sin(x), where a is a positive constant and -2pi <= x <= 2pi.</p></li>
</ol>
<p>a) Determine the x-coordinates of all points , -2pi <= x <= 2pi, where the line y = x + b is tangent to the graph of f(x) = x + a*sin(x).</p>
<p>b) Are the points of tangency described in part (a) relative maximum points of f? Explain.</p>
<p>c) For all values of a > 0, show that all inflection points of the graph of f lie on the line y = x.</p>
<p>That's a pretty comprehensive set of problems. Hint for 4: there are 2 general integration techniques, substitution and parts, choosing which to do first will solve the problem. Hint for 6: apply 5.</p>
<p><em>EDIT</em> If anyone is looking for a particular type of problem, I'll try my best to dig through and reproduce the problem with different symbols and wording.</p>
<p>OK, I realized I may have killed this thread with too many problems, some of which are more difficult than the usual kind. At the risk of triple posting, I'm going to post 3 Calc BC MC questions to revive this thread.</p>
<ol>
<li><p>If c is a positive integer, the lim(x -> +inf) of (x^c)/(e^x) = ?
A) 0 B) 1 C) e D) c! E) DNE</p></li>
<li><p>Suppose f and g are functions that are everywhere differentiable. Given that g is the inverse function of f and g(-2) = 5 and f '(5) = -1/2, then g '(-2) = ?</p></li>
</ol>
<p>A) 2 B) 1/2 C) 1/5 D) -1/5 E) -2</p>
<ol>
<li>If integral(1 to 4)[f(x)<em>dx] = 6, compute integral(1 to 4)[f(5-x)</em>dx]
A) 6 B) 3 C) 0 D) -1 E) -6</li>
</ol>
<p>^LOL.</p>
<p>Well, heres your first one: None of those series diverge. They all converge.</p>
<p>Ok, I will solve these out, when I get back from school today :)</p>
<p>I'm going to give limited explanation, let me know if you need more I'll be happy to hook u up :)</p>
<ol>
<li>E (it doesn't exist)</li>
<li>E (f-inverse = 1 / (f ' { f-inverse[x]} so it yields -2) </li>
<li>E (fundamental theorem of calc)</li>
</ol>
<p>I may be wrong about these, I did all 3 in like 45 seconds :)...and I HATE having 3 of the same letters in a row on tests, it drives me NUTS!!! :)</p>
<p>that is correct RESMonkey. vader, careful on 1 and 3. for number one, note that initially for any positive integer, the limit would approach +inf/+inf which should hint at which rule to apply (this one was tricky). For 3, note that 5-1 = 4 and 5-1 = 1, which are the limits of integration of the original integral. This should tell you how to approach the integral.</p>
<p>^wait, how is #3 not E?</p>
<p>fundamental theorem yields:</p>
<p>integral [1, 4] f(x)dx == F(4) - F(1) = 6 (given in question)
2nd part: integral [1, 4] == F(5-4) - F(5-1) == F(1) - F(4)...well, if you multiply the original equation from above step by -1, you get -F(4) + F(1) = -6...so E????</p>
<p>I AM CONFUSED????</p>
<h1>1 LOL...I misread it...I thought you meant INTEGRAL and not LIMIT...LOL..so I thought you meant Integral (x, infinity) of x^c / e^x, and given I have never seen anything like this, i just said DNE...lol...</h1>
<p>anyway #1..again...
well, its simple, just l'hopital it!!!
and eventually you'll get c! / e^x, so plug in infinity, and you get c! / infinity = 0</p>
<p>The answer is A. You have to be very careful if you are going to use the fundamental theorem here. Suppose that F(x) is the antiderivative of f(x). Then your first step is correct, F(4) - F(1) = 6. Note however that the antiderivative of f(5-x) = -F(5-x) + C because differentiating this (and keeping in mind the chain rule) gives you f(5-x). Therefore, applying the fundamental theorem again gives integral[1, 4] f(5-x) = -[F(5-4) - F(5-1)] = F(4) - F(1) = 6. </p>
<p>To avoid this trap, substitute u = 5-x and see what happens. I think substitution is the best way to relate two similar integrals.</p>
<p>AAAAAAAWWWWWWWWW THOSE SNEAKY COLLEGEBOARD ********!!!</p>
<p>Thanks snipez90, I better get tricky problems like this wrong now then on the real AP Test!!!! THANKS A LOT FOR THE HELP!!!!</p>
<p>Do you by any chance have anymore "tricky" problems like this? Thanks again :)</p>
<p>Exactly!</p>
<p>We need tricky problems!</p>
<p>Haha, no problem vader. Don't worry about it too much, it's pretty easy to forget to apply the chain rule sometimes. Anyways, two problems I found somewhat tricky.</p>
<p>Sigma<a href="1/3">k = n, +inf</a>^k = </p>
<p>A) 3/2 - (1/3)^n
B) 3/2[1 - (1/3)^n]
C) (3/2)<em>(1/3)^n
D) (2/3)</em>(1/3)^n
E) (2/3)*(1/3)^(n+1)</p>
<p>Which of the following series converge?</p>
<p>I. Sigma[n = 1, +inf] [(-1)^(n+1)] * 1/(2n+1)</p>
<p>II. Sigma<a href="1/n">n = 1, +inf</a>*(3/2)^n</p>
<p>III. Sigma[n = 2, +inf] 1/(n*ln(n))</p>
<ol>
<li>C [(3/2)*3^-n]</li>
</ol>
<p>snipez, in reply to this question:</p>
<p>
[quote]
Compute the derivative of y = [(x^2 + 8)^(1/3)] / [(2x + 1)^(1/4)] at x = 0
[/quote]
</p>
<p>I got an answer of -1.</p>
<p>snipez, in reply to this question:</p>
<p>
[quote]
Let f(x) = integral(-2 to x^2 - 3x) of e^(t^2) * dt. At what value of x is f(x) a minimum.
[/quote]
</p>
<p>x = 3/2 is my answer. Fundamental Rule of Calculus 1 for the win.</p>
<p>snipez, for your post with the first question with the x^2003 and x^(1/2003), is it a calc or noncalc? </p>
<p>And how do you do that? I know x(1/2003) > x^2003 from 0 to 1 and that you need 2 integrals.</p>