AP Calc BC...

<p>Hey guys, just wondering if you understood the 2 problems below. If so, can you please explain them? thanks! :)</p>

<ol>
<li>f(x)= (x^2-4)/(x-2) if x doesn't equal 2
f(x)= 1 if x=2</li>
</ol>

<p>Let f be the function defined above. Which of the following statements about f are true?</p>

<pre><code> I. f has a limit at x=2.
II. f is continuous at x=2.
III. f is differentiable at x=2.
</code></pre>

<ol>
<li>On the graph of y=f(x), the slope at any point (x,y) is twice the value of x. If f(2)=3, what is the value of f(3)?</li>
</ol>

<ol>
<li>I’d have to say only I. is right. II. is wrong because the function f(x) is essential x+2 (x=/= 2). At x= 2, it would be f(x)= 3 but as it’s stipulated that when x= 2, f(x)= 1, the function is discontinuous at x= 2.</li>
</ol>

<p>III. is wrong because II. is wrong. f(x) is discontinuous at x= 2 which means it’s not differentiable there.</p>

<p>This leaves I if that is the correct answer…</p>

<ol>
<li>The slope at any point (x,y) is twice the value of x. From this, we have dy/dx= 2x. Simple integration gives us y= x^2 + C. Plug in f(2)= 3 and you get C= -1. Thus, f(x)= x^2 -1 so f(3)= 8.</li>
</ol>

<p>thank you R3d3mpti0n!</p>

<p>Can somebody help me figure out a few more questions plz? thanks so much in advance! and good luck with everybody on their AP tests! :)</p>

<ol>
<li><p>What integral represents the area enclosed by the smaller loop of the graph of r=1+2sin(theta)?</p></li>
<li><p>The graph of the function f consists of four semicricles. (From x=-3 to x=-2, the semicircle is above x-axis, from x=-2 to x=0, semicircle is below x-axis, from x=0 to x=2, semicircle is above x-axis, from x=2 to x=3, semicircle is below x-axis.) If g(x)= integral of
f(t)dt with limits of integration of 0 to x, where is g(x) nonnegative?</p></li>
</ol>

<p>(Answer is supposed to be [-3, 3] but i don’t understand why.)</p>

<ol>
<li>A rectangle with one side on the x-axis as its upper vertices on the graph of y=cos s, as shown i nthe figure above. What is the minimum area of the shaded region? (graph is basically cos graph from x=-pi/2 to +pi/2…and rectangle is on the x-axis where the top right and top left corners touch cos graph and rectangle is symmetrical about y axis)</li>
</ol>

<p>(answer is supposed to be 0.878 but I don’t get why)</p>

<ol>
<li><p>A solid has rectnagular base that lies in the 1st quadrant and is bounded by the x- and y-axes and the lines x=2 and y=1. The height of the solid above the point (x,y) is 1+3x. What is a Riemann sum approximation for the volume of the solid?</p></li>
<li><p>If the function f is defined by f(x)=square root of (x^3+2) and g is an antiderivative of f such that g(3) =5, then g(1)= ?</p></li>
<li><p>Let g be the function given by g(x)=integral of (100(t^2-3t+2)e^-t^2) dt with limits of integration from 1 to x. Which of the following statements about g must be true?
I. g is increasing on (1,2)
II. g is increasing on (2,3)
III. g(3) is greater than 0.</p></li>
</ol>

<p>can somebody help me with at least some of these problems please? (it can also help you with your studying too :)</p>

<p>For 2007 part a, how did they find the area?</p>

<p><a href=“ap07_sg_calculus_bc.pdf”>ap07_sg_calculus_bc.pdf;

<p>@Staller: do you mean how did they get the integral for area or how did they get the decimal answer? Here’s how you get a decimal answer for an integral on a TI-83/84:</p>

<p>MATH -> 09: fnInt(function,X,lowerbound,upperbound)</p>

<p>So they did fnInt((20/(1+x^2)-2,X,-3,3)</p>

<p>@sporty:

  1. What integral represents the area enclosed by the smaller loop of the graph of r=1+2sin(theta)?</p>

<p>First set r to 0 and solve for theta to find the limits of integration. 0=1+2sin(theta) and you eventually get -pi/6 and 7pi/6
Then you use the formula 1/2 int r^2 dtheta
So it’s 1/2 int (1+2sin(theta)) dtheta from -pi/6 to 7pi/6</p>

<p>any good cram/review sheets, besides the one from scribd. </p>

<p>i like that review sheet; however, it does not hit some of the concepts that are tested on the college board exam, specially taylors . . .</p>