<p><a href=“Supporting Students from Day One to Exam Day – AP Central | College Board”>Supporting Students from Day One to Exam Day – AP Central | College Board;
<p>Page 36, number 21. Such a simple problem… yet I am not getting any of the choices. Can someone please explain?</p>
<p>Here’s how I got ‘B’:</p>
<p>1) Given: f(x)=sqrt(x^3 + 2)
g(x) = [integral] f(x)dx
g(3) = 5</p>
<p>2) They want to know g(1)=?, so set up a definite integral from 1 to 3
[def. int. from 1 to 3] f(x)dx = g(3)-g(1)
We know g(3), and we can solve the def. int., so all we have to do is solve for our unknown, g(1)</p>
<p>3) Rearrange to get: g(1) = g(3) - [def. int. of f(x) from 1 to 3]</p>
<p>4) Plug into calculator, answer is -1.585</p>
<p>^Noncalc portion.</p>
<p>EDIT: Actually, it is the calculator portion. (see page 33)
How would you do it without a calculator?</p>
<p>■■■■. I’m stupid. >.><br>
I did it by taking the antiderivative, then plugging g(3) to solve for C. Then plugging in 1, and kept on get 2.something.</p>
<p>D’oh.</p>
<p>But what did you get for the antiderivative of sqrt(x^3 + 2)? I’m not sure how to integrate it.</p>
<p>Well I obviously didn’t get the right answer… but here’s what I got: [(2/3)(x^3+2)6(3/2)]/(3x^2)</p>
<p>Can anyone also explain numbers 15 and 18 on the Calc BC portion of the above link?</p>
<p>Wait, what the ****? There is imaginary numbers on AP Calculus BC???</p>
<ol>
<li> For this problem, I look at it in two halves.
1st Half - Positive x’s
-Because g=[def int. from 0 to x] f(x)dx, for this half of the problem it’s just area under the curve from zero to ‘insert x value’. It’s obvious that the area from x=0 to x=2 is positive, and it’s always clear that the area from x=2 to x=3 is smaller from the area from x=0 to x=2, so we can conclude that for [0, 2], g(x) is positive.</li>
</ol>
<p>2nd Half - Negative x’s
-Because g=[def int. from 0 to x] f(x)dx, if we wan’t the area from x=0 to x=‘any negative number’, it can be rewritten as the NEGATIVE of the integral from x=‘negative number’ to x=0. The area from x= -2 to x=0 is negative, so multplied by -1 we get a positive answer. The area from x= -3 to x= -2 is smaller than the area from x= -2 to x= 0, so we can conclude that g(x) is positive from x= -3 to x= 0 (because from -3 to 0 the integral is negative, but from 0 to -3 the integral is positive (negative * negative) hope I explained it clearly enough)</p>
<p>Therefore, g is positive for [-3, 3]. I’ll check out #18 right now.</p>
<p>Also, lemone, the ‘i’ in problem 18 isn’t an sqrt(-1). I guess you could think of it as a variable - it’s the sum as ‘i’ goes from ‘1’ to ‘n’. Here’s the first few terms of answer a (just for an example) and I’ll keep ‘i’ in the parenthesis: SUM = [1 + 3(1)/n] + [1 + 3(2)/n] + [1 + 3(3)/n] + … + [1 + 3(n)/n]</p>
<p>Oh… I see. I thought about that afterward but never really worked it out since I already dismissed it (for #18). Thanks!!! </p>
<p>Also, I meant to write “[(2/3)(x^3+2)^(3/2)]/(3x^2);” but like I said, it’s still wrong :/</p>
<p>Okay I’ll try to explain #18… but when I get to the Reimann sums part I’ll probably mess up explaining it (I’m tired - feel free to correct me if I’m wrong).</p>
<p>I’ll split it up into steps:</p>
<p>1) Set up the graph
-This is pretty simple. Plot y=2, x=1, x=0, y=0, and you have your rectangular base.</p>
<p>2) Set up the integral
-The normal way to find the volume of this shape is to just integrate the cross-sectional area, so let’s do that. The general form is [integral from a to b] A(x)dx. Now we need to find A(x). Since the height is just straight (the height at each point is y=1+3x), each cross-section is going to be a rectangle. Area of a rectangle = base * height, so we need to identify those. The height they give us (y=1+3x), so we need to find the base. If you look at the graph, it’s pretty clear that the base is always 1, since 1 is the distance from y=1 to the x-axis. Now we plug back into A(x). A(x) = b * h = (1 + 3x) * (1) = 1 + 3x. Put this back in the integral to get:</p>
<p><a href=“1%20+%203x”>integral from 0 to 2</a> dx</p>
<p>4) Convert to Reimann Sum
-The general right-hand Reimann Sum is [Sum as i goes from 1 to n] f(Xi)*delta x (NOTE: Xi is read x sub i, not x * i. I just made the x capital to avoid confusion). Delta x = (b-a)/n = (2-0)/n = 2/n. Now we have</p>
<p>SUM f(Xi) * 2/n</p>
<p>Now let’s find f(Xi). When know f(x), so f(Xi) = 1 + 3Xi. Xi = 2i/n since Xn = 2, so put all this together to get</p>
<p>SUM (1 + 3<em>2</em>i/n) * 2/n</p>
<p>Rearrange:</p>
<p>SUM (2/n) (1 + 6i/n)</p>
<p>Gettysburg Address (AP Calculus version)</p>
<p>Five score and decade ago ETS brought forth on this continent, a new examination, conceived in College Board, and dedicated to the proposition that all men are created intelligently. </p>
<p>Now we are engaged in a great AP Calculus exam, testing whether that brain, or any brain so conceived and so dedicated, can long endure. We are met on a great classroom of that test. We have come to dedicate a portion of that desk, as a final resting place for those who here gave their knowledge that the student might pass. It is altogether fitting and proper that we should do this. </p>
<p>But, in a larger sense, we cannot dedicatewe cannot consecratewe cannot hallowthis classroom. The brave men, intelligent and stupid, who struggled here, have consecrated it, far above our poor knowledge to add or subtract. The teacher will little note, nor long remember what we thought here, but it can never forget what they wrote here. It is for us the intelligent, rather, to be dedicated here to the unfinished work which they who pass here have thus far so nobly advanced. It is rather for us to be here dedicated to the great examination remaining before usthat from these honored students we take increased devotion to that cause for which they gave the last full measure of knowledgethat we here highly resolve that these students shall not have studied in vainthat this exam, under God, shall have a new birth of intelligence and that AP program of College Board, by ETS, for the student, shall not perish from the classroom.</p>
<p>Diverge or Converge? If it converges, find the value it converges to. </p>
<p>sum(from 1 to infinity) of (-8)/((-3)^n)</p>
<p>Show how you get it.</p>
<p>Converges.</p>
<p>sum(from 1 to infinity) of (-8)/((-3)^n)
= sum(from 1 to infinity) of (-8) / [(-1)^n * (3)^n]
= sum(from 1 to infinity) of (-8)<em>(-1)^n / (3)^n
= sum(from 1 to infinity) of (8)</em>(-1)^(n+1) / (3)^n</p>
<p>Alternating series test</p>
<p>lim(n -> infinity) of 8/(3^n) = 0
8 / (3^(n+1)) < 8 / (3^n)</p>
<p>^
Yes it converges, but how do you find what it converges to?</p>
<p>= sum(from 1 to infinity) of (-8) * (-1/3) ^ n
= (-8) / ( 1 - (-1/3) )
= -6</p>
<p>Hold on, that’s from 0 to infinity. Whoops.</p>
<p>= -6 + 8
= 2</p>
<p>Ugh…so apathetic…don’t feel like studying</p>
<p>Why would it be 0 to infinity?</p>
<p>Because that’s the formula of a geometric series that begins at 0. However, the series he posted begins at 1, so you have to subtract the term n=0 from the sum from 0 to infinity to get the sum from 1 to infinity.</p>