<p>My AP Calc teacher has been extremely generous letting me slack on homework this year since I'm so busy w/ Academic Decathlon.</p>
<p>I'm stuck on some integral problems. I understand how to find em and have done most problems in this section...but I'm not sure how to solve ones like these:</p>
<p>The answer is supposed to be 4.</p>
<p>I'm sortof thinking I'm supposed to use the additivity rule for definite integrals...or some other rule... I'm so close... just not quite grasping it...</p>
<p>No idea. My precalc H. teacher is possessed by the devil. He's out to get everyone. I lost all respect for him. I don't respect him at all b/c he stands up in front of the class telling us what we should know, only that list happens to be the list of what he hasn't taught us thoroughly enough.</p>
<p>Ive never done this before, but I think its the 7-3=4. Since "0 to 4" is 7, and "0 to 3" is 3, you can hust subtract them to get "3 to 4". Does that make sense at all? I wish there was an integral key on the board...:(</p>
<p>The integral from the interval [a,b] is equal to the integral on the interval [a,c] + the integral on the interval [c,b], where c is in between a and b.</p>
<p>in this case, a=0, b=4, and c=3.</p>
<p>eq. 1 is on the interval [0,3] and eq. 3 is on the interval [3,4]. eq 2 is on the interval [0,4] and is therefore sum of eq. 1 and eq. 3</p>
<p>3 + (eq3) = 7</p>
<p>eq3 = 4</p>
<p>You're looking for the integral from 3 to 4, so you have to add. All you do is switch the bounds in the first integral to get -3, so that you have the proper sequence, then add the -3 and 7 and you get 4.</p>
<p>okeydokey, the answer is 4...
just reading from your photobucket image, the integral from 0 to 3 is 3, and the integral from 0 to 4 is 7, and you need to find the integral from 3 to 4. you're looking to find the integral of the big (7) minus the small (3). i wish i could draw a picture to show you because i'm horrible at completely verbal explanations. i'll try this way:
let's say you have this x axis
0 1 2 3 4
(0 1 2 3) 4: for the first integral from 0 to 3, the area between the curve of your function and the x axis is 3.
(0 1 2 3 4): for the second integral from 0 to 4, the area between the curve of your function and the x axis is 7.
0 1 2 (3 4): for the third integral from 3 to 4, you subtract your first integral from the second because you're finding the area between the curve of the function and the x axis between those 2 points on the x axis. so 7-3=4. you can do this because the second integral includes what you're trying to find, and it includes the 1st integral. sorry if this doesn't make sense, i hope it helps.</p>
<p>Wow.... thanks guys awesome! And squiggle, I totally get what you're saying... makes plenty of sense now! </p>
<p>I may be back for another question if I get stuck again though. :)</p>
<p>great!! glad to help :)</p>
<p>whoa I just subtracted. </p>
<p>Then again I got a similar question wrong on my Calc final.</p>
<p>Ahh!! Any tips for finding antiderivatives? </p>
<p>Problem is: integral from -1 to 1 of dx/(1+x^2) . </p>
<p>So now I'm <em>99% sure</em> I just need to take the antiderivative of 1/(1+x^2) to continue...Iknow how to do it... but just can't seem to figure it out for 1/(1+x^2)</p>
<p>It's an inverse trig function. int(1/(1+x^2)) = arctan(x), so you'd have: arctan(1) - arctan (-1)
pi/4 - (-pi/4)
pi/4 + pi/4</p>
<p>pi/2 is your final answer</p>
<p>Thanks again!! but once again... my final problem ever lol.</p>
<p>After finishing 5 days worth of assignments I get to the last one... only to get stumped. </p>
<p>how would I find <a href="http://i45.photobucket.com/albums/f100/chrisrog88/calc2.jpg%5B/url%5D">http://i45.photobucket.com/albums/f100/chrisrog88/calc2.jpg</a></p>
<p><em>sigh</em> I used to know how to do all of these lovely things....it was so much fun!</p>
<p>Alas, the perils of taking AP Calc as a junior.</p>
<p>Aha!!!!</p>
<p>All I had to do was plugin sqrt(x) for t, and multiply by sqrt(x) because of the chain rule. Yay all done!! :)</p>
<p>for the last one you make it sin(x) * 1/2root(x) because dt = deriv of x^1/2</p>
