Pre-Test: Applications of Integrals
Published on: Mar 4, 2016
Transcripts - Pre-Test: Applications of Integrals
<ul><li>Hello everyone. Kristina here, starting off the new cycle with another one of my slide presentations. No anime related opening picture this time. I wanted to put a panda or something instead but I thought this red panda was pretty good too XD. Don’t you love that mocking face it has on? Anyways, today was pre-test day, finally. Let’s get started shall we? </li></ul>
Multiple Choice This question is just another one of your simple washer questions. To make it easier, first take out a cross section to analyze it, like shown in the green diagram. Knowing that this is a washer question, you should remember that to get the area of the washer, you’ll have to subtract the inner circle from the outer circle.
Analysis! By analyzing the cross section and the graph, you should be able to see that the radius of the outer circle is f(x) + 1 since you are rotating about y = -1. As for the inner circle, since R is also enclosed by the x-axis, you’d just have to subtract 1 from y = 0 (the x-axis).
Solved! Now knowing all this, you can now input your newly found R and r into the area of the washer, π (R²-r²). Then you can integrate along the interval 0 to e. Why 0 to e you ask? Well that’s because ln(x+1) intersects the x-axis at 0 and since R is enclosed by the line x = e as well, that means that your next intersection point, b, will be e. Question solved.
Question DEUX! This one’s simple. All you have to do is take your average value theorem and plug in the values that you are given and solve for b. Or, you can do it the way I did by just plugging and chugging in the options given to you. Same deal, whatever floats your boat ya know?
Midpoint that Log! This one was kind of tricky since we didn’t know how to divide the log at first but after drawing an actual log and dividing it into 3, I’m guessing we all saw the light. So, like said, first draw an actual log like shown. Divide it into three pieces and make sure you label it from 0 to 6! Now after labeling, look for the values that are inbetween each section of the log. Those are your midpoint values you are going to use.
Midpoint that Log! Now make your Riemann’s sum using those midpoint values. If you notice, you’d see that ∆x is equal to 2. Do you see why? Well take a look at your sections on the table. Let’s explain using x = 1. There is one value on either side of that midpoint, each 1 unit apart. So add that up and you have 2. Voila.
Hello Cylindrical Shell :3 This question isn’t so bad. Like the title says, it’s a cylindrical shell since you have your shaded region being rotated around the y-axis. Like the other cylindrical shell questions we did, same goes for washers, first take a cross section out and analyze it. In the case of these shells, we “take apart” the shell and lay it out as a rectangle (though the tops are more like trapezoids >_>)
Hello Cylindrical Shell :3 With our rectangle, we fill in what we know about it. The width would be equal to the circumference of the cylindrical shell (NOTE: The radius is just equal to the x value), while the width is just that little change in x, which is really small so it just becomes dx. We should also note that the height is just equal to the difference between the top and bottom functions.
Hello Cylindrical Shell :3 We can now integrate to get our volume by using the formula for the volume of a rectangle. Of course, you first need to find your interval, so find those intersections! Finished.
Densiteasy! This one is simple. All you had to do was integrate d(x) from 0 to 2. It is to be noted that during the extra few minutes during fourth period, Mr. K told us that everything has a purpose when doing math since we were talking about that dx. Basically, in terms of dx, don’t just put it there just because its an integral. Think about it first and what its purpose is! In this case, its that change in miles to cancel out the mile in people/mile so you’re just left with people.
Nine Marks of Glory Part (a) was simple. It was just the basic use of the average value theorem. Just input your values (the function will again be top – bottom) and plug and chug into your calculator, which I’m guessing you shouldn’t do considering that this is a long answer question >_>. Yes…so integrate your function then multiply by ½ and consider number 1 fini!
Part B! Part (b) on the other hand was basically a stepping stone in order to move onto (c). If you didn’t understand (b), then you probably wouldn’t have gotten (c). Anywho, here we have another washer question. The only think I can see being the tricky part would be the fact that you are dealing with an unknown constant, k. Yes, so first start off with the usual taking out of a cross section.
Here we go… You should then see that your outer circle’s radius is g(x) and that your inner circle’s is f(x). Then after all this, you can again integrate your π (R²-r²) along the given interval and continue on with some algebraing with that tricky k.
Wahoo! An important note, since I’m sure I’m not the only one who could make this mistake. Remember that k is a constant, so that means when you anti-differentiate it, it will just be kx. Just a little tidbit that Mr. K went over with us during the extra time. And yes, once you worked everything out, you should end up with this.
I C You! This one is basically the same as (b), except your axis of rotation has changed to y = -2. Meaning, your radii will change by a factor of 2. Also, this time they’ve decided to spare us by letting us not solve it XD. Soo…basically since your axis of rotation is now y = -2, you’ll now have to subtract (or to put it simply, add) 2 to your outer and inner radii. Then the rest is the same dealio, just integrate but do not solve.
<ul><li>Well, that’s all. The scribe for the new unit shall be Lawrence . Don’t forget to BOB for tomorrow’s test as well! That is all, good luck everyone ;) </li></ul>Mudkip and Bidoof say good luck too :D