All that matters now is that I carry on. I'm in the mode right now and I'm ready to go. I've got my typing hands ready for blast off and I've got my 80's rock playlist already going. Don't ask why...for some people it's classical music that helps them concentrate, me, it's classic rock. CURRENT SONG: Carry On My Wayward Son - Kansas. oh Kansas... classic.... rock.... hit--- and the perfect song to keep you motivated, which is what we need lately in this class.From what I recall, we started Friday's class off with a little quiz. Or... it should have been little, based on what we've been learning lately. It's getting tense...and I still sense that we're behind. Really behind. We need to pick up the pace. Like Mr. K said, we can't keep continuing this way. Most of these things should be engraved in our head, and we should be able to get past this basic calculus. It should be "general knowledge" in a math sense. We're lagging behind, and that's not good. Because like he said, most of the stuff we're stumbling upon right now is assumed that we can do it by the time the AP exam comes along, and that's not very far. I know I'm one to talk [ha] but I think giving this little motivation kick in the noggin is necessary. Just a bit.

On to the quiz. I'm sure that I don't have to explain this too much, but I'll summarize it, and go over the most common mistakes that the majority of the class made. It's the least I can do for a scribe post like this.

For this first question, integration by parts is used. Two issues some people might have is assigning the values to the right variable correctly and also, the general formula of integration by parts, which is the sequence we should multiply by in order to integrate the function correctly.

To determine which value goes where, we use the.......the....thing we learned. LIATE.; which stands for, Logarithmic, Inverse Trigonometric, Algebraic, Trigonometric. zomg.. I got it now. Acronym. Right? To determine which function is " g' ", g' must be the most complicated function in which it can be antidifferentiated. In this case, g' = sin t.

That being said,---------------------------------------------------------------------------------------------------------------

This was the second question on the quiz, but we saved this one for last, since it was the hardest, or perhaps longest to antidifferentiate. This one was a special case where both substitution and integration by parts was used. The first thing we did was substitute and put it in terms of u, but we still wound up with z's, so we couldn't go far with that. With that dead end, this is where integration by parts comes in, of course using the same structure as the previous question. Once it is antidifferentiated in the sense of u, we resubstitute back in the values for u and put it back in terms of z. ONCE AGAIN, "+ C"!!!

There's a recurring phrase we use many-a-times in class; "There are many ways to skin a cat." Then of course, we always remember to mention the "PETA issue" regarding the actual skinning of a cat. This question, I think, can be done by either or, substitution or integration by parts. I think. Was this the question? I think. But from what I recall when I tried, I think it got messy using integration by parts. So we used substitution for this question. I think most of us got to the right answer for this one.Then there was this question that came before the last one, but I just got mixed up with the printscreen capture order. Haha. My bad. I never got that saying. that's just like saying "I own bad as if it were a conceivable object, yet I'm using the phrase while bashing it. But this question, if I guess, the fundamentals of integration by parts are engraved in your head, this should be no biggie.

As for the rest of the class, we picked up where we left off and reviewed with examples, antidifferentiating combining the rules of integration of trig functions with other rules of integration.

Then we whipped our way back into using approximations when trying to find integrals when we can't exactly determine the antiderivatives.

The class took a look at monotonic functions and how to find the areas under the curve. We compared the methods to see which was more accurate and we found out that:

- In MONOTONIC INCREASING fcns, taking the left hand sum is an underestimate and vice-versa, the right hand sum being an overestimate.
- In MONOTONIC DECREASING fcns, the left hand sum is an overestimate while the right hand sum is an underestimate.

But the reason why these conditions work the way depends on these main points:

- The TONICITY of the graph, how much of a curve there is, wide or steep, etc.
- Whether you take the LEFT or RIGHT sum.
- The CONCAVITY of the graph, whether it's concave up or concave down.

Once again... Carry On...My Wayward Son. [or just.. classmates.]

## 6 comments:

I was just listening to that song when I saw this haha.

So awesome.

what a coincidink.. argh... i'll have all the weird things going on with this scribe fixed.

Hi J + ME, so here's a question for you in regards to your reminder at the end of the post and definite integrals. As the number of subintervals approaches infinity in a Riemann sum on a given interval on a monotonic increasing function, is the left hand sum less than, equal to, or greater than the right hand sum? What about the midpoint sum?

GreyM

HI Grey-M! HA I just saw your post right now. First off, how's it been? University is treating you well?

As for your question...well to be honest, I forgot all about the size of the subintervals, and now that I think about it, I realize that subintervals play a factor in determining the magnitudes of the area estimates under a curve.

From what I recall, as there are more subintervals within the domain of the definite interval, the sizes of these subintervals get smaller and there are more of them. Because of this, this makes the estimate more accurate to the actual area and I think left hand sums are still underestimates under these conditions, and makes it less than the right hand sum but...perhaps equal to the midpoint sum at this point?

I'm not sure if that answers the question correctly or at all :S, but I took a stab at it.

Well that was a good stab at it. You have the right idea with the intervals getting smaller and smaller and therefore more accurate, BUT, these intervals aren't just getting small, they are getting infinitely small. So the left endpoint and the right endpoint of each interval become virtually the same point. Now think of an integral, it is the area under a curve with an infinite number of subintervals added up... well a Riemann sum is a bunch of subintervals added up, so if you have an infinite number of subintervals with infinitely small change in x that is an integral. A formal definition of a definite integral is actually a Riemann sum with infinite subintervals on some domain [a,b]. So does it matter if you take the left hand sum, right hand sum or midpoint sum with infinite subintervals? No, because they are all equal with the delta x being infinitely small.

And University is treating me well. Got my 4.3 GPA and I'm now cruising through most of my courses (there was a rough patch where I had mononucleosis and missed a few weeks, got seriously behind there, but caught up and now it's just fine and dandy)

4.3 GPA! Good on ya! ... and that was a great description of a definite integral. ;-)

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