Wednesday, January 10, 2007

6.1 Areas between curves


In this section we will be finding the area between curves that are not necessarily bound by just the x and y axis

Example 1:

Find the area of the bound by f (x) and g (x) on the interval [a,b]

To find the area, all you would need to do is subtract the integral of f (x) on the interval [a,b] from the integral of g (x) on the same interval.

Example #2

Now, find the area bound by the two graphs given below.

Graph it on the calculator to see it

Luckily, we already know that one of the intersection points comes at (0,0). We can use our graphing calculators to find the x-value for the other point. You can go to CALC then to INTERSECT, then magically we find out that the intersection is at the point 1.1807757. Immediately after you find this information out, store the value as the "a" value. Then, with more help form the calculator, you can calculate the area by calculating the integral.

It should look like this on your calculator:

fnInt (Y1 - Y2 , x , 0 , A) = .7853885505

This is because you are caluculating the integral of the difference between the two functions Y1 and Y2, in terms of x, on the interval 0 to A. (You already calculated and stored the value.)

When you have a problem that looks similar to a sinusoid, you have two options:
1. you can add the integrals comprised of one function subtracted from the other on the intervals of the intersection points. So, adding up the segments.

2. you can find the absolute value of the integral of the difference of two funtions, completely disregarding the segments. (Shown below)


Sometimes, merely subtracting one integral from the other is not the easiest way of doing things. When the rectangles that make the area and integral are horizontal and not vertical, it is easier to put things in terms of y instead of x.

For example, the equations y = x-1 and y^2 = 2x + 6

So, to work horizontally, rework the y equations to x.

x = y + 1

x = .5 y^2 - 3

Now we are left with the integral

Now you basically know how to find an area between curves. I bet you feel a lot smarter now that you know this. I know I do.

Quick link to check out:

Mr. French, here's one for you:

A friendly reminder that Grey's Anatomy is back on tomorrow, so you better all watch it!


man i'm excited already!!!!!!


Tuesday, January 09, 2007

Friday's Test Topics

Here’s a list of topics that will be covered on this Friday’s Chapter 5 Test.

Chapter 5 Test Topics
Fundamental Theorem of Calculus Part I (Sec. 5.3, #9,13,17)
Fundamental Theorem of Calculus Part II (Sec. 5.3, #13,41)
Substitution (Sec. 5.5, #13,23,278,31,37,41,57)
Evaluate an integral in terms of area (Sec. 5.2, #37)
Riemann sum: sketch, evaluate, explain and interpret (Sec. 5.1, #3,1,13,15)
Definite Integrals (Sec. 5.2, #33)
Net Change Theorem (Sec. 5.4, #47,48)

For additional practice problems, look at the chapter review (pp. 431-433)

That’s it! I’ll be around after school on Thursday and in early on Friday. Donut holes and OJ!

"It is inevitable that some defeat will enter even the most victorious life. The human spirit is never finished when it is is finished when it surrenders."
Ben Stein

Here’s someone who refused to surrender to incompetence – I admire his patience, but I probably wouldn’t be able to last this long. It’s a fairly long audio clip (about 25 minutes) and I just want to say how thankful I am that MY students understand the importance of units...

And on a lighter note:

Saturday, January 06, 2007

5.5 The Substitution Rule

Hello "Wild Calc Class!!!" I hope you have all enjoyed your break. I had an amazing time visiting my camp friends. Happy New Years (a little late, but late's better than never)!!! But to take you all off cloud 9, it really is time to go back to school. We currently have about 40 hours until class resumes, so relax while you can and do all that work you have been saving up, except for calculus is not due until after we get back. That is right, no calc homework over break...woo hooo!!!!!

But if you were planning on getting ahead, here are the notes from 5.5 just in case you need some help:

Substitution Rule = Reverse Chain Rule.

Yup instead of taking the derivative using the chain rule, we will be taking the antiderivative using the Substitution Rule. We will be working backwards. With many things they say working backwards is easier, but unfortunately, with this you are out of luck. The Substitution Rule is tedious and difficult, but with the brilliant minds we all have, completely do-able!

Hey look...the colors are like a watermelon (and that's how I think on vacation)

Let's dive in!

I am going to start with a problem and explain step by step because I think that's the easiest way to grasp the concept. With the reverse chain rule, you need to see it in action to master it.

Step 1:
Write out the integral

Step 2:
Identify u. Always choose u by figuring out which function is the nested function. Properly choosing u will make the whole problem much simpler

Step 3:
Identify du (don't forget to put dx along with the derivative because it is the derivative of u with respect to x)

Step 4:
Insert u and du into the original integral

Step 5:
Find the antiderivative using u

Step 6:
Insert Step 2 into Step 5

Here is a harder problem:
Step 1:

Step 2:

Step 3:
You will need to manipulate the equation of du so that it matches the initial integral. This makes finding du a little tricky sometimes. In the initial equation, there is xdx, but not 2xdx, so the right side of the equation must be what is in the initial integral equation.

Step 4:
When inserting du into the equation, make sure you also insert the coefficient of du, which you found in Step 3. The coefficient of du can be put on the outside of the integral sign.

Step 5:

Step 6:

That wasn't so bad now was it?! The key to those is making sure you take the antiderivative of the trig functions, not the derivative of them...I see that as being the likeliest place for me to mess up.

Speaking of trig functions, another necessary thing to be able to understand is how to use the substitution rule to find the integral of functions using tangent. This does not follow the 6 steps I broke the rest of the integral problems into. It follows its own pattern that will further our understanding of how to take integrals and also help teach us a new way to think about calculus. This is not impossible, it is just new and different and requires practice thinking in a new way.

Write out the integral and break it into usable pieces such as sine and cosine. Change the form to fit trig functions that are easier to work with.

Identify u and du.

Insert u and du into the integral

Find the antiderivative using u.

Insert u.


And "reduce" (well this is not so much reducing as putting in the simplest form, which is the form they are asking for.) You need to train your brain to think to finish this last step.

Did you know: a sneeze
travels out of a person's mouth at over 100 mph!!! I got a bad cold Friday night after my soccer game and thought I would share that fun little snapple cap fact.

Did you know (#2): Americans e
at 18 acres of pizza a day on average!!! (Snapple cap fact again)

Here's an awesome website you can use for some more help on the SUBSTITUTION RULE:

Substitution Rule for Definite Integrals: (I am so sorry I am using the problem you showed us in class, but I had trouble on this part of the homework and I don't want to show a wrong example, so once I understand it, I will change it if you would like...)
Step 1:
Write out the integral.

Step 2:
Find u and du

Step 3:

Insert u and du into integral

Step 4:
Adjust Bound
aries to u

u = 2x + 1
u = 2(4) + 1 = 9

u = 2(0) + 1= 1

Step 5:
Rewrite with the boundaries of u

Step 6:
Take antiderivative and solve.


Step 7:
Put x into antiderivative with initial boundaries and solve.

Even Functions: f(x) = f(-x)

Odd Functions: -f(x) = f(-x)

Lauren You Are Next!!!