### An Integral

The end of chapter three. That went by pretty fast if you ask me. A short chapter, but a concept somewhat hard to grasp, like derivatives.

So let me start of summarizing the chapter from the beginning.

Concept of finding slope of tangent lines at a point leads us from the:

Parent Function -- > First Derivative -- > Second Derivative

That was last chapter. Chapter 3 leads us from:

Second Derivative -- > First Derivative -- > Parent Function

This time it is the concept of finding area between the function and the x-axis.

In other subject areas, we have always known that area is a scale. So is speed, and mass. Unlike velocity, acceleration or displacement, those are vectors. So basically, we have not heard of an area that is negative. But within the concept that you are learning, an integral, is the area between the function and the x-axis, which can be a negative value. Why? Because technically an integral is not an area, its just the method.

So what is an

**INTEGRAL**?

**A number**. =D But to be able to say that you have to understand the whole concept of it, but in the end its just a number.

What is the

**concept**?

An integral is an

**infinite sum**of rectangles within an interval. The height is given by the output value of the function. The width is a very small number, close to 0, but not equal to 0.

{

*formally:**It is the total change in the parent function over the given interval.*}

** If the height is given by the output value of the function, and most functions go below the x-axis, the output may be negative. As for the width, will be positive. Therefore, the signed area given may be negative.

=D Gets so far? If so, please continue on reading. If not, try reading it again, but more slowly this time, and maybe try to visualize it in your head.

**Riemann Sums.**

We learned 4 types of Riemann sums, and I’ll go over each sum with you.

Let’s use x^2 as an example.

Let’s pretend we have a monotonic increasing function.

For x^2, it is monotonic increasing on the interval (0, infinity). But to make explaining easier, I’ll be using the interval [0,5] and with 5 subintervals.

The following table of values is resulted.

X Y

0 0

1 1

2 4

3 9

4 16

5 25

**Left Hand Sum:**

This sum is calculated using the left value per interval. Since we have a width of 1, we’ll factor it out to make calculations easier.

LHS = 1 ( 1+4+9+16)

= 30

So therefore, the interval [0,5], LHS is equal to 30.

**Right Hand Sum:**

This sum is calculated using the right value per interval. Since we have a width of 1, we’ll factor it out to make calculations easier.

RHS = 1 ( 4+9+16+25)

= 54

So therefore, the interval [0,5], RHS is equal to 54.

**Trapezoid Sum:**

If you recall, the formula for finding an area of a trapezoid is =

{b(short side + long side)} / 2

So therefore, this sum is calculated using the average of the left and right hand sum. Since we already multiplied the base or width within the calculations of right or left sum, we could skip that step. Then divide by 2 to get the trapezoid sum.

Trapezoid Sum = (30+54) / 2

= 42

So therefore, the interval [0,5], trapezoid sum is equal to 42.

**Midpoint Sum:**

Midpoint sum is found using the midpoint between each subinterval, and taking its output as the height, and width would still be the chosen width from the start.

In our example, lets first generate a table.

X Y

0 0

0.5 0.25

1 1

1.5 2.25

2 4

2.5 6.25

3 9

3.5 12.25

4 16

4.5 20.25

5 25

So in our calculations, we would be using the output we calculated from each midpoint. And our width would still be 1.

Midpoint = 1 ( 0.25 + 2.25 + 6.25 + 12.25 + 20.25)

= 41.25

So therefore, the interval [0,5], the midpoint sum is equal to 41.25.

**As for the programs we have in our calculators.**Once you know what each program does, and what each is actually calculating for you. You will want to use the most efficient program to get your answer faster.

**1)**I believe

**RIESUM**is the program on the T1`s that just calculate the RHS, LHS, and the Midpoint sum without showing graphs.

**2)**RIESUM is more efficient than

**RSUM**which shows the graphs as well.

**There are also built in functions in our calculators**that calculate the exact value of the integral.

**1)**

**Math 9**is the command called

**fnInt**. You use it on the home screen and going to the math menu and it is number 9 on the list.

**FnInt(funtion or Y1 , X, lower point, upper point)**

For example, going back to x^2, from the interval 0 to 1:

Command would be as follows: FnInt(X^2, X, 0, 1)

When you press enter it should show you the value = 1/3 or 0.3333...

**2)**There is also the command where you graph the function you want to integrate, then go

**2nd Calc menu and use option 7.**Option 7 is the command the

**integral of f(x) dx.**You choose your lower bound, then your upper bound. This command should shade the area within the graph and the x-axis from your stated lower and upper bound. It should then show you the exact value after shading.

Well sorry this is late guys. I finally got a new job, and I`ve been working

for the past weekend plus English homework *sigh*. Well hope you guys the best

of luck on the test. Hope everyone does good on the test. Did I mention we have

a very very very very very very very very SMALL class? If I haven`t , I shall.

We have a very very very very very very very very SMALL class. Lol. That`s all

for this unit. To the next unit. I think I like the next unit. *wink* *wink*

Till next time.

## 2 Comments:

Fantastic post Sarah!

I'm really impressed with the level of detail. In particular your knowledge of the TI-83 considering you use a different model from a different company. ;-)

Thank you for the information! I was working on homework and forgot how to do some Riemann sums, and your post cleared it up for me. Thank you again!

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