Showing posts with label Scribe Post. Show all posts
Showing posts with label Scribe Post. Show all posts
May 01, 2007
Scribe Post
Here are the answers for the exam review quiz. Just click on the images to make them bigger.
April 19, 2007
Scribe
Well we were supposed to have the pre-test today, but as a class we were not that ready. So instead we went into three different groups and worked on word problems. All the problems were practically the same, the steps were: 1) finding two points on the parent, 2)seperating varibles, 3) integrating, and 4) creating an equation. However we still had some difficulties or maybe it was just an off day. I don't know. But the more practice the better.
Remember everyone to keep up with the assigned homework. The AP exam is just around the corner. I also have to catch up and review. There's a homework sheet below for more practice.
The next scribe is Christian.
Remember everyone to keep up with the assigned homework. The AP exam is just around the corner. I also have to catch up and review. There's a homework sheet below for more practice.
The next scribe is Christian.
April 16, 2007
Scribe Post
Ahoy everyone! This is Jann, and I'll be your scribe for today.
We have finally finished the whole course. Today, we looked at some word problems on Differential Equations.
Here we go... (Note: I will be using Mr. K's slides... )
In order to solve this problem, we need to find the parent function. dy/dx is some constant multiplied by the parent function. We know that the temperature, T, varies with respect to time. (Basically, the temperature of the roast decreases over time) We need to find 2 points on the parent function...
At time, t=0, the initial temperature of the roast, which is 68 degrees F. Therefore, we have our first point, (0,68). For our second point, at time, t=2 hours, the temperature of the roast is down to 40 degrees F. This is our second point, (2,40).
Then, after acquiring the 2 points in the parent function, we integrate the formula. The Red Line shows that the calculus part of the problem ends. The following line is all Pre Calculus stuff. If we can recall our natural law from Grade 12 PreCal, "kt + C" is an exponent. Therefore, it is an exponent of the base "e". "e^C1" can be substituted into "C" because it is a constant. Then, we isol
ate the "T" in order to obtain the Parent Function.
Now that we have the structure of the parent function, we can solve for the missing terms such as "C" and "k".
Now that we have the parent function. We are ready to solve the problem...
Then, we looked at one more example...
That's all we did today in class. We were suppose to look at another example, but we ran out of time. Good day everyone! XD
Oh yeah... next scribe is Crystal.
We have finally finished the whole course. Today, we looked at some word problems on Differential Equations.
Here we go... (Note: I will be using Mr. K's slides... )
In order to solve this problem, we need to find the parent function. dy/dx is some constant multiplied by the parent function. We know that the temperature, T, varies with respect to time. (Basically, the temperature of the roast decreases over time) We need to find 2 points on the parent function...At time, t=0, the initial temperature of the roast, which is 68 degrees F. Therefore, we have our first point, (0,68). For our second point, at time, t=2 hours, the temperature of the roast is down to 40 degrees F. This is our second point, (2,40).
Then, after acquiring the 2 points in the parent function, we integrate the formula. The Red Line shows that the calculus part of the problem ends. The following line is all Pre Calculus stuff. If we can recall our natural law from Grade 12 PreCal, "kt + C" is an exponent. Therefore, it is an exponent of the base "e". "e^C1" can be substituted into "C" because it is a constant. Then, we isol
ate the "T" in order to obtain the Parent Function.
Now that we have the structure of the parent function, we can solve for the missing terms such as "C" and "k".
Now that we have the parent function. We are ready to solve the problem...
Then, we looked at one more example...
That's all we did today in class. We were suppose to look at another example, but we ran out of time. Good day everyone! XDOh yeah... next scribe is Crystal.
April 15, 2007
Scribe Post
So during Calculus class today we opened up the class reviewing Newton's method, which is used for approximating a nearby value using the slope of a tangent line on a differiential equation. e.g. approximating √(37) from using tangent line at √(36)
Then after that we programed our calculator with Euler's Method, which is just a mechanical process and is explained in the slides below. When you have a function that tells you the rate of change, we'd use this method to solve the differentiable equation. And a diffentiable equation is when one of the variables is a derivative. These problems can also be called initial value problems.
There's not much to explain for what we did in class, as you can see below in the slides.
Homework is chapter 9.3. Next class we'll tackle the newton's law of cooling question.
Next scribe will be Jann.
Then after that we programed our calculator with Euler's Method, which is just a mechanical process and is explained in the slides below. When you have a function that tells you the rate of change, we'd use this method to solve the differentiable equation. And a diffentiable equation is when one of the variables is a derivative. These problems can also be called initial value problems.
There's not much to explain for what we did in class, as you can see below in the slides.
Homework is chapter 9.3. Next class we'll tackle the newton's law of cooling question.
Next scribe will be Jann.
April 11, 2007
Scribe Post
THIS IS LINDSAY =)
In class, we learned how to draw slope fields.
Let's review from our previous calculus class:
The order of a differential equation is the highest-order derivative.
y'=y is a first order differential equation.
y"=-y is a second order differential equation.
The solution to a differential equation is any function that fits the eqatuion.
For example, to find the differential equation to y'=y you find that you have to find a function that is its own derivative. We know that y=ex is a solution.
When we think about derivatives, we think about slopes! For every input x, f''(x) is the slope of f(x) at (x,y). This is a first order differential equation: when it describes the slopes at a specific point.
Example: y'=y is a differential equation. At any point (x,y), the slope of the solution curve is the same as the y coordinate. If we use the point (1,2), the slope of the curve is 2. Whenever the y coordinate is y=3, the slope will be 3.
We were given a handout with some examples. We did this in class and the solutions are in the slide post.
1) y'=y; you would use y=ex because it is its own derivative.
2) y'=2x; the parent function is x2
3) y'=-x/y;
x2 + y2 = k
2x + 2yy' = 0
y'= -x/y
*parent function is a circle
4) y'=x+y
For homework, we need to create a new program to create slope fields. You can find how to create it in 9.2 or you can ask a calculus buddy to send it to your calculator! Also do all the odd questions!
In class, we learned how to draw slope fields.
Let's review from our previous calculus class:
The order of a differential equation is the highest-order derivative.y'=y is a first order differential equation.
y"=-y is a second order differential equation.
The solution to a differential equation is any function that fits the eqatuion.
For example, to find the differential equation to y'=y you find that you have to find a function that is its own derivative. We know that y=ex is a solution.
When we think about derivatives, we think about slopes! For every input x, f''(x) is the slope of f(x) at (x,y). This is a first order differential equation: when it describes the slopes at a specific point.
Example: y'=y is a differential equation. At any point (x,y), the slope of the solution curve is the same as the y coordinate. If we use the point (1,2), the slope of the curve is 2. Whenever the y coordinate is y=3, the slope will be 3.
We were given a handout with some examples. We did this in class and the solutions are in the slide post.
1) y'=y; you would use y=ex because it is its own derivative.
2) y'=2x; the parent function is x2
3) y'=-x/y;
x2 + y2 = k
2x + 2yy' = 0
y'= -x/y
*parent function is a circle
4) y'=x+y
For homework, we need to create a new program to create slope fields. You can find how to create it in 9.2 or you can ask a calculus buddy to send it to your calculator! Also do all the odd questions!
March 20, 2007
Scribe Post 90
First off, i would like to remind everyone to go and add to the marking rubric for our project. Rubric is due this Friday.
1) Oil is leaking at a rate of R(t) = 2000e-0.2t
, where t is measured in hours, how much oil has spilled in 10 hours?
We want to end up in gallons, so we multiply gallons per hour (which we are given) by some change in time. We want to find the total sum over the ten hours.
*When we are given a rate, such as the one above, we are trying to find a unit related to the rate. I.e. (gallons/hour)x(hour) = gallons.*
A) Density of an oil slick on a body of water is defined by
We cut a piece from the oil slick. If we drill down, the density of the oil slick is the same throughout.
Here we have cut a piece out of the oil slick. The ends of the strips (triangles) become smaller. So we take the triangle and move it to one side to form a rectangle. The area of the rectangle is what we need to find. Going back to our equation we can now get rid of m2
With our our information and our missing piece found, we can create an integral to solve for the mass of the oil slick. We take the limit of n as it approaches infinite and it becomes smaller. We can then take the sum, where n equals one. p(a) is our given equation so we multiply that by the rectangle piece we cut out, Am2. Then we get the integral from 0 to 1000 meters. 100/(1+r2) is kg/m2 multiplied by 2pi r dr m2, which will give us the mass of the oil slick 1000 m out. 2pi r is multiplied by dr because dr becomes infinitely smaller.
*check the slide for march 19 to view the solution to the integral*
B)What is the smallest radius that contains 75% of the oil slicks mass?
With our solution from A, all we do is take seventy percent of the value we found. We can then pull out 200pi. From here we just solve for r, by expanding the bracket notation first.
For the last part of class, we talked about how our physics formula can be derived. We also did a quick question before the bell rang. We were given an amount of cars per kilometer. instead of trying to cancel out a unit area we had to cancel a length to get the number of cars.
URGH, i got a cold.... next scribe is Suzanne... pre test tmrw?
1) Oil is leaking at a rate of R(t) = 2000e-0.2t
, where t is measured in hours, how much oil has spilled in 10 hours?
We want to end up in gallons, so we multiply gallons per hour (which we are given) by some change in time. We want to find the total sum over the ten hours.
*When we are given a rate, such as the one above, we are trying to find a unit related to the rate. I.e. (gallons/hour)x(hour) = gallons.*
A) Density of an oil slick on a body of water is defined by
Suppose the oil slick is extends from 0 to 1000 m. Determine the mass of the slick.
We have Kg/m2 and we want to get kg, so we have to get rid of the m2
We have Kg/m2 and we want to get kg, so we have to get rid of the m2 We cut a piece from the oil slick. If we drill down, the density of the oil slick is the same throughout.
Here we have cut a piece out of the oil slick. The ends of the strips (triangles) become smaller. So we take the triangle and move it to one side to form a rectangle. The area of the rectangle is what we need to find. Going back to our equation we can now get rid of m2
With our our information and our missing piece found, we can create an integral to solve for the mass of the oil slick. We take the limit of n as it approaches infinite and it becomes smaller. We can then take the sum, where n equals one. p(a) is our given equation so we multiply that by the rectangle piece we cut out, Am2. Then we get the integral from 0 to 1000 meters. 100/(1+r2) is kg/m2 multiplied by 2pi r dr m2, which will give us the mass of the oil slick 1000 m out. 2pi r is multiplied by dr because dr becomes infinitely smaller.
*check the slide for march 19 to view the solution to the integral*
B)What is the smallest radius that contains 75% of the oil slicks mass?
With our solution from A, all we do is take seventy percent of the value we found. We can then pull out 200pi. From here we just solve for r, by expanding the bracket notation first.
For the last part of class, we talked about how our physics formula can be derived. We also did a quick question before the bell rang. We were given an amount of cars per kilometer. instead of trying to cancel out a unit area we had to cancel a length to get the number of cars.
URGH, i got a cold.... next scribe is Suzanne... pre test tmrw?
March 16, 2007
SCRIBE POST: DAY 89
The plan was to have this up by Friday. Sorry guys. Anyway, I haven't been scribe for quite some time, so it took me awhile to get back into it...
We started class by talking about our due dates for our DEVELOPING EXPERT VOICES assignment. Remember guys: you can always move your due date earlier than expected. We also talked about the rubric for this assignment. Remember that changes must be made no later than Friday, March 23rd.
Let's get the math started...
To solve for a you simply plug in 1 for k. To solve for b, c, and d, you must integrate it from 1 to the number of items. We start with 1, because this is where the domain of the function begins. If you plugged 0 into the function, you would get 23 minutes. This doesn't make sense; it shouldn't take 23 minutes to make nothing. Once you know the integral, the rest is just punching it into the calculator. The answers can be found on the bottom of this page. https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKDzv2OOobRwicccwEDRidnFn0ZXsphtj3GaokY2BSlnEC23kRzA8syG4oaBXJoiiFeJXCaYzuylt93wtfbXQbCu861Hn1WdeEdoWRFt4goN56ek7EOiEEVv110iM7uUYrK5oCYxmsrcM/s1600-h/AppsOfTheIntegralExerc.jpg Note: This is a good exercise to practise for the unit test.
1. To find the area of S, we simply find the integral from 0 to 4. Because the function f(x) is on the bottom, we subtract it from g(x).
a)
b)
d) In this case, the solid is rotating around the y-axis, therefore the shape of the arbitrary slice is no longer a circle. It becomes a prism. The width of the prism is the circumference. (remember: the paper towel) Its thickness is dx, because we want the slice to be infinitesmally small. Its height is simply the top function, g(x), subtract the bottom function, f(x).
We started class by talking about our due dates for our DEVELOPING EXPERT VOICES assignment. Remember guys: you can always move your due date earlier than expected. We also talked about the rubric for this assignment. Remember that changes must be made no later than Friday, March 23rd.
Let's get the math started...
To solve for a you simply plug in 1 for k. To solve for b, c, and d, you must integrate it from 1 to the number of items. We start with 1, because this is where the domain of the function begins. If you plugged 0 into the function, you would get 23 minutes. This doesn't make sense; it shouldn't take 23 minutes to make nothing. Once you know the integral, the rest is just punching it into the calculator. The answers can be found on the bottom of this page. https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjKDzv2OOobRwicccwEDRidnFn0ZXsphtj3GaokY2BSlnEC23kRzA8syG4oaBXJoiiFeJXCaYzuylt93wtfbXQbCu861Hn1WdeEdoWRFt4goN56ek7EOiEEVv110iM7uUYrK5oCYxmsrcM/s1600-h/AppsOfTheIntegralExerc.jpg Note: This is a good exercise to practise for the unit test. 
With the help of Fooplot http://www.fooplot.com/, we were able to see what the graph looked like. Points of intersection tell you where to integrate.
1. To find the area of S, we simply find the integral from 0 to 4. Because the function f(x) is on the bottom, we subtract it from g(x).
2. To find the volume of S, we first find the volume of one piece. We take an arbitrary slice, and find its area. In this case, the shape of the slice is a semi-circle, where the radius is the top function, g(x), subtract the bottom function, f(x). Once you have the area, you must write the integral. You then apply this integral to find the volume of the whole thing.
3. Because we S is rotating, you can imagine a solid being formed with a hole in the middle. To find the area, we must take the area of the smaller circle and subract it from the area of the bigger circle. Questions a, b, and c are all the same question; the only difference between them is how the radii are formed.
a)
b)
c)
So that brings us to the end. If you're still having trouble, the slides from Thursdays class http://www.slideshare.net/gofull/30241/1 are up.. thank goodness for our super-di-duper smartboard!!! HMWK for the weekend is EXC 8.5 # 1,3,5,11,15. And by the way, MARK is the scribe for Monday! Enjoy the rest of your weekend guys =)
March 06, 2007
Scribe Post
Hi guys :) Yesterday we started unit 8.
First off we started with a few problems which you probably recognized from past units.
Find the area bound by the two curves:
To solve these, you must look at the graphs, and determine which function is above and which is below. Then you take the integral of the top function minus the bottom function. To find the interval, make the two functions equal each other and solve for x.
After that we discussed a problem where we are given a velocity function and asked to find the displacement, also known as net distance, and total distance travelled. The net distance is found by subtracting the negative integrals from the positive integrals. The total distance travelled is found by taking the integral of the absolute value of v(t).
That's all you need to know to do excercise 8.1.
Then we started to discuss how a 3 dimensional object is created by 'spinning' the function around the x axis (or y axis, but we didn't get into that just yet). We discussed how you can take many 'slices' of the solid, so thin that the width approaches zero. I guess we'll be expanding on this in class tomorrow.
The rest of the assigned homework includes the odd questions from 1-7 of exercise 8.2.
The scribe tomorrow shall be...... Danny.
First off we started with a few problems which you probably recognized from past units.
Find the area bound by the two curves:
To solve these, you must look at the graphs, and determine which function is above and which is below. Then you take the integral of the top function minus the bottom function. To find the interval, make the two functions equal each other and solve for x.
After that we discussed a problem where we are given a velocity function and asked to find the displacement, also known as net distance, and total distance travelled. The net distance is found by subtracting the negative integrals from the positive integrals. The total distance travelled is found by taking the integral of the absolute value of v(t).
That's all you need to know to do excercise 8.1.
Then we started to discuss how a 3 dimensional object is created by 'spinning' the function around the x axis (or y axis, but we didn't get into that just yet). We discussed how you can take many 'slices' of the solid, so thin that the width approaches zero. I guess we'll be expanding on this in class tomorrow.
The rest of the assigned homework includes the odd questions from 1-7 of exercise 8.2.
The scribe tomorrow shall be...... Danny.
February 27, 2007
Scribe
What did we do today?
We discussed how the pre-test went with the substitute and the multiple choice questions were pretty straight forward and no one really had any problems. What did leave us all puzzled was the long question in the back, the "open response".
The question was . . .
(1) Let f be a differentiable function such that f " is continuous and f and f ' have the values given in the table below.
(now I wished I had the smart board so I could draw the box like Mr. K had done in class..ah the wonders of technology, I honestly can't keep up with it.)
Don't forget the mnemonic " LIATE ".
L ogarithmic
I nverse trignometric
A lgebraic
T rignometric
E xponential
So an example problem that we did that even got Mr. K, who's a math gyro (uh I don't know if I am using that word right) thinking was . . .
So to review . . .
+ When the integral is a composite of functions, you are better off using substitution.
+ when the integral is a product of functions, you are better off using integration by parts.
Last but not least next scribe is Suzanne.
We discussed how the pre-test went with the substitute and the multiple choice questions were pretty straight forward and no one really had any problems. What did leave us all puzzled was the long question in the back, the "open response".
The question was . . .
(1) Let f be a differentiable function such that f " is continuous and f and f ' have the values given in the table below.
(now I wished I had the smart board so I could draw the box like Mr. K had done in class..ah the wonders of technology, I honestly can't keep up with it.)
Don't forget the mnemonic " LIATE ".
L ogarithmic
I nverse trignometric
A lgebraic
T rignometric
E xponential
So an example problem that we did that even got Mr. K, who's a math gyro (uh I don't know if I am using that word right) thinking was . . .
So to review . . .
+ When the integral is a composite of functions, you are better off using substitution.
+ when the integral is a product of functions, you are better off using integration by parts.
Last but not least next scribe is Suzanne.
February 14, 2007
Scribe Post: Day 82
It's Danny! The day started off with a multiple choice quiz which we will be having every class in preparation for the final exam.
After that Mr. K gave us three integrals to antidifferentiate. When solving these three questions you must consider these two derivative rules:
After that Mr. K gave us three integrals to antidifferentiate. When solving these three questions you must consider these two derivative rules:
Now onto the questions and solutions:
QUESTIONS:
SOLUTIONS:
After that Mr. K talked about "Trapezoid Sums" and "Midpoint Sums". I'll sum it all up in four simple diagrams.
For Trapezoid Sums when the curve is concave down, the sum would be an under estimate. When the curve is concave up, the sum would be an over estimate.
For Midpoint Sums, if you rotate it and make it a line tangent to the curve you'll see that when the curve is concave down, the sum would be an over estimate. When the curve is concave up, the sum would be an under estimate.
Well that's all I was able to muster up for this class, as you all know Charlene was the next scribe, her post is already up. Enjoy!
I also have a very important announcement to make. Even though I was convinced to stay in the course by Mr. K himself I truly believe now that it's best if I don't continue on. I experienced something in our last class that couldn't be a better reason for my leaving. I was having trouble with completing a chart, not only because I was unaware of the fact I was doing my task improperly, but also because there was something in there that I SHOULD'VE KNOWN from past pre-calculus courses that I completely forgotten. That explains to me that I haven't done my past math courses to the best of my ability, therefore hindering my learning experiences in my present math courses. I've already begun trying to fix up my past by taking my PRECAL 40S course again. This decision hopefully will allow me to take Calculus again in University, and this time understand things completely with the mind set of thinking I know material I should know and that'll make things easy for me. One last thing, I know Mr. K you tried to keep me in your class and you succeed. However, this time I've decided and no matter how disappointed you may be, I feel this is the right thing for me to do in my life and my soon to be career in education. Please I'm not asking you to agree with me on my decision, but I'd like to know you're always there for me when I need you. I'm terribly sorry Mr. K, but I'm not going to ask for your forgiveness on my decision.
QUESTIONS:
SOLUTIONS:
After that Mr. K talked about "Trapezoid Sums" and "Midpoint Sums". I'll sum it all up in four simple diagrams.
For Trapezoid Sums when the curve is concave down, the sum would be an under estimate. When the curve is concave up, the sum would be an over estimate.
For Midpoint Sums, if you rotate it and make it a line tangent to the curve you'll see that when the curve is concave down, the sum would be an over estimate. When the curve is concave up, the sum would be an under estimate.
Well that's all I was able to muster up for this class, as you all know Charlene was the next scribe, her post is already up. Enjoy!
I also have a very important announcement to make. Even though I was convinced to stay in the course by Mr. K himself I truly believe now that it's best if I don't continue on. I experienced something in our last class that couldn't be a better reason for my leaving. I was having trouble with completing a chart, not only because I was unaware of the fact I was doing my task improperly, but also because there was something in there that I SHOULD'VE KNOWN from past pre-calculus courses that I completely forgotten. That explains to me that I haven't done my past math courses to the best of my ability, therefore hindering my learning experiences in my present math courses. I've already begun trying to fix up my past by taking my PRECAL 40S course again. This decision hopefully will allow me to take Calculus again in University, and this time understand things completely with the mind set of thinking I know material I should know and that'll make things easy for me. One last thing, I know Mr. K you tried to keep me in your class and you succeed. However, this time I've decided and no matter how disappointed you may be, I feel this is the right thing for me to do in my life and my soon to be career in education. Please I'm not asking you to agree with me on my decision, but I'd like to know you're always there for me when I need you. I'm terribly sorry Mr. K, but I'm not going to ask for your forgiveness on my decision.
February 07, 2007
Scribe Post: 81
Hello everyone! Today, Mr. K talked about Gr.9's from other schools who are going to take a tour of Daniel Mac and how we're going to be teaching them how to do several games. We decided on 5 games but one of the games is yet to be changed.
Here is the draft of whose in what group and whose in what game:
1)Game: Magic
Group:
-Jann
-Mark (a maybe)
2)Game: River
Group:
-Lindsay
-Danny
-Christian
3)Game: Hanoi
Group:
-Suzanne
-Manny
-Katrin
4)Game: Painted Faces **Has yet to be decided on**
Group:
-Linger
-Danny
-Mark
5)Game: Tangrams
Group:
-Charlene
-Anh
-Ashlynn
At the end of Class Mr. K talked about the rules of the Calculus Exam.
**keep in mind that nobody is NOT allowed to say a word about the exam EVER
**remember in the exam it's better to skip and not answer a multiple choice question than to just totally guess on a multiple choice question. ***you will NOT lose any marks if you skip a multiple choice question***...but if you totaly guess on a m.c.q. and you get it wrong, marks WILL be deducted.***
NOTE:
-starting next class we will be doing timed multiple choice quizes at the beginning of class to prepare us for the exam! DO NOT BE LATE FOR CLASS b/c we're going to be timed!
Homework: read the next exercise: 7.6 (it will help you for the next class)
NEXT SCRIBE: DANNY (for friday's class)
Here is the draft of whose in what group and whose in what game:
1)Game: Magic
Group:
-Jann
-Mark (a maybe)
2)Game: River
Group:
-Lindsay
-Danny
-Christian
3)Game: Hanoi
Group:
-Suzanne
-Manny
-Katrin
4)Game: Painted Faces **Has yet to be decided on**
Group:
-Linger
-Danny
-Mark
5)Game: Tangrams
Group:
-Charlene
-Anh
-Ashlynn
At the end of Class Mr. K talked about the rules of the Calculus Exam.
**keep in mind that nobody is NOT allowed to say a word about the exam EVER
**remember in the exam it's better to skip and not answer a multiple choice question than to just totally guess on a multiple choice question. ***you will NOT lose any marks if you skip a multiple choice question***...but if you totaly guess on a m.c.q. and you get it wrong, marks WILL be deducted.***
NOTE:
-starting next class we will be doing timed multiple choice quizes at the beginning of class to prepare us for the exam! DO NOT BE LATE FOR CLASS b/c we're going to be timed!
Homework: read the next exercise: 7.6 (it will help you for the next class)
NEXT SCRIBE: DANNY (for friday's class)
February 05, 2007
Scribe Post
Hey everyone! I tried making my scribe as a stop-action animation but I couldn't get it to work right so I guess I'll make it the old way...
With this definite integral, we can find its antiderivative and find the area.
Let's take a look at the graph...
Solving problems like this are not always this easy as this one. We can't always find the exact value of the definite integral.
Take a look at this problem and its graph.
We could use Riemann sums to find the area under the curve.
____________________________________________________________________
These are the differentiation and integration formulas. We use these to help us determine the problems below.
Determine:
1)
Use substitution. (I FORGOT TO PUT dx INTO THE PICTURE)
u= x 3
du= 3x2
1/3du= x2 dx
If your wondering why u = x3 its because
(x3)2 = x6. So u is u2. We are trying to make the original function look like d/dx Arctan(x) integration formula.
Substitute ....................................................Resubstitute
You don't need " + C " in the intermediate steps because the C are not the same while your substituting.
(There's supposed to be an equal (=) sign in between the pictures)
_____________________________________________________________
2)
If the 8 was a 9 in the polynomial we could complete the square.
So we do this,
6x-x2 -8 Factor out the negative.
-[(x2 -6x+8 +1) -1]
-[(x2 -6x+9) -1]
-[(x-3)2 -1]
1 - (x-3)2 We can replace this into the integral.
Next we use substitution.
Let u = x+3
du = dx
Then RE-substitute..
And we're done
Homework: finish up arctrig exercises (7.5)
Wednesday's scribe will be Katrin.
With this definite integral, we can find its antiderivative and find the area.
Let's take a look at the graph...
Solving problems like this are not always this easy as this one. We can't always find the exact value of the definite integral.
Take a look at this problem and its graph.
We could use Riemann sums to find the area under the curve.
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These are the differentiation and integration formulas. We use these to help us determine the problems below.
Determine:
1)
Use substitution. (I FORGOT TO PUT dx INTO THE PICTURE)u= x 3
du= 3x2
1/3du= x2 dx
If your wondering why u = x3 its because
(x3)2 = x6. So u is u2. We are trying to make the original function look like d/dx Arctan(x) integration formula.
Substitute ....................................................Resubstitute
You don't need " + C " in the intermediate steps because the C are not the same while your substituting.
(There's supposed to be an equal (=) sign in between the pictures)
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2)
If the 8 was a 9 in the polynomial we could complete the square.So we do this,
6x-x2 -8 Factor out the negative.
-[(x2 -6x+8 +1) -1]
-[(x2 -6x+9) -1]
-[(x-3)2 -1]
1 - (x-3)2 We can replace this into the integral.
Next we use substitution.Let u = x+3
du = dx
Then RE-substitute..
And we're done
Homework: finish up arctrig exercises (7.5)
Wednesday's scribe will be Katrin.
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