Okay so this week .FOR ME. was tough!
First we learned volume by disks
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..WHAT DO YOU DO???
Next we learned volume by washers
so you know that the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the inbetween number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. REMEMBER TO GRAPH!
just a reminder if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
we also learned about area...it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
examples:
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
hope this helps...[question is at the top]
~ELLIE~
Friday, December 4, 2009
ABOUT THE COMMENTS!
for some reason my computer didnt wanta let me get on the site..much less comment last night..so today i put them in...hope it's not too late!
Thursday, December 3, 2009
Ahhhh hmwk help!
Okay, I did part of last night's homework
Checked it today on Calcchat
Got it wrong!
I have no idea what I'm doing!!
Can someone PLEASE help with numbers 12-14??
Maybe 15 and 19 too, I haven't gotten there yet, I've spent forever on this dumb problems that refuse to solve -.-
Also, what in the world are the rectangles that calcchat uses??? I don't understand that and it makes me even more confused
I'm soooo lost! Can someone explain it?
Checked it today on Calcchat
Got it wrong!
I have no idea what I'm doing!!
Can someone PLEASE help with numbers 12-14??
Maybe 15 and 19 too, I haven't gotten there yet, I've spent forever on this dumb problems that refuse to solve -.-
Also, what in the world are the rectangles that calcchat uses??? I don't understand that and it makes me even more confused
I'm soooo lost! Can someone explain it?
HELP !
ok, i need some serious help on the homework... i do not understand how to find the bounds? If anyone can help me it would be a big help.. thanks
Integration and Volume
Here is a great site with visuals for the disk and washer method.
http://www.intmath.com/Applications-integration/4_Volume-solid-revolution.php
Also if anyone finds a copy of the program for mram, lram and rram let me know where.
Wednesday, December 2, 2009
Comments?
Ok, so the blog's being stupid and not letting me post comments?
I posted two comments on Ryan G.'s post, and none of them showed up
I posted two comments on Ryan G.'s post, and none of them showed up
Monday, November 30, 2009
week 14 and 15
this week was crunch time. we had the intense holiday packet to do and while doing it i was able to pick up on substitution.
substitution
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Also something that i understand in calculus is intergration.
intergration is basically setting the derivative back to the origianl equation
you add one to the exponent and then multiply the term by the reciprocal of your exponent.
for instance - x^2 is 1/3x^3
one thing that i dont understand fully is mRam. I understand the process for the most part for lRAM and rRAM but mRAM i just dont get. I get lost and am not to certain of the steps.
substitution
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Also something that i understand in calculus is intergration.
intergration is basically setting the derivative back to the origianl equation
you add one to the exponent and then multiply the term by the reciprocal of your exponent.
for instance - x^2 is 1/3x^3
one thing that i dont understand fully is mRam. I understand the process for the most part for lRAM and rRAM but mRAM i just dont get. I get lost and am not to certain of the steps.
14 andddd 15
man i just suck at internet stuff, i always forget.
holiday's are overr.. sad. but one step closer to graduationnn!
so for week 14 i'll just talk about riemans sum because that's really the only thing i remember before i left.
RRAM: right handed aprroximation
formula= delta x [ f(a + delta x) ... + f(b)]
EX: calculate the right tiemann sum for a(x)=3x on the interval [-2,3]
divided into 4 sub intervals. (sub intervals are just how many intervals you're setting up/how many numbers you're plugging in.
deltaX=5/4.. use (b-a)/# of subintervals to find this.
5/4[f(-3/4)+f(1/2)+f(7/4)+f(3)]
5/4[(9/4)+(3/2)+(21/4)+(9)= 81
LRAM: left handed approximation
formula= delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
MRAM: is to calculate the middleeee
formula= delta x [ f(mid) + f(mid) + .... ]
MMMM.. let's talk integrals!
yeahh, the fun S stuff.
it is basically the reverse action of a derivative.
.. there are two types: definite and indefinite.
Indefinite: the result of indefinite integration is an equation.
S(x^n)dx = [x^(n+1)]/[n+1] +C.
Definite: the result of definite integration is a number.
Definite - is just F(a) +F(b)= #
-take the intergral of a and b (x,y) and then plug it in to get an answer.
... mmm for #15 i wasnt there the week before thanksgiving so we're just going to pull some stuff from wayy back when, but it's okay because at the end i'll have tonnnss of questions. one thing i know i am is LOST. ohhh, and not to forget to put +C!
related rates:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
TANGENT LINES! MY FAVORITE
Steps:
Take f ‘(x).
Plug in x to find your slope: m.
Plug x into f(x) to find y (if not already given).
Using m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1).
okay so for what i don't understand..
any new things with "graphs" i don't have a clue..
it'd be great if someone would like to go over what i missed those 4 days i was absent, you'd be a life saver and i'll bring you candy.
umm, and on the packet we had over the break.. i don't know what to do with 16-25..
anybody wanna help?
holiday's are overr.. sad. but one step closer to graduationnn!
so for week 14 i'll just talk about riemans sum because that's really the only thing i remember before i left.
RRAM: right handed aprroximation
formula= delta x [ f(a + delta x) ... + f(b)]
EX: calculate the right tiemann sum for a(x)=3x on the interval [-2,3]
divided into 4 sub intervals. (sub intervals are just how many intervals you're setting up/how many numbers you're plugging in.
deltaX=5/4.. use (b-a)/# of subintervals to find this.
5/4[f(-3/4)+f(1/2)+f(7/4)+f(3)]
5/4[(9/4)+(3/2)+(21/4)+(9)= 81
LRAM: left handed approximation
formula= delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
MRAM: is to calculate the middleeee
formula= delta x [ f(mid) + f(mid) + .... ]
MMMM.. let's talk integrals!
yeahh, the fun S stuff.
it is basically the reverse action of a derivative.
.. there are two types: definite and indefinite.
Indefinite: the result of indefinite integration is an equation.
S(x^n)dx = [x^(n+1)]/[n+1] +C.
Definite: the result of definite integration is a number.
Definite - is just F(a) +F(b)= #
-take the intergral of a and b (x,y) and then plug it in to get an answer.
... mmm for #15 i wasnt there the week before thanksgiving so we're just going to pull some stuff from wayy back when, but it's okay because at the end i'll have tonnnss of questions. one thing i know i am is LOST. ohhh, and not to forget to put +C!
related rates:
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
TANGENT LINES! MY FAVORITE
Steps:
Take f ‘(x).
Plug in x to find your slope: m.
Plug x into f(x) to find y (if not already given).
Using m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1).
okay so for what i don't understand..
any new things with "graphs" i don't have a clue..
it'd be great if someone would like to go over what i missed those 4 days i was absent, you'd be a life saver and i'll bring you candy.
umm, and on the packet we had over the break.. i don't know what to do with 16-25..
anybody wanna help?
15th post
holidays!!!!!! are over and reality is unfortunaltly back so.....here we go again with the almighty calculus blog and yes i am doing on monday because i forgot....
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
so I am still not understanding e integration and even though I have an example up im still having problem working them.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
so I am still not understanding e integration and even though I have an example up im still having problem working them.
Calculus Blog - Makeup
So the last week was amazing...I completely forgot about school, and it felt great :-)
Looks like in the process, I completely forgot about doing my blog as well :-)
Anyway, so I guess what I can explain is... e^x integration
For example:
integral of e^2x is just (1/2)e^2x+C
The derivative of e^x is just e^x times derivative of x.
So the integral is the opposite of its derivative times e^x. Kind of hard to explain...it just works....
Like for example...The integral of
(2x+4)e^(x^2+4x) is just e^(x^2+4x) + c because the derivative of u was already there. You only have to put the opposite whenever its not there...like the problem before.
Had it been 2e^2x then the integral would just be e^2x + c. You can check your answer by taking the derivative of that...which it works out.
The other thing I understand pretty well is ln integration...
1/x is ln|x|
So 2/x is 2ln|x|+C and 2/(1-x) is 2ln|1-x|+c
So anyway, ln integration is easy...
What else is there...
Today we learned how to finally do those ones with d/dx in front of the integral symbol...it turned out really easy actually...
You just cancel the integration and the d/dx and you plug in the top x or x^2 or whatever it is. So that's easy.
Overall the integration packet was not that bad...just really long. The rram, lram, mram, and trapezoidal ram took the longest because of crazy fractions and stuff though.
Anyway, that wraps up my post.
-John
Looks like in the process, I completely forgot about doing my blog as well :-)
Anyway, so I guess what I can explain is... e^x integration
For example:
integral of e^2x is just (1/2)e^2x+C
The derivative of e^x is just e^x times derivative of x.
So the integral is the opposite of its derivative times e^x. Kind of hard to explain...it just works....
Like for example...The integral of
(2x+4)e^(x^2+4x) is just e^(x^2+4x) + c because the derivative of u was already there. You only have to put the opposite whenever its not there...like the problem before.
Had it been 2e^2x then the integral would just be e^2x + c. You can check your answer by taking the derivative of that...which it works out.
The other thing I understand pretty well is ln integration...
1/x is ln|x|
So 2/x is 2ln|x|+C and 2/(1-x) is 2ln|1-x|+c
So anyway, ln integration is easy...
What else is there...
Today we learned how to finally do those ones with d/dx in front of the integral symbol...it turned out really easy actually...
You just cancel the integration and the d/dx and you plug in the top x or x^2 or whatever it is. So that's easy.
Overall the integration packet was not that bad...just really long. The rram, lram, mram, and trapezoidal ram took the longest because of crazy fractions and stuff though.
Anyway, that wraps up my post.
-John
Sunday, November 29, 2009
Ash's 15th Post
First off: I rename Holidays to Stressingdays
I.
Hate.
"Holidays"
Now that that's half-way out of my system...let's move on
Wow...I just got majorly distracted on another site..I'm going to stay focused this time
Let's explainnn...umm...Riemann Sums! Okay, yes, this is a bit over-done, but I don't know what else to do! =P
First step!!
Find delta.
How do you find delta? Why, like this :)
b
S (insert equation here)
a
With N subintervals
Here's the formula: (b-a)/n
Simple right?
Now, take your delta and plug it into your equation.
What? There's more to plug in than just that, you say? Well, let's get to it!
So, there's all of these crazy letters and shapes in parenthesis right? Right!
But what do they mean?
Well, let's provide a vocab list
Trianglex = Delta
f(a) = plug in the number on the bottom of your integration symbol
f(b) = plug in the number on the top of your integration symbol
f(Trianglex) = plug in delta to your equation
f(a+trianglex) = solve a+delta and then plug it into your equation
f(b-trianglex) = find b-delta and plug it in (usually the last number to plug in on LRAM)
f(mid) = find the middle number of two numbers and then plug it into your equaion
Now see? Easy peasy!
After you know your delta, a, and b, plug it all into the formulas!
LRAM: delta(f(a)+f(a+delta)+f(a+2delta)...+f(b-delta))
RRAM: delta(f(a+delta)+f(a+2delta)+...f(b))
MRAM: delta(f(mid)+f(mid)...)
Trapezoidal: delta/2(f(a)+2f(a+delta)+2f(a+2delta)....+f(b))
Okay, so I've got a lot of questions
But
I've posted some of them three times already on here, so I'm NOT going to retype them AGAIN. -.-
But
For Trapezoidal, is the next term 2f(a+3delta)??
And I don't understand how to graph using Trapezoidal
In fact
I don't get graphs at all
Have you noticed? =]
I.
Hate.
"Holidays"
Now that that's half-way out of my system...let's move on
Wow...I just got majorly distracted on another site..I'm going to stay focused this time
Let's explainnn...umm...Riemann Sums! Okay, yes, this is a bit over-done, but I don't know what else to do! =P
First step!!
Find delta.
How do you find delta? Why, like this :)
b
S (insert equation here)
a
With N subintervals
Here's the formula: (b-a)/n
Simple right?
Now, take your delta and plug it into your equation.
What? There's more to plug in than just that, you say? Well, let's get to it!
So, there's all of these crazy letters and shapes in parenthesis right? Right!
But what do they mean?
Well, let's provide a vocab list
Trianglex = Delta
f(a) = plug in the number on the bottom of your integration symbol
f(b) = plug in the number on the top of your integration symbol
f(Trianglex) = plug in delta to your equation
f(a+trianglex) = solve a+delta and then plug it into your equation
f(b-trianglex) = find b-delta and plug it in (usually the last number to plug in on LRAM)
f(mid) = find the middle number of two numbers and then plug it into your equaion
Now see? Easy peasy!
After you know your delta, a, and b, plug it all into the formulas!
LRAM: delta(f(a)+f(a+delta)+f(a+2delta)...+f(b-delta))
RRAM: delta(f(a+delta)+f(a+2delta)+...f(b))
MRAM: delta(f(mid)+f(mid)...)
Trapezoidal: delta/2(f(a)+2f(a+delta)+2f(a+2delta)....+f(b))
Okay, so I've got a lot of questions
But
I've posted some of them three times already on here, so I'm NOT going to retype them AGAIN. -.-
But
For Trapezoidal, is the next term 2f(a+3delta)??
And I don't understand how to graph using Trapezoidal
In fact
I don't get graphs at all
Have you noticed? =]
Post 15
This week we were off of school. I got way to used to it and I definately do not want to go to school tomorrow. The integration packet was pretty difficult, but we got through the most of it. The packet included things we've learned over the past weeks with integration. The packet included problems that delt with definite integrals, indefinite integrals, Riemann Sums, substitution, substitution with e, natural logs, and space between a curve and a line.
First of all, an integral is the opposite of a derivative. Instead of multiplying the exponent to the coefficient in front then subtracting from the exponent, you will first add one to the exponent then divide by it.
A indefinite integral is the most basic type of integral. It is just doing the opposite of a derivitive, and the answer comes out in the form of an expression.
A definite integral is an integral with a range. For definite integrals, you first integrate the problem, then you plug in for your range. The number at the top is b and the number at the bottom is a. Once plugged in, the sum of a is subtracted from the sum of b, leaving you with a number for an answer. If there is ever an absolute value in the problem, the integral must be split. This can be done by solving for x.
Substitution is used when there seems to be a chain rule, a product rule, or a quotent rule. These rules do not apply in integration, so substitution must be used. Substitution is basically forcing a problem to work, using one rule for all problems. In the problem, a derivative and non derivitive must be located. The nonderivitive will be noted with u and the derivitive with du.
Substitution with e is the same. The derivite of e is e, so the answer to any integral with e will basically be e with it's exponent + c
Tonight working on the packet, I had a few problems. One of them delt with finding the area between a parabola and a line. I just don't understand these problems. Also, in the packet there were natural log problems with ln in the integral. I don't know what to do with these.
First of all, an integral is the opposite of a derivative. Instead of multiplying the exponent to the coefficient in front then subtracting from the exponent, you will first add one to the exponent then divide by it.
A indefinite integral is the most basic type of integral. It is just doing the opposite of a derivitive, and the answer comes out in the form of an expression.
A definite integral is an integral with a range. For definite integrals, you first integrate the problem, then you plug in for your range. The number at the top is b and the number at the bottom is a. Once plugged in, the sum of a is subtracted from the sum of b, leaving you with a number for an answer. If there is ever an absolute value in the problem, the integral must be split. This can be done by solving for x.
Substitution is used when there seems to be a chain rule, a product rule, or a quotent rule. These rules do not apply in integration, so substitution must be used. Substitution is basically forcing a problem to work, using one rule for all problems. In the problem, a derivative and non derivitive must be located. The nonderivitive will be noted with u and the derivitive with du.
Substitution with e is the same. The derivite of e is e, so the answer to any integral with e will basically be e with it's exponent + c
Tonight working on the packet, I had a few problems. One of them delt with finding the area between a parabola and a line. I just don't understand these problems. Also, in the packet there were natural log problems with ln in the integral. I don't know what to do with these.
Post #15
Okay so, since we didn't have school this week and didn't learn anything new I will review Riemann sums because I just did about 30 problems using it on the packet.
Riemann sums is an approximation of area using rectangles or trapezoids.
First of all, delta x is found by subtracting b from a and dividing by n subintervals.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
I am having trouble with:
1. Problems numbers 52-55 on our packet which give you an equations such as A(x) = the integral of b=x and a=-2 sin^3 (t) dt and you are asked to find A'(pi/2). I'm not exactly sure what they are asking for or where to even start.
2. Integrating trig functions raised to exponents such as the integral of (csc^6 x) (cot^2x)dx
3. The whole last page of the packet which asks to find the area of the region enclose by two equations between two x or y values.
This is kind of late so I'm not expecting answers before we turn our packets in, but answers will still be appreciated for future problems.
Posting...#15
Week 15 was my favorite week of school so far because we had nothing to do for school. WEll i hope everyone had a nice thanksgiving this week; my thanksgiving was alright i had a bad thing happen but overall it was a good week and the food was great. And my birthday is in 2 days yeah thats right december first. WEll now lets get to the rundown about mathmatics.
Welll monday we were off tuesday we were off wendsday we were also off thursday was no exception we were off that day too and friday so we didn't do anything in calculus the whole week.
So integration you have two types
indefinite- your answer is an equation
and definite- your awnser is a number
indefinite is basically the backwards version of a derivative
and definite intervals are a little different
If you have b/s\a f(X)dx=f(b)-F(a)= your awnser
first you take the indefnite and then you plug in your b and subtract it by your a pluged in to get your awnser.
For what i don't understand and what i have the most trouble with is substitution so if anyone can help me with that thanks much obliged.
As for that packet i got wednesday because i missed friday i have no idea what im doing which is a big problem and i finally did my wiki the link finally worked. yeah
Welll monday we were off tuesday we were off wendsday we were also off thursday was no exception we were off that day too and friday so we didn't do anything in calculus the whole week.
So integration you have two types
indefinite- your answer is an equation
and definite- your awnser is a number
indefinite is basically the backwards version of a derivative
and definite intervals are a little different
If you have b/s\a f(X)dx=f(b)-F(a)= your awnser
first you take the indefnite and then you plug in your b and subtract it by your a pluged in to get your awnser.
For what i don't understand and what i have the most trouble with is substitution so if anyone can help me with that thanks much obliged.
As for that packet i got wednesday because i missed friday i have no idea what im doing which is a big problem and i finally did my wiki the link finally worked. yeah
post 14 and 15
So this past week was our Thanksgiving holidays, which was a much needed break. But before this break we learned about many different things. One thing was substitution. The steps for this are
1. Pick out your u
2. Take the derivative of u
3. Substitute it back in to original
4. Then integrate like normal
5. Take all of the u's you get and put them back into the original problem
For old stuff that I know limit rule is not hard at all.
1. If the biggest exponent on the top and bottom are equal then the limit is the coefficient of the highest exponent on the top over the the coefficient of the highest exponent on the bottom.
2. If the biggest exponent is on the top then the limit is infinity.
3. If the bigger exponent is on the bottom then the limit is zero.
The steps for related rates are
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
Some things I do not know are ones like numbers 16, 46, 54, and 91 on the packet we were given to do over the holidays.
1. Pick out your u
2. Take the derivative of u
3. Substitute it back in to original
4. Then integrate like normal
5. Take all of the u's you get and put them back into the original problem
For old stuff that I know limit rule is not hard at all.
1. If the biggest exponent on the top and bottom are equal then the limit is the coefficient of the highest exponent on the top over the the coefficient of the highest exponent on the bottom.
2. If the biggest exponent is on the top then the limit is infinity.
3. If the bigger exponent is on the bottom then the limit is zero.
The steps for related rates are
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
Some things I do not know are ones like numbers 16, 46, 54, and 91 on the packet we were given to do over the holidays.
post 14 and 15
ok, i didn't know i had to do two blogs either, so like a lot of other people, i am going to combine them on this blog.
we learned about substitution, substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
So there are four different methods of integration, LRAM, RRAM, MRAM, and trapezoidal.
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
i suck at integration, that's my only problem.
we learned about substitution, substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
So there are four different methods of integration, LRAM, RRAM, MRAM, and trapezoidal.
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
i suck at integration, that's my only problem.
post 14 and 15 like everyone else
here we go.
There were two types of integration. the first type was indefinite and the second was definite. you do the same thing for both except that for definite you see two numbers and at the end you have to plug them in and then subtract what you get by each other in order to get your answer. Remember that if it's indefinite then you have to put the plus 'c' but if it's definite then you just should have a number.
Then we learned about the change of variables which means that there is at least two different variables in the equation you are solving.
Then we learned about the area between the curves of graphs and the formula is bSa top equation minus the bottom equaion. in order to find a and b you need to set the equations equal to each other and then solve. what you solve for depends on what you're looking for or have in the equations. for instance if the area is on the y axis, then a and b need to be y values so the equations can be solved for x and opposite if the values need to be x values.
i kinda like the e integration the most. when you have the e integration, whatever it's raised to the e power will be your 'u' and 'du' will be the derivative of 'u'.
substitution
Okay, so I know the steps. They are extremely simple to understand; however, when I begin to actually do these steps I get thrown off by one thing: balancing. But anyway, here are the steps in my terms:
1. Identify your u (I identify this by looking inside the parentheses...and voila!)
2. Take the derivative of u to find your DU!!! (and don't forget the dx at the end)
3. Go back to original integration problem and substitute (ha! get it...substitution).
4. Then Integrate normal
5. Put all your u's back in to the original problem.
i dont remember the LRAM RRAM and MRAM. if someone wants to help me out w/ that
There were two types of integration. the first type was indefinite and the second was definite. you do the same thing for both except that for definite you see two numbers and at the end you have to plug them in and then subtract what you get by each other in order to get your answer. Remember that if it's indefinite then you have to put the plus 'c' but if it's definite then you just should have a number.
Then we learned about the change of variables which means that there is at least two different variables in the equation you are solving.
Then we learned about the area between the curves of graphs and the formula is bSa top equation minus the bottom equaion. in order to find a and b you need to set the equations equal to each other and then solve. what you solve for depends on what you're looking for or have in the equations. for instance if the area is on the y axis, then a and b need to be y values so the equations can be solved for x and opposite if the values need to be x values.
i kinda like the e integration the most. when you have the e integration, whatever it's raised to the e power will be your 'u' and 'du' will be the derivative of 'u'.
substitution
Okay, so I know the steps. They are extremely simple to understand; however, when I begin to actually do these steps I get thrown off by one thing: balancing. But anyway, here are the steps in my terms:
1. Identify your u (I identify this by looking inside the parentheses...and voila!)
2. Take the derivative of u to find your DU!!! (and don't forget the dx at the end)
3. Go back to original integration problem and substitute (ha! get it...substitution).
4. Then Integrate normal
5. Put all your u's back in to the original problem.
i dont remember the LRAM RRAM and MRAM. if someone wants to help me out w/ that
Post Number Fourteen and Fifteen
This break flew by as usual. But seniors the year's about to fly by! :) So, i thought i was going to remember this after the break, wow i thought wrong. I'm gonna post fourteen and fifteen together since the holidays overwhelmed me and i completely forgot last week. Let's get this going.
Integration is what we've learned lately. The first thing we talked about was Riemann Sums. It is the approximation of area using rectangles oor trapezoids. I did okay with these when we first learned them but now i'm kind of clueless. I can put together how to do LRAM and RRAM but get completely lost with MRAM and TRAM if anyone would like to further explain these to me. I have to formulas it's just hard for me to follow exactly what to do by looking at them..
Now on to Integration itself. Remember integration is the opposite of a derivative. We never use the rules of derivatives inside integration so don't try to do product and quotient rule! The problem will always be manipulated to be integrated. I've finally got down definite and indefinite integration of polynomials, so I'll do one of those as my example problem:
S x^2 + 4x + 9 dx
= 1/3x^3 + 4/2x^2 + 9x + C
= 1/3x^3 + 2x^2 + 9x + C
simple enough, right? It's like working backwards. Just don't forget the plus C in all indefinite integrals!
Definite integrals are also pretty simple. You are given an interval to integrate with. All you do is integrate regularly and then after plug in the numbers you are given. The answer is always a number, f(b) - f(a).
As for the things i need help with, which are a lot may i add..
Substitution- just in general, i need serious guidance
Riemann sums- more or less MRAM and TRAM
change of variables- i caught on well to this in class, i think i just need a reminder
e and ln integration
Now a specific question i have is a problem like
S -4x^-3 - 5x^2 + 5x^-1dx
i was integrating it normally but i got confused at the last part.
adding 1 to -1 gives you 0 and i didn't know what to do.
I think this is when you do natural log integration but i can't remember exactly, so if anyone can answer this for me I'd love you forever :)
Well i need to go finish working on this MONSTER packet. If anyone is up for helping me catch on a little better my number is (504) 919-7250. See yall tomorrow!
Integration is what we've learned lately. The first thing we talked about was Riemann Sums. It is the approximation of area using rectangles oor trapezoids. I did okay with these when we first learned them but now i'm kind of clueless. I can put together how to do LRAM and RRAM but get completely lost with MRAM and TRAM if anyone would like to further explain these to me. I have to formulas it's just hard for me to follow exactly what to do by looking at them..
Now on to Integration itself. Remember integration is the opposite of a derivative. We never use the rules of derivatives inside integration so don't try to do product and quotient rule! The problem will always be manipulated to be integrated. I've finally got down definite and indefinite integration of polynomials, so I'll do one of those as my example problem:
S x^2 + 4x + 9 dx
= 1/3x^3 + 4/2x^2 + 9x + C
= 1/3x^3 + 2x^2 + 9x + C
simple enough, right? It's like working backwards. Just don't forget the plus C in all indefinite integrals!
Definite integrals are also pretty simple. You are given an interval to integrate with. All you do is integrate regularly and then after plug in the numbers you are given. The answer is always a number, f(b) - f(a).
As for the things i need help with, which are a lot may i add..
Substitution- just in general, i need serious guidance
Riemann sums- more or less MRAM and TRAM
change of variables- i caught on well to this in class, i think i just need a reminder
e and ln integration
Now a specific question i have is a problem like
S -4x^-3 - 5x^2 + 5x^-1dx
i was integrating it normally but i got confused at the last part.
adding 1 to -1 gives you 0 and i didn't know what to do.
I think this is when you do natural log integration but i can't remember exactly, so if anyone can answer this for me I'd love you forever :)
Well i need to go finish working on this MONSTER packet. If anyone is up for helping me catch on a little better my number is (504) 919-7250. See yall tomorrow!
Post 14 and 15
I was completely unaware that I had to do TWO blogs this week (didn't check edline, thanks to Ashley who now has me worrying about these), so here's all TWO of them...no wait..im only doing one:
One of them.
Substitution:
Okay, so I know the steps. They are extremely simple to understand; however, when I begin to actually do these steps I get thrown off by one thing: balancing. But anyway, here are the steps in my terms:
1. Identify your u (I identify this by looking inside the parentheses...and voila!)
2. Take the derivative of u to find your DU!!! (and don't forget the dx at the end)
3. Go back to original integration problem and substitute (ha! get it...substitution).
4. Then Integrate normal
5. Put all your u's back in to the original problem.
So for an example that I seriously get thrown of in (hint hint..if anyone would like to EXPLAIN this to me.....GREATLY APPRECIATED...just so you know...because I'm going to go insane).
Given:
S 6x(3x^2 + 4)^3 dx
Now I know my u is 3x^2 + 4.
My du is then 6x.
Subsitute the u and du
S du(u)^3
Now, in my notes I don't know what happens to this du..does it disappear?
Continuing the problem.
WHAT DO I DO WITH THE BALANCING THING!!! am I balancing the du or the u or WHAT???? I'm literally going insane...just to let you know...
I put a 1/6
so 1/6 [1/4(u)^4]
Final answer: 1/24(3x^2 + 4)^4
Is that right?? Can someone help me?
Also, as I'm sitting here to the packet what does the initial problems entail? I'm sitting here looking through my notebook for something that vaguely resembles these problems and am totally stumped. They are no where to be found....AMAZING! POOF!!! GONE!!!! pulled a flipping Houdini on me...apparently..
Now back to doing some other homework...by the way...I actually get some things in Calculus like basic Definite and indefinite...
OH!! I just remembered..what do you do when a number...say 7...is rasied to an exponent...say x^2...do I just substitute or what..and then how do i integrate that!??
HELP!!!!!!!!!!!!!!!!!!
One of them.
Substitution:
Okay, so I know the steps. They are extremely simple to understand; however, when I begin to actually do these steps I get thrown off by one thing: balancing. But anyway, here are the steps in my terms:
1. Identify your u (I identify this by looking inside the parentheses...and voila!)
2. Take the derivative of u to find your DU!!! (and don't forget the dx at the end)
3. Go back to original integration problem and substitute (ha! get it...substitution).
4. Then Integrate normal
5. Put all your u's back in to the original problem.
So for an example that I seriously get thrown of in (hint hint..if anyone would like to EXPLAIN this to me.....GREATLY APPRECIATED...just so you know...because I'm going to go insane).
Given:
S 6x(3x^2 + 4)^3 dx
Now I know my u is 3x^2 + 4.
My du is then 6x.
Subsitute the u and du
S du(u)^3
Now, in my notes I don't know what happens to this du..does it disappear?
Continuing the problem.
WHAT DO I DO WITH THE BALANCING THING!!! am I balancing the du or the u or WHAT???? I'm literally going insane...just to let you know...
I put a 1/6
so 1/6 [1/4(u)^4]
Final answer: 1/24(3x^2 + 4)^4
Is that right?? Can someone help me?
Also, as I'm sitting here to the packet what does the initial problems entail? I'm sitting here looking through my notebook for something that vaguely resembles these problems and am totally stumped. They are no where to be found....AMAZING! POOF!!! GONE!!!! pulled a flipping Houdini on me...apparently..
Now back to doing some other homework...by the way...I actually get some things in Calculus like basic Definite and indefinite...
OH!! I just remembered..what do you do when a number...say 7...is rasied to an exponent...say x^2...do I just substitute or what..and then how do i integrate that!??
HELP!!!!!!!!!!!!!!!!!!
Post #14 & #15
Sorry I didn't do the last blog. I guess the excitement of the holidays made me forgot.
Anyways post #14, I missed class two days this week so I have many questions.(*S is for the integral)
We learned about substitution, and I looked over my notes and examples to see if I could understand it. The only thing I see is that du is the derivative of u. Could someone explain the steps to a substitution problem?
Like what is the u of these problems? Then what do you do after the u and du step?
S(x^2+1)(2x)dx
Sx(x^2+1)^2dx
Also with e integration and ln integration, you have to use substitution in some problems?
I do know indefinite integration though. Sx^ndx=(x^n+1/n+1)+C (EQUATION) Examples:
1. Sx^3dx= (1/4)x^4+C 2. Sx^2+4x+9dx
(1/3)x^3+(4x^2/2)+9x+C
=(1/3)x^3=2x^2+9x+C
Now for post #15...we have been off a whole week!
I think I'm going to discuss definite integrals this time.
The formula for definite integration: bSa f(x)dx= F(b)-F(a) =NUMBER. First you integrate and put the little line with the b and a, and then you would just plug in to your formula. Example:
1. 3S0 x^2dx= (1/3)x^3 (and then you would put the little line with 3 and 0, but it is hard to type this on here)
(1/3)(3)^3-(1/3)(0^3)= 9
Now something that does confuse me is something like 4S0 absolute value of(x^2-9)dx I checked out calcchat.com but I don't understand what they did. Anyone can help with that?
Anyways post #14, I missed class two days this week so I have many questions.(*S is for the integral)
We learned about substitution, and I looked over my notes and examples to see if I could understand it. The only thing I see is that du is the derivative of u. Could someone explain the steps to a substitution problem?
Like what is the u of these problems? Then what do you do after the u and du step?
S(x^2+1)(2x)dx
Sx(x^2+1)^2dx
Also with e integration and ln integration, you have to use substitution in some problems?
I do know indefinite integration though. Sx^ndx=(x^n+1/n+1)+C (EQUATION) Examples:
1. Sx^3dx= (1/4)x^4+C 2. Sx^2+4x+9dx
(1/3)x^3+(4x^2/2)+9x+C
=(1/3)x^3=2x^2+9x+C
Now for post #15...we have been off a whole week!
I think I'm going to discuss definite integrals this time.
The formula for definite integration: bSa f(x)dx= F(b)-F(a) =NUMBER. First you integrate and put the little line with the b and a, and then you would just plug in to your formula. Example:
1. 3S0 x^2dx= (1/3)x^3 (and then you would put the little line with 3 and 0, but it is hard to type this on here)
(1/3)(3)^3-(1/3)(0^3)= 9
Now something that does confuse me is something like 4S0 absolute value of(x^2-9)dx I checked out calcchat.com but I don't understand what they did. Anyone can help with that?
post 14 & 15
first off, i'd like to say HAPPY BIRTHDAY TO ME & MAMIE! :) :) :)
secondly, sorry i forgot about my blog last weekend, so i'm gonna do just like steph and do two in one.
POST #14
so basically this past week we've just been reviewing integration. with definite and indefinite integrals, how to not use product and quotient rule, how to integrate something with a trig function, and all that good stuff. we also learned substitution. this is when you set something in the problem equal to u and something else in the problem equal to du. typically the part of the problem that looks like the derivative of the other half is du.
i was getting really good at substitution right before the holidays, but lucky me, i left my binder at school over thanksgiving break, and now it's almost completely left my mind. i'll try and explain it the best i can though.
after fiding u and du, you then plug in u and du back in. let's say you had f(x) = (cos^2(X))(-sin(x)) dx
u = cos(x) du=-sin(x)
f(x) = u^2 du dx
then you would plug cos(x) back into u, and du dx cancels out
f(x)=c0s^2(x)
and i'm pretty sure that's how you do it.. if i'm wrong please correct me! that will be what i need help with for this blog
POST # 15
OK, so it's the last day of our holiday's and my birthday :) so obviously since we weren't at school we didn't learn anything new this week.
i'm gonna review limit rules, since we still have problems with limits sometimes.
if a limit is approaching any number, plug that number into the equation at x. solve for x, and that is your answer. if you can cancel out things in the equation, do that before you plug in for x and start solving. also, if the bottom is equal to 0, your answer is undefined. sometimes, when you can't plug in for x, you have to plug the entire equation into your calculator and go to the table function and see what number is approaching your limit. if you are given a limit approaching infinity, you have three basic rules:
1. if top degree is larger than bottom degree, it is equal to +/- infinity.*****
2. if top degree is smaller than bottom degree, it is equal to 0.
3. if top degree is euqal to bottom degree, then divide the leading coefficients.
*****to find out whether it is positive or negative infinity, you plug it into the graph on your calculator.
as easy as limits seem, they are still hard. especially whenever you have to use your table function in the calculator. also if do use your table function and the limit is approaching something with E in it, the answer is undefined. you can always use your table function to double check yourself but you don't have to.
the thing i still don't understand is LRAM, MRAM, RRAM. i guess i never really grasped it in the first place to ever understand it. we kinda learned it fast, and never really dealt with it after that. i'm just nervous because it's gonna be on the exam! i don't understand any of it, so if you do please give me a brief explanation, thanks! :)
secondly, sorry i forgot about my blog last weekend, so i'm gonna do just like steph and do two in one.
POST #14
so basically this past week we've just been reviewing integration. with definite and indefinite integrals, how to not use product and quotient rule, how to integrate something with a trig function, and all that good stuff. we also learned substitution. this is when you set something in the problem equal to u and something else in the problem equal to du. typically the part of the problem that looks like the derivative of the other half is du.
i was getting really good at substitution right before the holidays, but lucky me, i left my binder at school over thanksgiving break, and now it's almost completely left my mind. i'll try and explain it the best i can though.
after fiding u and du, you then plug in u and du back in. let's say you had f(x) = (cos^2(X))(-sin(x)) dx
u = cos(x) du=-sin(x)
f(x) = u^2 du dx
then you would plug cos(x) back into u, and du dx cancels out
f(x)=c0s^2(x)
and i'm pretty sure that's how you do it.. if i'm wrong please correct me! that will be what i need help with for this blog
POST # 15
OK, so it's the last day of our holiday's and my birthday :) so obviously since we weren't at school we didn't learn anything new this week.
i'm gonna review limit rules, since we still have problems with limits sometimes.
if a limit is approaching any number, plug that number into the equation at x. solve for x, and that is your answer. if you can cancel out things in the equation, do that before you plug in for x and start solving. also, if the bottom is equal to 0, your answer is undefined. sometimes, when you can't plug in for x, you have to plug the entire equation into your calculator and go to the table function and see what number is approaching your limit. if you are given a limit approaching infinity, you have three basic rules:
1. if top degree is larger than bottom degree, it is equal to +/- infinity.*****
2. if top degree is smaller than bottom degree, it is equal to 0.
3. if top degree is euqal to bottom degree, then divide the leading coefficients.
*****to find out whether it is positive or negative infinity, you plug it into the graph on your calculator.
as easy as limits seem, they are still hard. especially whenever you have to use your table function in the calculator. also if do use your table function and the limit is approaching something with E in it, the answer is undefined. you can always use your table function to double check yourself but you don't have to.
the thing i still don't understand is LRAM, MRAM, RRAM. i guess i never really grasped it in the first place to ever understand it. we kinda learned it fast, and never really dealt with it after that. i'm just nervous because it's gonna be on the exam! i don't understand any of it, so if you do please give me a brief explanation, thanks! :)
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