I will start by explaining the area between curves. The formula is bSa top equation-bottom equation
1. draw the picture of the graphs
2. subtract the two equations from each other
3. put the like terms together and integrate the result
Next I will talk about trig inverse integration. The trig inverse integration formulas are: (sr=square root)
1. S du/sr(a^2-u^2)=-1/sr(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sr(u^2-a^2)=1/du(a)arcsec lul/a +C
Another thing I will talk about is substitution. Substitution takes the position of derivative rules when integrating. The steps for substitution are:
1. Find a derivative of something else inside of the integral.
2. Set u equal to the non derivative found in the integral
3. Then take the derivative of u
4. Substitute back in
5. Solve
Example problem:S (2x^2+5) (4x) dx u=2x^2+5 du=4x dx
S u du
1/4 u^2+C
1/4 (2x^2+5)^2+C
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
Sunday, February 21, 2010
Post 26
Finally now that I have working internet I can do my blogs.
If the word approximate is used then you should know to use linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Example problem:(sr=square root)
Use differentiability to approximate sr(4.5)
f(x)=sr(x) sr(4)+(1/2 sr(4) )(.5)=1.125sr(x)+(1/2 sr(x) ) dx
error=.005
dx=.5
x=4
The next topic I will talk about is integration. Integration finds the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]delta x=b-a/number of subintervals
Example problem: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.
delta x=2-0/4=1/2
LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]
1/2[-3+-5/2+-2+-3/2]1/2[-9]= -9/2
RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]
1/2[-5/2+-2+-3/2+-1]1/2[-7]= -7/2
MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]
1/2[-11/4+-9/4+-7/4+-5/4]1/2[-8]= 4
Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]
1/4[-3+-10/2+-4+-6/2+-1]1/4[-16]= 4
If the word approximate is used then you should know to use linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Example problem:(sr=square root)
Use differentiability to approximate sr(4.5)
f(x)=sr(x) sr(4)+(1/2 sr(4) )(.5)=1.125sr(x)+(1/2 sr(x) ) dx
error=.005
dx=.5
x=4
The next topic I will talk about is integration. Integration finds the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]delta x=b-a/number of subintervals
Example problem: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.
delta x=2-0/4=1/2
LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]
1/2[-3+-5/2+-2+-3/2]1/2[-9]= -9/2
RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]
1/2[-5/2+-2+-3/2+-1]1/2[-7]= -7/2
MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]
1/2[-11/4+-9/4+-7/4+-5/4]1/2[-8]= 4
Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]
1/4[-3+-10/2+-4+-6/2+-1]1/4[-16]= 4
post 27
here we go again.
implicit derivatives.
1. Take the derivative of both sides (implicit derivatives usually have an equals sign)
2. When you take a derivative of a y term, state that. Do so by putting y prime or dy/dx. Like ( taking the derivative of y^2 would be 2y(dy/dx) )
3. Move all of the terms that don't have a dy/dx in them to one side. Factor out a dy/dx out of all the terms that do have it, then divide to finish solving for dy/dx.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph)
i dont really understand the particle and bacteria problems. yeerd.
implicit derivatives.
1. Take the derivative of both sides (implicit derivatives usually have an equals sign)
2. When you take a derivative of a y term, state that. Do so by putting y prime or dy/dx. Like ( taking the derivative of y^2 would be 2y(dy/dx) )
3. Move all of the terms that don't have a dy/dx in them to one side. Factor out a dy/dx out of all the terms that do have it, then divide to finish solving for dy/dx.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph)
RELATED RATES:
1. identify all variables and equations
2. identify what you are looking for
3. sketch and label
4. write an equations involving your variables (you can only have one unknown so a secondary equation may be given)
5. take the derivative (with respect to time)
6. substitute in derivative and solve
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
i dont really understand the particle and bacteria problems. yeerd.
post 27
examples examples examplesssssssss.
EXAMPLE 1:Given the equation y = (x-3)/(2-5x). Find dy/dx.
dy/dx = ((2-5x)(1) -(-5)(x-3))/(2-5x)^2
= (2-5x + 5x -15)/(2-5x)^2
= - 13/(2-5x)^2
take the derivative of the function using the quotient rule
EXAMPLE 2: What is the maximum value for the following: f(x) = xe^-x
because i finally understand e's!
Take derivative using product rule (multiplying two things).
x(-e^-x) + e^-x(1)remember that the der. e is e^w/e times the der. of the exponent
Simplify:
-xe^-x + e^-x
You can factor out an e^-x
(-x+1)e^-x
To find the critical values (possible maxima) set (-x+1) equal to zero and solve for x.
This yields x = 1. Now plug in that one to e^-x.
This then gives you 1/e====>your maximum value.
Example 3: The average value of f(x)= -1/x^2 on [1/2, 1].
average value means take the integral!
1/b-a 1S1/2 (-1/x^2)
1/(1-1/2)S x^-2
2[(-x^-1)/-1] = 2/x still have to integrate from 1/2 to 1
(2/1)-(2/(1/2) = -2
i could use a refresher on anything and every thing that has to do with ln.
EXAMPLE 1:Given the equation y = (x-3)/(2-5x). Find dy/dx.
dy/dx = ((2-5x)(1) -(-5)(x-3))/(2-5x)^2
= (2-5x + 5x -15)/(2-5x)^2
= - 13/(2-5x)^2
take the derivative of the function using the quotient rule
EXAMPLE 2: What is the maximum value for the following: f(x) = xe^-x
because i finally understand e's!
Take derivative using product rule (multiplying two things).
x(-e^-x) + e^-x(1)remember that the der. e is e^w/e times the der. of the exponent
Simplify:
-xe^-x + e^-x
You can factor out an e^-x
(-x+1)e^-x
To find the critical values (possible maxima) set (-x+1) equal to zero and solve for x.
This yields x = 1. Now plug in that one to e^-x.
This then gives you 1/e====>your maximum value.
Example 3: The average value of f(x)= -1/x^2 on [1/2, 1].
average value means take the integral!
1/b-a 1S1/2 (-1/x^2)
1/(1-1/2)S x^-2
2[(-x^-1)/-1] = 2/x still have to integrate from 1/2 to 1
(2/1)-(2/(1/2) = -2
i could use a refresher on anything and every thing that has to do with ln.
Post Number Twenty Seven
So these holidays went by way too fast. I am NOT ready to come back to school but ummmmm here’s my blog.
Limit rules since I always miss those.
When the limit is approaching infinity:
Top degree is less than the bottom degree = 0
Top degree is greater than the bottom degree = infinity
Top degree is equal to the bottom degree = top coefficient over the bottom coefficient
For example:
Lim 4x – 2/ x^2 = 0
x>infinity
Lim 3x^4 – 2x + 5/ 6x = infinity
x>infinity
Lim 2x^2 – 1/ 3x^2 = 2/3
x>infinity
When the limit is approaching a number:
Three steps-
1. Factor out the top and the bottom
2. Look for cancellations
3. Plug in the number
Riemann Sums
Ever since that worksheet we did in class this is a LOT easier.
Delta x = b – a /# of subintervals
LRAM- left hand approximation –obviously finds the area on the left side of the curve : deltax[f(a)+f(a+deltax)+...f(b)]
RRAM- right hand approximation – finds the area on the right side of the curve: deltax[f(a+deltax)+...f(b)]
MRAM- middle approximation – finds the area on top of the curve: deltax[f(mid)+f(mid)+...]
Trapezoidal- most accurate, uses trapezoids instead of rectangles: deltax/2[f(a)+2f(a+deltax)+2f(a+2deltax)+...f(b)]
Hmmm stuff I need help with?
Related rates (and I already know the steps so please don’t post those), linearization (knew how last week but forgot again), normal/tangent lines, and problems like the ones with a ladder leaned up on a house or whatever..
Obviously I can use help on a lot of things so feel free to be nice (:
Sooo back to school tomorrow, can’t wait.
SIKEEEEEEE, kbye.
Limit rules since I always miss those.
When the limit is approaching infinity:
Top degree is less than the bottom degree = 0
Top degree is greater than the bottom degree = infinity
Top degree is equal to the bottom degree = top coefficient over the bottom coefficient
For example:
Lim 4x – 2/ x^2 = 0
x>infinity
Lim 3x^4 – 2x + 5/ 6x = infinity
x>infinity
Lim 2x^2 – 1/ 3x^2 = 2/3
x>infinity
When the limit is approaching a number:
Three steps-
1. Factor out the top and the bottom
2. Look for cancellations
3. Plug in the number
Riemann Sums
Ever since that worksheet we did in class this is a LOT easier.
Delta x = b – a /# of subintervals
LRAM- left hand approximation –obviously finds the area on the left side of the curve : deltax[f(a)+f(a+deltax)+...f(b)]
RRAM- right hand approximation – finds the area on the right side of the curve: deltax[f(a+deltax)+...f(b)]
MRAM- middle approximation – finds the area on top of the curve: deltax[f(mid)+f(mid)+...]
Trapezoidal- most accurate, uses trapezoids instead of rectangles: deltax/2[f(a)+2f(a+deltax)+2f(a+2deltax)+...f(b)]
Hmmm stuff I need help with?
Related rates (and I already know the steps so please don’t post those), linearization (knew how last week but forgot again), normal/tangent lines, and problems like the ones with a ladder leaned up on a house or whatever..
Obviously I can use help on a lot of things so feel free to be nice (:
Sooo back to school tomorrow, can’t wait.
SIKEEEEEEE, kbye.
27th post
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Mean Value Theorem
The Mean Value Theorem states that If the function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f'(c)=(f(b)-f(a))/(b-a).
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection.
linerazation is something im having problems with
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Mean Value Theorem
The Mean Value Theorem states that If the function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f'(c)=(f(b)-f(a))/(b-a).
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph!!!)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection.
linerazation is something im having problems with
Post #27
So, these holidays zoomed by..umm, this time i'll explain some frequently missed things..
For one, i'll explain the chain rules.
cos(4x^2)
you have to start with the outtermost first, then innermost..
so take the derivative of cos, which is -sin, and then the derivative of the inside quantity, 4x^2, 8x
= -sin(8x)
Now, lets review some limit rules.
WHEN THE LIMIT IS APPROACHING INFINITY:
top degree < bottom degree = zero
top degree > bottom degree = positive or negative infinity
top degree = bottom degree = divided coefficiants
WHEN THE LIMIT IS APPROACHING A NUMBER:
1. factor out the top and bottom
2. look for cancellations
3. plug in the number
Hopefully this was a good review for everyone..its seems to the the things i commonly miss..
For one, i'll explain the chain rules.
cos(4x^2)
you have to start with the outtermost first, then innermost..
so take the derivative of cos, which is -sin, and then the derivative of the inside quantity, 4x^2, 8x
= -sin(8x)
Now, lets review some limit rules.
WHEN THE LIMIT IS APPROACHING INFINITY:
top degree < bottom degree = zero
top degree > bottom degree = positive or negative infinity
top degree = bottom degree = divided coefficiants
WHEN THE LIMIT IS APPROACHING A NUMBER:
1. factor out the top and bottom
2. look for cancellations
3. plug in the number
Hopefully this was a good review for everyone..its seems to the the things i commonly miss..
Sara-Rachelle Alyssin Robinson
Sara-Rachelle has arrived! She was born on Wednesday at 12:59 weighing 7lbs 1oz. Some people had asked to see pictures and here they are! Also, Blogger won't show the whole picture but if you click on it it will show you in another window.
Post 27
Yay...school tomorrow.
Anyway, here's the steps for implicit derivatives for those of you who have forgotten.
1. Take the derivative of both sides (implicit derivatives usually have an equals sign)
2. When you take a derivative of a y term, state that. Do so by putting y prime or dy/dx. Like ( taking the derivative of y^2 would be 2y(dy/dx) )
3. Move all of the terms that don't have a dy/dx in them to one side. Factor out a dy/dx out of all the terms that do have it, then divide to finish solving for dy/dx.
That's basically all it is. Just make sure when you are doing these is like...say you are doing product rule with like sin's and cos's...make sure you put cos(x) where its supposed to and cos(y) where its supposed to. It would really mess up problems if you mix this up. So take these longer derivatives slow and just avoid making silly mistakes.
As for what else I can explain...hmm
Riemann Sums, yay.
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 this is the most accurate of the Riemann summs. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
all of the above is assuming that x is delta x which is (b-a)/(subintervals)
Anyway, just wanted to mention two things people still miss.
Have a good day :-P
Anyway, here's the steps for implicit derivatives for those of you who have forgotten.
1. Take the derivative of both sides (implicit derivatives usually have an equals sign)
2. When you take a derivative of a y term, state that. Do so by putting y prime or dy/dx. Like ( taking the derivative of y^2 would be 2y(dy/dx) )
3. Move all of the terms that don't have a dy/dx in them to one side. Factor out a dy/dx out of all the terms that do have it, then divide to finish solving for dy/dx.
That's basically all it is. Just make sure when you are doing these is like...say you are doing product rule with like sin's and cos's...make sure you put cos(x) where its supposed to and cos(y) where its supposed to. It would really mess up problems if you mix this up. So take these longer derivatives slow and just avoid making silly mistakes.
As for what else I can explain...hmm
Riemann Sums, yay.
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 this is the most accurate of the Riemann summs. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
all of the above is assuming that x is delta x which is (b-a)/(subintervals)
Anyway, just wanted to mention two things people still miss.
Have a good day :-P
Post #27
I CAN'T BELIEVE WE HAVE SCHOOL TOMORROW!...HOPE EVERYONE HAD A GREAT HOLIDAY!
TO TAKE A DERIVATIVE take the exponent and multiply it by the coeffient to make a new coeffient, then subtract one from the exponent for the new exponent and that's your derivative!
TO TAKE AN INTEGRAL take the exponent and add one, then take it's reciprical and multiply it by the coeffient to make the new coeffient. that's the integral!
LIMITS:
~If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
~If the degree on top is bigger than the degree on the bottom, then the limit is infinity.~If the degree on top is smaller than the degree on the botton, then the limit is zero.
PRODUCT RULE:
the product rule is used when two things are being multiplied together.
copy the first times the derivative of the second PLUS copy the second times the derivative of the first
for example:
Find the derivative of (2x)sin(x)
2(cos(x)) + sin(x)(2) =
4cos(x)+sin(x)
QUOTIENT RULE:
the quotient rule is used when there's something being divided by something and you need to take a derivative you copy the bottom times the derivative of the top MINUS the top times the derivative of the bottom all over the bottom squared
for example:
Find the derivative of (2x^2) over (x)
(x)(4x)-(2x^2)(1) all over (x^2)
(4x^2)-(2x^2) all over (x^2)
therefore: (2x^2) over (x^2) givis you 2 for the derivative
CHAIN RULE:
the chain rule is used when you are finding the derivative with exponents.
*work from the outside in.
Example: sin^2 (x^2)
2(sin (x^2))(cos (x^2)(2x)
4x sin(x^2)cos(x^2)
~ElliE~
TO TAKE A DERIVATIVE take the exponent and multiply it by the coeffient to make a new coeffient, then subtract one from the exponent for the new exponent and that's your derivative!
TO TAKE AN INTEGRAL take the exponent and add one, then take it's reciprical and multiply it by the coeffient to make the new coeffient. that's the integral!
LIMITS:
~If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
~If the degree on top is bigger than the degree on the bottom, then the limit is infinity.~If the degree on top is smaller than the degree on the botton, then the limit is zero.
PRODUCT RULE:
the product rule is used when two things are being multiplied together.
copy the first times the derivative of the second PLUS copy the second times the derivative of the first
for example:
Find the derivative of (2x)sin(x)
2(cos(x)) + sin(x)(2) =
4cos(x)+sin(x)
QUOTIENT RULE:
the quotient rule is used when there's something being divided by something and you need to take a derivative you copy the bottom times the derivative of the top MINUS the top times the derivative of the bottom all over the bottom squared
for example:
Find the derivative of (2x^2) over (x)
(x)(4x)-(2x^2)(1) all over (x^2)
(4x^2)-(2x^2) all over (x^2)
therefore: (2x^2) over (x^2) givis you 2 for the derivative
CHAIN RULE:
the chain rule is used when you are finding the derivative with exponents.
*work from the outside in.
Example: sin^2 (x^2)
2(sin (x^2))(cos (x^2)(2x)
4x sin(x^2)cos(x^2)
~ElliE~
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