This week in Calculus BC was pretty good.
Tuesday we re-learned Integration by-parts.
Integration by-parts:
Sudv = uv - Svdu
Example:
Sxe^(x)dx
Pick a u and dv and derive/integrate:
u = x du = 1 dx
dv = e^x(dx) v = e^x
You then plug these into your equation:
xe^(x) - Se^(x)dx
And solve:
xe^(x) - e^(x) + C
We also reviewed "Chasing the Rabbit"
This method of integration happens when you have an e term and a trig term in an integration problem and you basically keep integrating by-parts until you arrive at an integration term the same as you started off with, you them set everything you have equal to the original problem.
We learned something brand new this week that I'm having a little trouble with: Trig Integration.
I get the basic problems, but I was having issues with the homework over the weekend.
Two examples would be:
Ssec^3(pix)dx
and
Stan^5(x/2)dx
Sunday, August 29, 2010
Steph's First Calc BC Post
So, lets get this thing going again. First i'll give a little summary of things I did this past week. We went over lo'hospital's rule, integration, substitution, bi-parts, trig integration, and i think there's one more thing that just isn't clicking this moment.
So since this week was basically a review, let me touch on everything we did. Kinda like a re-review.
So, lets do some reviewing...
LO'HOSPITAL'S RULE:
1. This deals only with limits..so if you don't have a limit, dont do this!
2. Plug in the number the limit is approaching into the equation and make sure you get an infinitave..such as infinity/infinity, zero/zero, infinity/zero, zero/infinity.
3. Next, you need to take the derivative of the top, and the derivitave of the bottom.
******NOT QUOTIENT RULE, JUST TWO DERIVATIVES.
4. Plug in the number the limit is approaching
5. If you get another infinitive, repeat.
INTEGRATION:
*So, i know this is a little review, but don't laugh...you mess up on these sometimes too!
1. You know it's an integral when you see the "s" looking thinggyy.
2. Add one to the exponent
3. Multiply the coefficient by the recriprocal of the new exponent.
If indefinite, be sure to include +c, if not, solve.
So, these are the two topics that i was most confortable with my quizzes.
Now, the things i'm pretty sure i didn't do good on, bi-parts and trig substitution..
These two topics just don't click in my head. Any suggestions on how i should study thesee things?
Also, how do you tell if its biparts or trig sub..?
What would you do for S xarctanx?
So since this week was basically a review, let me touch on everything we did. Kinda like a re-review.
So, lets do some reviewing...
LO'HOSPITAL'S RULE:
1. This deals only with limits..so if you don't have a limit, dont do this!
2. Plug in the number the limit is approaching into the equation and make sure you get an infinitave..such as infinity/infinity, zero/zero, infinity/zero, zero/infinity.
3. Next, you need to take the derivative of the top, and the derivitave of the bottom.
******NOT QUOTIENT RULE, JUST TWO DERIVATIVES.
4. Plug in the number the limit is approaching
5. If you get another infinitive, repeat.
INTEGRATION:
*So, i know this is a little review, but don't laugh...you mess up on these sometimes too!
1. You know it's an integral when you see the "s" looking thinggyy.
2. Add one to the exponent
3. Multiply the coefficient by the recriprocal of the new exponent.
If indefinite, be sure to include +c, if not, solve.
So, these are the two topics that i was most confortable with my quizzes.
Now, the things i'm pretty sure i didn't do good on, bi-parts and trig substitution..
These two topics just don't click in my head. Any suggestions on how i should study thesee things?
Also, how do you tell if its biparts or trig sub..?
What would you do for S xarctanx?
first calc BC post..
alright, so back to another year of blogs... YAY :D
haha, anyways. this week we took 3 tests. oh my lord, yes i know.
one on l'hopital's rule, one on basic integration, and one on integration by parts. i did great on the first one, hopefully i did good on the other two, too. :)
ok, so let's go over some integration by parts.
first you need to find your u and your dv.
usually your u is whatever can be reduced.
then after that, you find your du and your v.
*remember sometimes your dv can be your dx and your v would then become x.
then you simply plug into the formula and integrate.
oh .. & the formula is
uv - S vdu
*p.s. - S is the integration symbol
EXAMPLE:
S xsin2x dx
u = x v = -1/2cos2x
du = 1 dx dv=sin2x
x-1/2cos2x - S -1/2cos2x
=> your answer would be...
-(x)1/2cos(2x) + 1/4sin(2x) + c
easy right? yeah. just wait til you get to the hard ones. lol OH & there is chasing the rabbit. this happens whenever you integrate something using by parts twice, and you end up with the same thing you started with in the original problem. then you simply set it equal to the original and solve it like that.
Another aspect of calculus bc that we reviewed this week, which i'll review briefly, is l'hopitals rule.
this is whenever you try to find a limit of something, but it's in indeterminate form... then you have to use l'hopitals rule.
which means you take the derivative of the top of the fraction and the derivative of the bottom of the fraction... SEPARATELY! no quotient rule. then plug in & solve.
but, you must make sure that it is in fraction form.. because you cannot use l'hopitals rule if it is not.
alright, well that's all for now.
haha, anyways. this week we took 3 tests. oh my lord, yes i know.
one on l'hopital's rule, one on basic integration, and one on integration by parts. i did great on the first one, hopefully i did good on the other two, too. :)
ok, so let's go over some integration by parts.
first you need to find your u and your dv.
usually your u is whatever can be reduced.
then after that, you find your du and your v.
*remember sometimes your dv can be your dx and your v would then become x.
then you simply plug into the formula and integrate.
oh .. & the formula is
uv - S vdu
*p.s. - S is the integration symbol
EXAMPLE:
S xsin2x dx
u = x v = -1/2cos2x
du = 1 dx dv=sin2x
x-1/2cos2x - S -1/2cos2x
=> your answer would be...
-(x)1/2cos(2x) + 1/4sin(2x) + c
easy right? yeah. just wait til you get to the hard ones. lol OH & there is chasing the rabbit. this happens whenever you integrate something using by parts twice, and you end up with the same thing you started with in the original problem. then you simply set it equal to the original and solve it like that.
Another aspect of calculus bc that we reviewed this week, which i'll review briefly, is l'hopitals rule.
this is whenever you try to find a limit of something, but it's in indeterminate form... then you have to use l'hopitals rule.
which means you take the derivative of the top of the fraction and the derivative of the bottom of the fraction... SEPARATELY! no quotient rule. then plug in & solve.
but, you must make sure that it is in fraction form.. because you cannot use l'hopitals rule if it is not.
alright, well that's all for now.
Abbey's first BC blog!
Well first 2 week of school done!
One thing I'm comfortable with is L'Hopital's Rule. It is used when an indeterminate form occurs with a limit. It could be in any form of: infinity-infinity, 0/0, infinity/infinity, 0(infinity), 0^0, 1^infinity, and infinity^0.
When you get an indeterminate form, you take the derivative top then derivative of bottom. If you still get an indeterminate form just repeat the second step.
EXAMPLES:
lim e^2x - 1/x = e^2(0) - 1/1 = 1-1/0 = o/o <---indeterminate form
x-->0
So, take derivative of top then bottom.
e^2x - 1/x = 2e^2x/1 = 2(1)/1 = 2
lim lnx/x = infinity/infinity <---indeterminate form
x-->infinity
1/x/1 = 1/x = 0
With this example you would use your limit rules because it is as x approaches infinity. Since the degree of bottom is larger than the degree of top it equal zero.
Another thing we have covered is substitution. I can usually pick out my u and du, but sometimes I have trouble completing the problem. I guess I just need more practice.
EXAMPLE:
S tsint^2dt
u=t^2 du=2t
1/2 S sinudu
-1/2cost^2 +C
By parts is also something I can usually get my u, du, dv, and v. I just sometimes have a hard time finishing the problem. The formula is: uv - Svdu. Remember that the u is something that you want to reduce and dv can be the dx.
EXAMPLE:
S xe^x
u=x v=e^x
du=1 dv=e^xdx
xe^x- S e^xdx
xe^x-e^x +C
What I need help on! The whole sin, cos, secant, tan stuff with the formulas. I really didn't get the homework...
How would you work this?
S sec^4 5xdx
One thing I'm comfortable with is L'Hopital's Rule. It is used when an indeterminate form occurs with a limit. It could be in any form of: infinity-infinity, 0/0, infinity/infinity, 0(infinity), 0^0, 1^infinity, and infinity^0.
When you get an indeterminate form, you take the derivative top then derivative of bottom. If you still get an indeterminate form just repeat the second step.
EXAMPLES:
lim e^2x - 1/x = e^2(0) - 1/1 = 1-1/0 = o/o <---indeterminate form
x-->0
So, take derivative of top then bottom.
e^2x - 1/x = 2e^2x/1 = 2(1)/1 = 2
lim lnx/x = infinity/infinity <---indeterminate form
x-->infinity
1/x/1 = 1/x = 0
With this example you would use your limit rules because it is as x approaches infinity. Since the degree of bottom is larger than the degree of top it equal zero.
Another thing we have covered is substitution. I can usually pick out my u and du, but sometimes I have trouble completing the problem. I guess I just need more practice.
EXAMPLE:
S tsint^2dt
u=t^2 du=2t
1/2 S sinudu
-1/2cost^2 +C
By parts is also something I can usually get my u, du, dv, and v. I just sometimes have a hard time finishing the problem. The formula is: uv - Svdu. Remember that the u is something that you want to reduce and dv can be the dx.
EXAMPLE:
S xe^x
u=x v=e^x
du=1 dv=e^xdx
xe^x- S e^xdx
xe^x-e^x +C
What I need help on! The whole sin, cos, secant, tan stuff with the formulas. I really didn't get the homework...
How would you work this?
S sec^4 5xdx
Mal's First BC Post...
So, I've got to get back into the routine of doing these things. Luckily my dad's got my back and remembered for me.
Calculus. Right. Here I go...
A brief overview of basic integration (a few tips):
1. Remember that normally your u is whatever is being raised to a power or under the square root or the bottom of a fraction, etc. For example:
∫x/√(x^2+1)
u=x^2+1
du=2x
Now balance the 2 by putting a ½ in front and you have:
1/2∫du/(√u)
=√(x+1) + C
2. You have to be able to recognize basic derivatives (i.e. trig ones) So:
∫secx tanx
We know that the derivative of sec x is sec tan, so obviously the integral equals:
secx+C
3. Also remember that if you see 1/x and your integrating..that’s natural log integration.
4. Okay. By parts. You must remember that the formula is as follows:
uv- ∫(v)du
That, my friends is a crucial part. Also remember that your u is usually that which will decrease more. So if you have x and x^3, your u would be x and dv x^3. Got it? Good. HOWEVER, if you have a ln in the equation, that needs to be your u because you cannot integrate a ln…It’s just not conducive.
5. Next subject: Trig Substitution. I know we had some homework over the weekend, and for a lot of it, I was unsure. I did find out, I believe, that the integral of sec x is always going to be:
∫secx = ln(secx + tanx) + C
Don’t ask me why, that’s just what I found. You have to memorize it I believe? Now my main problem is trying to figure out when to substitute in certain trig identities. Does anyone know a way to remember it ? In desperate need of help if I’m going to pass that quiz on Wednesday. Muchas Gracias!
Calculus. Right. Here I go...
A brief overview of basic integration (a few tips):
1. Remember that normally your u is whatever is being raised to a power or under the square root or the bottom of a fraction, etc. For example:
∫x/√(x^2+1)
u=x^2+1
du=2x
Now balance the 2 by putting a ½ in front and you have:
1/2∫du/(√u)
=√(x+1) + C
2. You have to be able to recognize basic derivatives (i.e. trig ones) So:
∫secx tanx
We know that the derivative of sec x is sec tan, so obviously the integral equals:
secx+C
3. Also remember that if you see 1/x and your integrating..that’s natural log integration.
4. Okay. By parts. You must remember that the formula is as follows:
uv- ∫(v)du
That, my friends is a crucial part. Also remember that your u is usually that which will decrease more. So if you have x and x^3, your u would be x and dv x^3. Got it? Good. HOWEVER, if you have a ln in the equation, that needs to be your u because you cannot integrate a ln…It’s just not conducive.
5. Next subject: Trig Substitution. I know we had some homework over the weekend, and for a lot of it, I was unsure. I did find out, I believe, that the integral of sec x is always going to be:
∫secx = ln(secx + tanx) + C
Don’t ask me why, that’s just what I found. You have to memorize it I believe? Now my main problem is trying to figure out when to substitute in certain trig identities. Does anyone know a way to remember it ? In desperate need of help if I’m going to pass that quiz on Wednesday. Muchas Gracias!
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