We meet again bloggers.
Well it is time for another blog. I always feel like I'm talking about the same thing over and over again. You think it would help me remember it but sadly no. Tomorrow, I think, we take another ap test. I'm so not looking foward to it because of my past scores. I am trying though..I wish they gave points for that in multiple choice. Okay now I should probably talk about the stuff I learned this week:
We have been talking about key words lately.
key word stuff:
1. linearization-->equation of a tangent line
2. if given a table(the sets of points)-->dealing with slope or rram/lram/ect.
3. how many-->integrate
4.(in calculator part)volume problems-->find intersection
We have learned some new formulas.
some formula stuff:
1. Volume of a sphere: V=4/3pi(r^3)
2. dealing with inverses: g^1(y)=1/f^1(x)
Example: Let f(x)=x^5+1 and let g be the inverse function of f. What is the value of g^1(0)?
g^1(y)=1/f^1(x)
Solve for x: x^5+1=0
x^5=-1
x=-1
plug in x to problem: 5(-1)^4=5
take reciprocal: 1/5
3. words like bacteria, birth rate, contaminates, you will probably use the exponential growth formula: A(subscript 0)e^kt
For what I don't know:
Substitution, once again. I think I need an example problem shown. Any takers?
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So, integrate 2x/(x^2+9)^2
ReplyDeleteSo you see that the bottom is being squared..which is your cue to plug in u for whatever's being raised.
u = x^2 + 9
du = 2x
Now you're going to substitute those back in::::you have.
S du/ u^2
So technically you can separate this and make negative exponents...
S u^-2 du
Integrate Normal!!!
-1/u
Before we said u = x^2 +9 so...SUBSTITUTE
-1/(x^2 + 9)
Then you'd just plug in your given bounds....and that's it.
Go look at the video...it really does help..
Let's try this one:
ReplyDeleteintegrate from pie/2 to pie/4 sin^3 cos dx.
your u=sinx du=cosxdx
So you have u^3 du.. so integrate like normal which equals
1/4u^4.. now plug sin back in which equals 1/4sin^4x.
Now plug in the bounds:
1/4(sin pie/2)^4- 1/4(sin pie/4)^4
1/4(1)^4-1/4(the square root of 2/2)^4
Your answer is 3/16.. hope this helps.
I know you're a visual and step-by-step person, so if you still don't understand it, I can help at school...I think it's a lot easier to see it on a notebook than a computer
ReplyDeleteA simpiler example would be S cosx / sinx.
ReplyDeleteOk, so since you can't integrate fractions, we have to use substitution.
u = sinx
du = cosx
So you get S 1/u * du
And when you integrate that you get
ln|u| = ln |sinx|
hope it helps a little
You use substitution when you are multiplying or dividing something in an integral that you can't break apart or bring up. You can't use product or quotient rule in in integral, so substitution is a way of forcing a problem to work. The non-derivative in the integral is u and the derivative of u is du. You then integrate udu + c if it is indefinite, and solve.
ReplyDeleteSubstitution is strictly integrals with a product or a quotient in them. You cannot use product rule or quotient rule like in derivatives, so substitution is the way to go. Find the non derivative and make that u. du is the derivative of u. Integrate udu or and remember plus c if the problem is an indefinite integral.
ReplyDelete