« Reply #902 on: September 24, 2018, 05:11:03 PM »

For the first one start by writing 9x^4 as (3x^2)^2. ∫x/((3x^2)^2+4)dx Now let u=3x^2, the du=6x. 1/6∫1/(u^2+4)du Factor 1/4 out of the denominator. 1/6∫1/((1/4)u^2+1)du Rewrite (1/4)u^2 as ((1/2)u)^2 1/6∫1/(((1/2)u)^2+1)du Now use the substitution v=(1/2)u, then dv=1/2du. 1/12∫1/(v^2+1)dv Now integrate. (1/12)tan^-1(v)+C=(1/12)tan^-1((1/2)u)+C=(1/12)tan^-1(1/2(3x^2))+C

For the second integral we will start by completing the square of the quadratic under the square root, to do this we start by factoring the minus sign. ∫1/sqrt(-(x^2-6x)dx Rewrite the term in the parentheses as x^2-6x+9, since this would be multiplied by negative 1 we'll have to add 9 outside the parentheses. ∫1/sqrt(-(x^2-6x+9)+9)dx We now use the fact that x^2-6x+9=(x-3)^2 to rewrite the integrand as ∫1/sqrt(-(x-3)^2+9)dx. Now use the substitution u=x-3, then du=dx. ∫1/sqrt(9-u^2)du Now use the substitution u=3sinθ, then du=3cosθdθ. Applying this substitution we have ∫3cosθ/sqrt(9-(3sinθ)^2)du=∫3cosθ/sqrt(9-9sin^2θ)dθ=∫3cosθ/sqrt(9(1-sin^2θ))dθ Apply the identity sin^2θ+cos^2θ=1, and simplify. ∫3cosθ/sqrt(9cos^2θ)dθ=∫3cosθ/3cosθdθ=∫dθ Lastly, integrate and back substitute. θ+C=sin^-1(u/3)+C=sin^-1((x-3)/3)+C

« Reply #903 on: September 24, 2018, 05:17:15 PM »

This is for one of my young neighbors. They told me they needed so much help with this, so I decided to post it here. They don't know what the problem is looking for and neither do I. I don't know what this is all about, but whatever. It's seems very abstract though.

Recall that lim f(x)=L x->c means: For all ϵ>0 there is a δ>0 such that for all x satisfying 0<|x−c|<δ we have that |f(x)−L|<ϵ. What if the limit does not equal L? Think about what the means in ϵ,δ language. Consider the following phrases: 1. ϵ>0 2. δ>0 3. 0<|x−c|<δ 4. |f(x)−L|>ϵ 5. but 6. such that for all 7. there is some 8. there is some x such that

Order these statements so that they form a rigorous assertion that: lim f(x) =/= L. x->c

« Reply #904 on: September 24, 2018, 05:23:10 PM »

That's the formal definition of a limit. It's used to formally prove the value of a limit. It is more likely to be seen in a Real Analysis class than in a calculus class.

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That's the formal definition of a limit. It's used to formally prove the value of a limit. It is more likely to be seen in a Real Analysis class than in a calculus class.

So, what would be the answer anyways? Now, I got another question from them and I don't know why they're asking me and not someone else?

The function is (sin(3x))/(x) and the table asked them to find the output of these values: -0.1, -0.01, -0.001, -0.0001, 0.0001, 0.001, 0.01, 0.1. They already figured these out.

However, they want to know this: (c) Graph the function to see if it is consistent with your answers to parts (a) and (b). By graphing, find an interval for x near zero such that the difference between your conjectured limit and the value of the function is less than 0.01. In other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom. What is the window?

« Reply #907 on: September 24, 2018, 06:39:03 PM »

In mathematics you normally encounter two types of problems. The first type is given certain information find another piece of information. This information can be any kind of mathematical object (numbers, functions, sets, vectors, etc.). This is the kind of problem you would see in grade school. The second type of problem, and the type of problem you are asking, is given a fact, prove or disprove the fact. Any logical argument would be a suitable answer for this type of problem. Some facts you might be asked to prove are: If x is odd then x^2 is odd, sqrt2 is irrational, etc. All of these facts are trivial though their proof (namely the proof for the second fact) may not be trivial. A fact you might be asked to disprove is there exists an even prime number greater than 2. Again this fact is trivial, in fact you may already intuitively understand why it is false.

The formal definition of a limit can be intuitively understood using rectangles. Given any width you can center a rectangle of the given width centered at the limit point (the (x,y) coordinate the limit approaches) with a height of twice the functions value. The formal definition of a limit says such a rectangle will always exist if the limit exists.

Now use the substitution u=3sinθ, then du=3cosθdθ. Applying this substitution we have ∫3cosθ/sqrt(9-(3sinθ)^2)du=∫3cosθ/sqrt(9-9sin^2θ)dθ=∫3cosθ/sqrt(9(1-sin^2θ))dθ Apply the identity sin^2θ+cos^2θ=1, and simplify. ∫3cosθ/sqrt(9cos^2θ)dθ=∫3cosθ/3cosθdθ=∫dθ Lastly, integrate and back substitute. θ+C=sin^-1(u/3)+C=sin^-1((x-3)/3)+C

He's been saying this for a week just because his teacher asked a student to pretend to be their spouse. Not very professional but it doesn't mean he's a pedo.

« Reply #913 on: September 25, 2018, 04:38:19 AM »

If he got accepted, then it's clearly obvious. It doesn't matter when he gets accepted. Not every school offers Early Action/Early Decision or some other dumb stuff, just a regular decision, for most of the part. I don't know how his school works, so I can't say anything else.