HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

[braintic]

Re: MX2 2015 Integration Marathon

Is that necessary when your doing an exam?

Is what necessary?

 

[hypermax]

Re: MX2 2015 Integration Marathon

A little easy compared to the standard set here

A question fit for 3unit

 

[integral95]

Re: MX2 2015 Integration Marathon

 

[hypermax]

Re: MX2 2015 Integration Marathon

is it

 

[integral95]

Re: MX2 2015 Integration Marathon

is it

nope lol

Hint: use a substitution

 

[braintic]

Re: MX2 2015 Integration Marathon

Let the Integrand be f(x).

Squaring both sides:
f²(x) = x + f(x)
f²(x) - f(x) - x = 0

Using the quadratic formula:
f(x) = [1+√(1+4x)]/2 (ignoring the -ve since the integrand is clearly +ve)

Integrating:
x/2 + [ (1+4x)^(3/2) ] / 12

 

[FrankXie]

Re: MX2 2015 Integration Marathon

interesting question, when x=0, f(0)=(1+\sqrt{1+4(0)})/2=1?

 

[braintic]

Re: MX2 2015 Integration Marathon

interesting question, when x=0, f(0)=(1+\sqrt{1+4(0)})/2=1?

Clearly there is something happening there in the limit, because there is no way the answer can't be zero.
But fiik what that something is.
Perhaps one of the the uni maths freaks can suggest what is happening.

Also, it gives a value for values between -1/4 and 0.
My mind is spinning, wondering how subbing -1/4 could give an answer of +1/2.

 

[InteGrand]

Re: MX2 2015 Integration Marathon

Clearly there is something happening there in the limit, because there is no way the answer can't be zero.
But
fiik what that something is.
Perhaps one of the the uni maths freaks can suggest what is happening.

Also, it gives a value for values between -1/4 and 0.
My mind is spinning, wondering how subbing -1/4 could give an answer of +1/2.

Here was my attempt at the Q:

Assume . Assume that is real, so is only defined for x that makes the expression under the radical non-negative and real. This is any non-negative x. Also, assume the dx isn't inside the radical!

Let .

Then

So (by the quadratic formula).

i.e. .
For and , and . So we cannot take this solution for x > 0, so .

Now,



.

 

[InteGrand]

Re: MX2 2015 Integration Marathon

So for positive x, the function is always greater than 1/2 I think, but at x = 0, we need the solution with the negative radical for c?? (Since c is surely 0 when x = 0)

 

[braintic]

Re: MX2 2015 Integration Marathon

.

The issue at the moment is not with the answer (we had it), but with how to interpret the answer for x = 0 and for -1/4 < x < 0.

 

[InteGrand]

Re: MX2 2015 Integration Marathon

Well for negative x, the function is not real-valued, and the HSC Q's usually only deal with real-valued functions for integration?

 

[InteGrand]

Re: MX2 2015 Integration Marathon

Also, I haven't considered non-real values of x yet (or other negative values, where it's less than 1/4).

 

[FrankXie]

Re: MX2 2015 Integration Marathon

So for positive x, the function is always greater than 1/2 I think, but at x = 0, we need the solution with the negative radical for c?? (Since c is surely 0 when x = 0)

yes, I already figured it out, when x=0, f(x)=0, when x>0, f(x) is what you derived, and for x<0, f(x) is not defined. besides, for any positive x, we can use monotonic convergence to prove f(x) is well defined.
Interesting question and good dicussion

 

[InteGrand]

Re: MX2 2015 Integration Marathon

For what values of x does that function converge?

 

[InteGrand]

Re: MX2 2015 Integration Marathon

This is out of curiosity (won't be needed for HSC level), but is it possible to prove the function is undefined for negative and non-real x?

 

[InteGrand]

Re: MX2 2015 Integration Marathon

Just experimented a bit on Wolfram trying f(-1), where f(z) is that infinite nested square root thing for complex z.

The more terms under the radical I include, the closer the answer seems to get closer to the quadratic solution to this of .

With three terms: http://www.wolframalpha.com/input/?i=root(-1+root(-1+i))

With five terms: http://www.wolframalpha.com/input/?i=root(-1+root(-1+root(-1+root(-1+root(-1)))))

With seven terms: http://www.wolframalpha.com/input/?i=root(-1+root(-1+root(-1+root(-1+root(-1+root(-1+root(-1)))))))

 

[FrankXie]

Re: MX2 2015 Integration Marathon

This is out of curiosity (won't be needed for HSC level), but is it possible to prove the function is undefined for negative and non-real x?

well, i would say x is automatically treated as nonnegative. for negative and even imaginary x, that involves multi-valued function, which I don't think talking about convergence or divergence is appropriate. like what is \sqrt{i+\sqrt{i}}? there are four values for it! unlike square root of positive real number, the square root is defined to the positive number, but square root of imiginary numbers have two values.

 

[InteGrand]

Re: MX2 2015 Integration Marathon

well, i would say x is automatically treated as nonnegative. for negative and even imaginary x, that involves multi-valued function, which I don't think talking about convergence or divergence is appropriate. like what is \sqrt{i+\sqrt{i}}? there are four values for it! unlike square root of positive real number, the square root is defined to the positive number, but square root of imiginary numbers have two values.

At least one of these may, assuming convergence, be one of the solutions of the quadratic equation?

 

[InteGrand]

Re: MX2 2015 Integration Marathon

Also, isn't the principle square root of a complex number a+bi (a, b real, a not negative if b = 0) defined as the one with positive real part (this definition fails if the complex number is a negative real number).