Prove or disprove a statement about the integral.

Now we're gonna see that we can use some properties of the integral To prove that A statement about the integral Needs to be verified. So we know that the integral is linear. We now also that we can multiply the integral by a constant and we can take it outside. We're now there, how to do the substitution how to do integration by part and how to change the cosine, the sign and tangent in dealing with substitution. We also have seen Weirstrauss substitution.
So now we have to approve, a statement is true or false Which approved that statement is false. We can find one single example that with violate the truthness of that statement In our case, we want to find one function. That would tell us that the statement here is false, so to disprove this one order, if you did, we can find one simple function. And in this case, the simple function, the simplest function that we can find is f of x. Equals one. In this case, we don't have the truth. Net of the statement. Okay, that's when we want to prove that statement is false. We can simply find one example. In that case, we call that example a counter example, that will refuse the statement that we want to prove, Okay? So to find that we simply find one simple function that doesn't satisfy the, the statement that we want to have. And our example, we're gonna have F of x equals one. And in that case we're gonna plug it in and see that it doesn't satisfy the statement or the integral that we are Computing. And therefore, since we found one example, we can
say that the statement is false for all the others
#maths #education #probability #python

[@f.herumusu8341]

I would say: Disproved by simply taking a look at it. That statement obviously wrong.

[@Archimedes_Notes]

@@f.herumusu8341 That was easy free 10 points. Some people got it wrong.

[@f.herumusu8341]

@@Archimedes_Notes I would rather say: If you just pull an x before the integral you have a serious problem understanding calculus. That's a bad mistake. Even worse if you think that that's a theorem.

[@eliaperli2485]

The statement actually can't be neither false nor true because it's not a well formed formula

[@luciangv3252]

Can u derivate both sides?
D(I {xf(x)}) = D(x*I{f(x)})
xf(x) = I{f(x)} + x*f(x)
I{f(x)} = 0

[@Archimedes_Notes]

@@luciangv3252 You msy do that. Assume that f is not zero. On one side you will get 0 and on tbe other you will get integral of f dx between a and b whch is not zero (a not = b) is equal to zero .This is a contradiction

[@BedrockBlocker]

This doesn't work, you don't know if it's even differentiable.

[@Archimedes_Notes]

@@BedrockBlocker I did write to you instead of somebody else. The inyegral is differentiable. You will have to prove it in 2 lines.

[@BedrockBlocker]

There is a problem with free and bound variables here, no? The x before the integral has no meaning.

[@Archimedes_Notes]

@@BedrockBlocker The question is :
Is that statement true or false? That is what we want. If you think thst is wrong, you can correct it.

[@kappasphere]

I don't think this is a good problem, because it uses x in two different ways to purposely make it harder to read than it needs to be. A maths problem should be difficult because of its contents, not because it uses bad notation.
The problem becomes a lot clearer when you keep the variable in the integral as x, but name the global variable as c. This gives:
int_a^b x f(x) dx = c int_a^b f(x) dx
But as a general equation, this is so obviously wrong that it becomes a boring problem to even think about.
What I think is more interesting instead is to look for choices of f for which, given arbitrary a,b,c, the equation becomes true.
For this, use the linearity of the integration operator and get:
int_a^b (x-c) f(x) dx = 0
Now use the power series approach f(x)=sum_{k=0}^infinity v_k x^k, and insert into the equation:
int_a^b (x-c) sum_{k=0}^infinity v_k x^k = 0
int_a^b sum_{k=0}^infinity (c v_k x^k - v_k x^{k+1}) = 0
int_a^b sum_{k=0}^infinity c v_k x^k - sum_{k=1}^infinity v_{k-1} x^k = 0
int_a^b sum_{k=0}^infinity (c v_k - v_{k-1}) x^k = 0
[sum_{k=0}^infinity (c v_k - v_{k-1}) x^{k+1}/(k+1)]_a^b = 0
sum_{k=0}^infinity (c v_k - v_{k-1}) (b^{k+1}-a^{k+1})/(k+1) = 0
What's notable about this result is that depending on how you choose v_k, you can make each component of this sum equal to any arbitrary number you want. So as long as you have a series that sums to 0, say sum_{k=0}^infinity d_k = 0, you can get a choice of v_k that leads to the total sum being 0 by equating the earlier series' individual summands with d_k.
You get:
(c v_k - v_{k-1}) (b^{k+1}-a^{k+1})/(k+1) = d_k
c v_k - v_{k-1} = d_k (k+1)/(b^{k+1}-a^{k+1})
c v_k = v_{k-1} + d_k (k+1)/(b^{k+1}-a^{k+1})
v_k = (v_{k-1} + d_k (k+1)/(b^{k+1}-a^{k+1}))/c
which fully defines v_k as a recursive relationship, together with v_{k-1}=0.
A simple choice of d_k would be that of d_0=1, d_1=-1, d_k=0 otherwise. The sum of this is clearly 0. The recursion gives:
v_0 = (0 + 1 (0+1)/(b-a))/c = 1/(c(b-a))
v_1 = (1/(c(b-a)) - 2/(b²-a²))/c = 1/(c²(b-a)) - 2/(c(b²-a²))
and after that:
v_k = v_{k-1}/c + 0
So every increase in k does nothing but add another factor of 1/c.
This gives v_{1+i}=v_1 * 1/c^i, implying:
v_k = (1/(c²(b-a)) - 2/(c(b²-a²)))/c^(k-1)
= (1/(c(b-a)) - 2/(b²-a²))/c^k
So, going back to f(x)=sum_{k=0}^infinity v_k x^k, inserting the known values for v_k, you get:
f(x)=1/(c(b-a)) + (1/(c(b-a)) - 2/(b²-a²)) sum_{k=1}^infinity (x/c)^k
which is clearly a geometric series with convergence radius c, so this example is only guaranteed for |a|,|b|

[@zinzhao8231]

f = 1 disproves it

[@Archimedes_Notes]

That problem was asked by professor Stewart to his class many times. It was a free 10 points. If you need the complete reference, you may contact me by email. There is nothing wrong with the notation. Hard problems are coming. I am waiting for a good camera and decent microphone. Space engineering and astronomy problems will be here soon.Thank you.