Integrals giving me a hard time

[Rectifier]

Hey!
It is the first time I post on this subform. Please forgive me if I do something wrong.

Homework Statement

F(x)=\int^x_0f(t)dt
for
R \in t \rightarrow f(t)

Homework Equations

Is it true that

0 \leq f(x)\leq 3 for 0<x<1

\int^x_0 tf(t)dt \leq x^2

for all x \in (0, 1)? "

The Attempt at a Solution

\int^x_0 tf(t)dt \leq x^2 \Leftrightarrow t \int^x_0 f(t)dt \leq x^2 \Leftrightarrow t F(x) \leq x^2
I don't know how to continue from here.

 

[kontejnjer]

If I'm understanding this correctly, you could start by writing x^{2} as \int^{x}_{0}2tdt and then comparing this to the original integral. Will the inequality be valid for any value of f(x) on x\in (0,1)?

 

[Rectifier]

If I'm understanding this correctly, you could start by writing x^{2} as \int^{x}_{0}2tdt and then comparing this to the original integral. Will the inequality be valid for any value of f(x) on x\in (0,1)?

Thank you for your reply!

\int^x_0 tf(t) dt \leq \int^x_0 2t dt

Should I insert x=1(I can't insert x=0 since integrals must have bounds that are not equal)? What should I do with the f(t)?

EDIT:
Sorry, I just started this chapter and I am not so sure what to do.

 

[kontejnjer]

You could rewrite the inequality as

\int^{x}_{0}[f(t)-2]tdt\leq0

Now, since x \in (0,1), the t will clearly not be negative on the interval, meaning f(t)-2 will either oscillate between positive and negative values, or will just be negative (since their product must be negative on at least one subset of (0,1) for the whole integral to be negative or zero). Now, since f(t) can be ANY function (as long as f(t)\in(0,3)), you can sometimes pick it to be the simplest one, in this case a constant. Can you then choose a constant f to simultaneously satisfy the above inequality and the initial assumption?

 

[Rectifier]

Thank you for your reply once again!

I could pick f(t)=2 that will satisfy the inequality. But if I choose f(t)=3 it will mean that the inequality is wrong since the integral will be postitive. I guess that means that the initial inequality is wrong.

Is that right?

 

[Mark44]

Hey!
It is the first time I post on this subform. Please forgive me if I do something wrong.

Homework Statement

F(x)=\int^x_0f(t)dt
for
R \in t \rightarrow f(t)

I don't understand what the above is supposed to mean. Can you explain in words what you're trying to convey here?

Homework Equations

Is it true that

0 \leq f(x)\leq 3 for 0<x<1

\int^x_0 tf(t)dt \leq x^2

for all x \in (0, 1)? "

The wording here is somewhat confusing, IMO. What part is given (or we are supposed to assume) and what part are we to show? IOW, are we supposed to verify that 0 ≤ f(x) ≤ 3 when 0 < x < 1?

And then show that the inequality is true?

The Attempt at a Solution

\int^x_0 tf(t)dt \leq x^2 \Leftrightarrow t \int^x_0 f(t)dt \leq x^2 \Leftrightarrow t F(x) \leq x^2
I don't know how to continue from here.

You can't pull t outside the integral like you have done above.

You are giv

 

[kontejnjer]

Thank you for your reply once again!

I could pick f(t)=2 that will satisfy the inequality. But if I choose f(t)=3 it will mean that the inequality is wrong since the integral will be postitive. I guess that means that the initial inequality is wrong.

Is that right?

Yup, you found a counterexample, so the initial theorem doesn't hold, that's all there is to it.

 

[Rectifier]

I don't understand what the above is supposed to mean. Can you explain in words what you're trying to convey here?

The wording here is somewhat confusing, IMO. What part is given (or we are supposed to assume) and what part are we to show? IOW, are we supposed to verify that 0 ≤ f(x) ≤ 3 when 0 < x < 1?

And then show that the inequality is true?

You can't pull t outside the integral like you have done above.

You are giv

Thank you for your post.
I am having a hard time translating the problem to English from Swedish. We were supposed to verify if the inequality is true having (0 ≤ f(x) ≤ 3 when 0 < x < 1) as our starting conditions.

 

[Rectifier]

Yup, you found a counterexample, so the initial theorem doesn't hold, that's all there is to it.

Thank you :)