Question

Can someone just help me confirm these?
I'll do the question in the next part

Answer 1

What is it

Answer 2

Assuming that \[\int\limits_{1}^{\infty}g(x)dx\] converges
a. What can you say about the convergence of \[\int\limits_{1}^{\infty}f(x)dx\]?
It converges/diverges/can't say anything
b. What can you say about the convergence of \[\int\limits_{1}^{\infty}h(x)dx\]?
It converges/diverges/can't say anything

Answer 3

Omg i have no idea :/ i thought this would be an easier question im only in sixthgrade

Answer 4

good luck when you get to calculus :)

Answer 5

Lol thanks :) not looking forward to it :P

Answer 6

Note that \(h(x)\le g(x) \le f(x)\), hence, taking the integral on all sides:
\[
\int h<\int g<\int f
\]Where all operators and functions are defined on \(x\).
Hence, the convergence of \(\int f\) on \([a, \infty]\) implies the convergence of \(\int g\) on the same interval and that of \(\int h\). But, the converse is *not* necessarily true.

What did you answer for these?

Answer 7

(Replace the \(<\) with \(\le\) ), sorry, mate.

Answer 8

I put can't say anything for a and converges for b

Answer 9

That is correct.

Answer 10

ok so then the next part is Assuming that \[\int\limits_{1}^{\infty}g(x)dx\] diverges

Answer 11

All right, so, we know that that is the case, now what's the contrapositive of the statement?

Answer 12

I'm sorry I don't understand what you're asking me

Answer 13

Contrapositive means that, if \(P\to Q \iff \lnot Q\to \lnot P\).
So, in our case, we know that if \(\int f\) converges, then \(\int g\) converges.

So its contrapositive is:
If \(\int g\) does not converge (it diverges), then \(\int f\) cannot converge (it diverges). But the same is not necessarily true for \(\int h\).

Answer 14

so for f(x) it would be divergent and h(x) would be can't say?