Calculas 1 problem on converging and diverging?

Question

If 0
Is that true or false? and most importantly whats the reason? explanation?

Answer 1

I'm not sure what the question is. You write "is known to diverge" and then "also converges"

If 0 < g(x) < f(x) then if the integral over [a,b] of f(x) dx converges, then so does the integral of g(x) dx over the same interval.

This is obvious if you think of the integral of a sum of boxes. The f boxes are always bigger than the g boxes, so if the sum of the f boxes is well defined, so is that of the g boxes.

Similarly, if the integral of g(x) dx diverges, so does that of f(x) dx.

But just because I(f(x) dx) diverges, doesn't mean that I(g(x) dx) diverges.

Consider f(x) = 1/x and g(x) = 1/x^2.

For all x > 1, 0 < g(x) < f(x)

The integral of f(x) dx from 1 to X is just ln X. This clearly has no limit as X goes to infinity.

The integral of g(x) is -1/x so the definite integral from 1 to X is (1 - 1/X). As X goes to infinity, this converges very nicely to 1.

Answer 2

Diverges. think of with regard to the habit of the expression for super ok. The numerator behaves like ok^2, the denominator like ok^{5/2} so the fraction behaves like a million/ok^{a million/2}. because of the fact the sequence for a million/ok^p diverges for p<=a million, and right here p=a million/2, your sequence diverges. Thats the tough thank you to think of roughly it, yet its no longer a ideal evidence. you ought to apply a assessment attempt and teach that your expression exceeds C/ok^{a million/2} for some consistent C. As a million/ok^{a million/2} diverges, then so does your sequence.