Is the integral of $x^n$ unbounded for $n<0$, over the finite interval $(0,1)$?

Question

I have solved the question above and found that the Option C is correct. But I am unable to understand 2nd option. Can you help me with it?

My Attempt at solving the question:-

$$ I_n= \int_0^1 x^n dx $$ $$ I_n = \frac{1^{n+1} - 0^{n+1}}{n+1} $$ Option A: For $n = -1$ the integral does not exists.

Option C: For $n = 1,2,3... m$, the integral becomes $ \frac{1}{n+1}$ and consequently the product $I_1 I_2.... I_m$ becomes $\frac{1}{(m+1)!}$

Option D: We can see from above that $I_m ...... <I_3< I_2 < I_1$

Can you please help for Option B? Thank you!

Depends on the author what "finite" refers to. It could be that "The term finite .. is stressing the fact that it is 'defined', 'well defined' or 'exists as a real number'". Or, "finite on a set sometimes meant bounded on the set.". See What does “finite but unbounded” mean?. — Its_me, Sep 04 '20 at 16:01
@Its_me okay so i just saw that and thought about it and i believe the option B it to be correct since for n<0, the integrals are going to ln(x) , -1/x, -1/2x^2 and so on... and all these are not bounded but are finite in the interval (0,1). So am I right in my reasoning? — Asad Ahmad, Sep 04 '20 at 18:13
Have you noticed that B talks about the integrands and not the integral. — Its_me, Sep 04 '20 at 18:51

score 1 · Answer 1 · answered Sep 04 '20 at 23:38

Its_me is correct: The B question talks about the integrand, not the integral. The integrand is the function that is being integrated, in this case, $f(x) = x^n$ with $n < 0$. By unbounded, it means

For every $M > 0$, there is some $x \in (0,1)$ such that $|f(x)| > M$.

By finite, they just mean

For every $x \in (0,1), |f(x)| < \infty$, or more appropriately, that $f(x)$ is defined for every $x \in (0,1)$.

I trust you can show both, though if you haven't encountered a full definition of $x^n$ for irrational $n$ yet, you may have to skip those details you've never been taught.

Is the integral of $x^n$ unbounded for $n<0$, over the finite interval $(0,1)$?

1 Answers1