(a) The improper integral ∫1^∞(1)/(x^p) dx converges if and...

(a) The improper integral $\int_{1}^{\infty} \frac{1}{x^{p}} dx$ converges if and only if _____. Select the correct answer. (1) $p > 0$ (2) $p > 1$ (3) $p \ge 1$ (4) $p < 0$ (5) $p < 1$ (6) $p \le 1$ (b) The improper integral $\int_{0}^{1} \frac{1}{x^{p}} dx$ converges if and only if _____. Select the correct answer. (1) $p > 0$ (2) $p > 1$ (3) $p \ge 1$ (4) $p < 0$ (5) $p < 1$ (6) $p \le 1$

Transcript

00:01 So in this question, we want to know what are all the values of p for which the improper integral converges.

00:07 Now, i claim that the answer is going to end up being p is greater than 1, but let's see why that is the case.

00:16 So if i have an improper integral like this, the first step is to rewrite it as a limit.

00:24 So i'm going to say that i have the limit as b approaches infinity, the integral from 1 to b of 5 over x to the p power d x.

00:40 Now, in order to find my anti -derivative this time, what am i going to do? well, i'm going to rewrite 5 over x to the p.

00:50 I'm going to write this as 5x to the negative p power d x.

01:00 Now i'm ready to find my anti -derivative.

01:03 How do i get my anti -derivative? well, i'm going to add one to the exponent and divide by the new exponent.

01:12 So 5x to the power of negative p plus 1.

01:18 And now i'm dividing by that new exponent negative p plus 1.

01:25 And this is being evaluated on the interval from 1 to b.

01:31 Now, 5 is a constant as is negative p plus 1.

01:37 And so what i can do is i can slide those through my limit symbol, just to clean this up a little bit.

01:44 I can say this is 5 over negative p plus 1 times the limit as b approaches infinity of x to the negative p plus 1 power being evaluated on the interval from 1 to b.

02:06 That's 5 over negative p plus 1 times the limit...

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