Mathematics: Show that ∫^[infinity]_1 1/x^p dx converges if 1 and diverges if p≤1.

Show that ∫^[infinity]_1 1/x^p dx converges if 1 and

hairystrong1915

15.01.2022

Step-by-step explanation:

Consider the integral

$\int\limits^1_{-infty} \frac{1}{x^p} \, dx$

for various values of p

CaseI:p =1 Then integral is ln x and for infinity this value diverges so diverges

Case 2: p<1

The integrated value would be $\frac{x^{-p+1} }{-p+1}$

Since P <1 numerator x will have positive exponent and when infinity is substituted this will diverge

Case 3: p >1

$\frac{x^{-p+1} }{-p+1}$ will have x with negative exponent. So when x = infinity this value becomes 0 thus making the integral to converge

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yol123

17.04.2022

The answer is 2

Step-by-step explanation:

because the slope, when these 2 are graphed, is 6/3, which simplifies to 2/1, and then to 2

hope this helps

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