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

Step-by-step explanation:

Consider the integral

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

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

Case 3: p >1

will have x with negative exponent. So when x = infinity this value becomes 0 thus making the integral to converge

x-3y=4

step-by-step e

c

step-by-step explanation: