- #1
Dell
- 590
- 0
given the improper integral from 0 to 1 of
[tex]\int[/tex]dx/[tex]\sqrt[3]{x}[/tex](ee-e-x)
i am asked if it comverges or diverges,
what i have learned is that if i can find :
-a similar function that is bigger than my function and that i know converges, then my function also converges
-a similar function that is smaller than my function and that i know dicerges, then my function also diverges
-a function g(x) such that
lim f(x)/g(x)=K, (K not 0, not infinity) then f(x) acts like g(x)
x->inf
my similar function g(x) is 1/3[tex]\sqrt[3]{x}[/tex] which is smaller than f(x), BUT my g(x) converges, so that doesn't prove anything according to the 1st 2 conditions,
generally can i use the 3rd condition even though my integral is from 0 to 1 and not to infinity??
in this case that too doesn't work. what can i do from here??
if not how would you solve this?
[tex]\int[/tex]dx/[tex]\sqrt[3]{x}[/tex](ee-e-x)
i am asked if it comverges or diverges,
what i have learned is that if i can find :
-a similar function that is bigger than my function and that i know converges, then my function also converges
-a similar function that is smaller than my function and that i know dicerges, then my function also diverges
-a function g(x) such that
lim f(x)/g(x)=K, (K not 0, not infinity) then f(x) acts like g(x)
x->inf
my similar function g(x) is 1/3[tex]\sqrt[3]{x}[/tex] which is smaller than f(x), BUT my g(x) converges, so that doesn't prove anything according to the 1st 2 conditions,
generally can i use the 3rd condition even though my integral is from 0 to 1 and not to infinity??
in this case that too doesn't work. what can i do from here??
if not how would you solve this?