Question

int from 1 to infinity of (1/ sqrt(x+4))

Answer 1

\[\int\limits_{1}^{\infty} 1/(\sqrt[?]{x+4})\]

Answer 2

try \(u=x+4\)

Answer 3

\[\int\limits_{1}^{t} 1/(\sqrt{x+4}) dx\]

Answer 4

oops forgot lim as t approaches infinity

Answer 5

\[\int\limits_{1}^{\infty}du/\sqrt{u}\]

Answer 6

like @satellite73 said

Answer 7

you get
\[2\sqrt{x+4}|_1^t\] now fairly clearly this will not converge as \(t\to\infty\)

Answer 8

with a little experience you will be able to say "no way" before even starting. if you want to integrate to infinity, the power in the denominator has to be bigger than one, and you only have a power of \(\frac{1}{2}\)

Answer 9

\[\lim t \rightarrow \infty \int\limits_{1}^{t} (1/\sqrt{u}) du\]

Answer 10

sorry i'm reading the comments now

Answer 11

\[\lim_{t\to \infty}\sqrt{t+4}\] is not finite

Answer 12

why doesn't it converge when 2 sqrt(1+4) = 2 sqrt(5)

Answer 13

oh you can plug in1 alright, that is no problem, one is a number

Answer 14

but you also have to plug in \(t\) and then take the limit as \(t\to\infty\)

Answer 15

that is infinite, so no integral

Answer 16

which is an infinitly big number...true

Answer 17

so I should be able to look at the equation and see that the denominator becomes infinitely big and assume it doesn't converge?

Answer 18

one more question plz

Answer 19

so the equation \[\sum_{1}^{\infty} 1/n^5\]
convergest because the x in the numerator?

Answer 20

converges i mean

Answer 21

\[\int\limits_{1}^{\infty} 1/x^5 dx\]

Answer 22

converges because \(5>1\)