Question

Determine whether the improper integral diverges or converges

Answer 1

\[\int\limits_{0}^{4}dx/\sqrt{x}\]

Answer 2

its converging i think

Answer 3

2 x^1/2
2(4)^1/2 - 2(0)^1/2

I might be wrong

Answer 4

because as x increases 1/sqrt(X) decreases and so the area bounded is also converges to a limit

Answer 5

0 is not within

Answer 6

you need to compute
\[\lim_{l\rightarrow 0}\int_l^4\frac{dx}{\sqrt{x}}\]

Answer 7

you get
\[2\sqrt{x}\]evaluate at 4, get 4, evaluate at l get
\[2\sqrt{l}\] and then take
\[\lim_{l\rightarrow 0}2\sqrt{l}\] which is pretty clearly 0

Answer 8

@dipaner the problem is at 0, because the integrand is not defined there. that is why it is improper. it is not because as x increases the integrand decreases

Answer 9

\[\int\limits_{l}^{n}\frac{dx}{\sqrt x}=\int\limits_{\sqrt l}^{\sqrt n} 2dy=2(\sqrt n-\sqrt l)\]
\[\int\limits_{l}^{n+1}\frac{dx}{\sqrt x}=\int\limits_{\sqrt l}^{\sqrt {n+1}} 2dy=2(\sqrt {n+1}-\sqrt l)\]
\[\lim_{n \rightarrow \infty}\frac{\int\limits_{l}^{n} \frac{dx}{\sqrt x}}{\int\limits_{l}^{n+1} \frac{dx}{\sqrt x}}=\frac{\sqrt n-\sqrt l}{\sqrt {n+1}-\sqrt l}=1\]