Question

Improper integrals,

My first time doing improper integrals
, could you help me out on the first one please ?
https://www.dropbox.com/s/wridon5503p8r3z/Screenshot%202013-11-03%2020.24.16.jpg

Answer 1

\[\int_{12}^{+\infty}\frac{dx}{\sqrt x(x+4)}\]

Let \(u=\sqrt x\)
\(du=\)

\[x=+\infty\quad \to\quad u=+\infty\]\[x=12\qquad \to\quad u=\]

Answer 2

u = 0 ??

Answer 3

du = 1/sqrt(x)1

Answer 4

1/sqrt(x)2 *

Answer 5

what is missing

Answer 6

du = 1/aqrt(x)2 and u = 0 ?? I may be wrong though

Answer 7

your just missing the dx at the end

\[du=\frac1{2\sqrt x}dx=\tfrac12x^{1/2}dx\]

Answer 8

\(du=\frac1{2\sqrt x}dx=\tfrac12x^{-1/2}dx\)*

Answer 9

ok and this http://screencast.com/t/j4wthX1Hq ?

Answer 10

we let u= √x

so the limit , for when x=12

becomes u=√12

Answer 11

just as √∞=∞ in the other limit

Answer 12

and and what do we do then ?

Answer 13

substitute into the integral

Answer 14

I can substitute du but I can't find another sort for u

Answer 15

\[du=\tfrac12x^{-1/2}dx\]
\[2\sqrt x du=dx\]

Answer 16

x=u^2

Answer 17

I am kind of confused :S

Answer 18

\[\int_{12}^{+\infty}\frac{dx}{\sqrt x(x+4)}\]

\[\int_{\sqrt{12}}^{+\infty}\frac{2du}{(u^2+4)}\]

Answer 19

arc tan sqrt(inf) - arc tan sqrt(12) ?

Answer 20

basically ½ arctan(inf) - ½ arctan(sqrt(12) / 2 ??

Answer 21

what happened to the 2?

Answer 22

(your answer is very close i think you just need to check the details, and you will be able to find the mistake)

Answer 23

½ arctan(inf) - ½ arctan(sqrt(12)/ 2) ??

Answer 24

There is this step here which I don't understand http://screencast.com/t/ResGOQGk

Answer 25

how did he measure the arc tan of inf ?

Answer 26

the arctangent of infinity is π/2 or 90°

think of a triangle