Please help me solve this integral

[Cha0s]

It is said that it converges. And i need to get the answer = 6. Please help me with steps.
I found that if it was infinite, it would = -2*SQRT(9-x) . I need to take the limit, but how?
http://img152.exs.cx/img152/2094/x5nd.png

 

[Atomicus]

Find an integral table and look for the one with *integral* 1/ sqrt(a^2 - x^2)

 

[Kyteland]

*integral* 1/ sqrt(a^2 - x^2) = sin-1(x/a)+C

Edit: Just noticed that's not what his equation is asking. 😱

Cha0s, You got the correct integral (-2*sqrt(9-x)+C), and the final step is that you need to evaluate it between the supplied intervals, 0 and 9.

=-2*sqrt(9-x)+C |0,9
=-2*sqrt(9-9)+C - (-2*sqrt(9-0)+C)
=-2*sqrt(0)+C+2*sqrt(9)+C
=0+2*3
=6

 

[DrPizza]

How about a simple u-substitution?
Let u = 9-x
Then, you have - integral of u^-(1/2)du

Edit: wait a second, wasn't looking carefully at the limits of integration.
Hang on.

 

[Mathlete]

Originally posted by: DrPizza
How about a simple u-substitution?
Let u = 9-x
Then, you have - integral of u^-(1/2)du

Edit: wait a second, wasn't looking carefully at the limits of integration.
Hang on.

yes and just change the limits

0 to 9 changes to 9 to 0

and you get = -2SQRT(u) on the interval 9 to 0

 

[DrPizza]

okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

 

[Whitecloak]

6?

let 9-x = t
therefore, -dx = dt

the eqn now becomes integral (0,9) 1/sqrt(t) (-dt)
= integral (9,0) 1/sqrt(t) dt
= 2 {sqrt(t) } (9,0)
= 6

gimme a cookie!

 

[DrPizza]

Originally posted by: Mathlete

Originally posted by: DrPizza
How about a simple u-substitution?
Let u = 9-x
Then, you have - integral of u^-(1/2)du

Edit: wait a second, wasn't looking carefully at the limits of integration.
Hang on.

yes and just change the limits

0 to 9 changes to 9 to 0

Not quite so simple, although it works out that way this time...
Notice that the original function isn't continuous at x = 9. There's a vertical asymptote at x = 9.

 

[Mathlete]

Originally posted by: DrPizza
okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

no need for a limit here

 

[DrPizza]

Originally posted by: whitecloak
6?

let 9-x = t
therefore, -dx = dt

the eqn now becomes integral (0,9) 1/sqrt(t) (-dt)
= integral (9,0) 1/sqrt(t) dt
= 2 {sqrt(t) } (9,0)
= 6

gimme a cookie!

no cookie for you!
Read the chapter on evaluating improper integrals 😉 😛

 

[DrPizza]

Originally posted by: Mathlete

Originally posted by: DrPizza
okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

no need for a limit here

Yes there is.

 

[Whitecloak]

Originally posted by: DrPizza

Originally posted by: Mathlete

Originally posted by: DrPizza
okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

no need for a limit here

Yes there is.

no there isnt

 

[Mathlete]

Originally posted by: DrPizza

Originally posted by: Mathlete

Originally posted by: DrPizza
okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

no need for a limit here

Yes there is.

Srry I missed the asymptote(sp?) in the original function. Tried to go to quickly.

 

[DrPizza]

Originally posted by: Mathlete

Originally posted by: DrPizza

Originally posted by: Mathlete

Originally posted by: DrPizza
okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

no need for a limit here

Yes there is.

Srry I missed the asymptote(sp?) in the original function. Tried to go to quickly.

no problem... same mistake I made at first. My class won't get to those for 2 weeks... I was caught off guard for a second.

 

[Mathlete]

Originally posted by: DrPizza

Originally posted by: Mathlete

Originally posted by: DrPizza

Originally posted by: Mathlete

Originally posted by: DrPizza
okay, that works just fine. Except you'll have to replace the 9 with L, and take the limit as L -> 9
Answer = 6.

The indefinite integral becomes
-2 sqrt(9-x) after the u substitution. (+c)

To evaluate it from 0 to 9, you have to evaluate
Limit as l -> 9 from 0 to l

no need for a limit here

Yes there is.

Srry I missed the asymptote(sp?) in the original function. Tried to go to quickly.

no problem... same mistake I made at first. My class won't get to those for 2 weeks... I was caught off guard for a second.

I have been teaching algebra and probability and stats so I get rusty on the calc.

 

[DrPizza]

🙂 2 weeks until I ask the class to evaluate integral from 0 to 2 of dx/(x-1)^2

 

[Mathlete]

Originally posted by: DrPizza
🙂 2 weeks until I ask the class to evaluate integral from 0 to 2 of dx/(x-1)^2

Hope they don't make the same mistake I did

 

[Cha0s]

thanks a lot i get it now