integration (1 Viewer)

[want2beSMART]

integral from 1 to 49 of 1/ (sqrt(x) sqrt(1+sqrt(x))) dx using substitution of u = 1+ sqrt(x)

cheers

 

[Dreamerish*~]

integral from 1 to 49 of 1/ (sqrt(x) sqrt(1+sqrt(x))) dx using substitution of u = 1+ sqrt(x)

cheers

can you write the question again?
i don't get the "(sqrt(x) sqrt(1+sqrt(x)))" part...

 

[want2beSMART]

integral from 1 to 49 of 1/ {sqrt(x) sqrt[1+sqrt(x)]} dx using substitution of u = 1+ sqrt(x)

 

[Dreamerish*~]

is this it?

write it on paint and post it up.

 

[want2beSMART]

yes...................... ^^

 

[Dreamerish*~]

that was a messy load of confusion.

but i managed to grab a most-likely-to-be-incorrect answer

0.44 (2 d.p.)

 

[want2beSMART]

how did you do it?

 

[acmilan]

If the question is what i think it is use substitution:

Make u = root(x)
du = 1/(2root(x)) dx

when x = 49, u = 8
when x = 1, u = 2

 

[king_of_boredom]

If the question is what i think it is use substitution:

Make u = root(x)
du = 1/(2root(x)) dx

when x = 49, u = 7
when x = 1, u = 1

i thought it was:

when x = 49, u = 8
when x = 1, u =2

becuase u = 1+ sqrt(x)

 

[Dreamerish*~]

was my answer right?

 

[acmilan]

possibly i didnt actually take a look at the question

 

[shafqat]

no acmilan was right
both substitutions work i think
wait...mayb not, its too messy
no it works fine lol

 

[acmilan]

yeah using u = root(x) works. Answer 4root2?

 

[shafqat]

really? i just get 1/sqrt(1+u)
edit: yep acmilan edited. he's method works fine

 

[want2beSMART]

you should leave it in exact form

 

[acmilan]

haha yeah you can turn it into an inverse sign but its undefined from 1 to 7. Better stick to normal integration

 

[want2beSMART]

but the question says to sub u= 1+ sqrt(X)

 

[acmilan]

Oh then do it using that substitution. But either way it gives the same answer

u = 1 + root(x)
du = 1/(2rootx)dx
when x = 49, u = 8
when x = 1, u = 2

so you can get rid of the dx and the root(x) from the bottom by substituting u and du into it and adding a 2 in the front.

youll get du/root(1+u)

the integral of this is 4root(1+u). Then you have to just sub in u = 8 and u = 2 and subtract them. Answer will be 4root2

 

[want2beSMART]

umm yeah i know all this...

but can you actally show me the working out...

 

[acmilan]

int(49 to 1){dx/(root(x)root(1 + root(x)))}

= 2int(8 to 1){du/root(u)}
= 2[2root(u)] from 8 to 1

the integration of 1/root(u) is 2root(u)