Definite Integral problems

[rj686]

I'm having trouble figuring out the Integral from 0 to 1 of (36 dx)/(2x+1)^3, i know the answer is 8 from my calculator but i cant figure out how to get it manually, i keep getting .8888888

 

[Daniel_Feldman]

u=2x+1

du=2dx

(1/2)du=dx

You have to change your limits.

At x=0, u=1 and at x=1, u=3

So you evaluate the integral, from 1 to 3, of 9/u^3 du. You should get 8.

 

[rj686]

thx for the reply i totally forgot you had to change the domain... lol

can u show me how u got 9 in the numerator?

shouldnt you be evaluating it of U^4/4?

nvm... is there anyway you could work out teh whole problem?

 

[Daniel_Feldman]

My apologies. I was typing too fast. It should be 18, not 9. Here's why:

Remember: 36 is a constant, so you can take it out.

You have 36 Integral (1/(2x+1)^3)dx

u=2x+1
du=2dx
dx=(1/2)du

So now we have


36(1/2) Integral from 1 to 3 of (1/u^3)du

1/u^3 is u^-3, so integrating that yields u^-2/-2


So we have

(18/-2)u^-2=-9u^-2=-9/u^2.....and we evaluate from 1 to 3

=(-9/3^2)-(-9/1^2)

=(-1)-(-9)

=8

 

[rj686]

w00t ty vm

 

[rj686]

one more sorry Integral(5 Sin(x)^3/2 Cos (x) dx)......

 

[stapel]

w00t ty vm

I'm sorry, but I have no idea what this means. Please clarify. Thank you.

one more...

Please post new questions as new threads, not as replies to old threads, where they tend to be overlooked.

When you repost this, please clarify the integrand. Is it either of the following?

. . . . .[5 sin(x³)] / [2 cos(x)]

. . . . .5 sin^(3/2)(x) cos(x)

Or is it something else? Please also inclue limits (if any), the instructions for this exercise, and a clear listing of the steps you have tried thus far.

Thank you.

Eliz.

 

[rj686]

sorry for my aim speech
w00t- is similar to "heck yeah" - i was happy because i had gotten the problem answered
ty-thank you
vm-very much

5 sin(3/2)(x) cos(x) <-- that was the integral the limits are (0,*pi*/2)

The instructions are only to solve the problem analytically