`int_3^4 1/(x-3)^(3/2) dx` Explain why the integral is improper and determine whether it diverges or converges. Evaluate the integral if it converges

Expert Answers

An illustration of the letter 'A' in a speech bubbles

The integral is improper because the function under the integral `f(x)=1/(x-3)^(3/2)` is not defined at 3 (for `x=3` denominator is equal to zero). 

`int_3^4 1/(x-3)^(3/2)dx=`

Substitute `u=x-3` `=>` `du=dx,` `u_l=3-3=0,` `u_u=4-3=1.`

`u_l` and `u_u` denote new lower and upper bound respectively.

`int_0^1 1/u^(3/2)du=int_0^1 u^(-3/2)du=-2u^(-1/2)|_0^1=-2cdot1^(-1/2)+lim_(u to 0)2u^(-1/2)=`

`-2+infty=infty`

As we can see the integral diverges.

The...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

The integral is improper because the function under the integral `f(x)=1/(x-3)^(3/2)` is not defined at 3 (for `x=3` denominator is equal to zero). 

`int_3^4 1/(x-3)^(3/2)dx=`

Substitute `u=x-3` `=>` `du=dx,` `u_l=3-3=0,` `u_u=4-3=1.`

`u_l` and `u_u` denote new lower and upper bound respectively.

`int_0^1 1/u^(3/2)du=int_0^1 u^(-3/2)du=-2u^(-1/2)|_0^1=-2cdot1^(-1/2)+lim_(u to 0)2u^(-1/2)=`

`-2+infty=infty`

As we can see the integral diverges.

The image below shows the graph of the function and area under it corresponding to the integral. Both axis are asymptotes of the function.                                                                                                                   

Images:
This image has been Flagged as inappropriate Click to unflag
Image (1 of 1)
Approved by eNotes Editorial Team