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...

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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.