Question

How do you evaluate the definite integral `int (u-2)/sqrtu` from [1,4]?

Answer 1

`phi= int_1^4(u-2)/sqrtu du =2/3`

Explanation:

Let,
`phi= int_1^4(u-2)/sqrtu du`

`u=t^2` `=>` `du=2tdt`

When, `u=1` `=>` `t=1` and `u=4` `=>` `t=2`

`phi= int_1^2(t^2-2)/t 2tdt`

`phi=2 int_1^2t^2-2dt`

`phi=2 [t^3/3-2t]_1^2`

`phi=2 [8/3-4-1/3+2]`

`phi=2 [7/3-2]`

`phi=2 [1/3]`

`phi=2/3`

`phi= int_1^4(u-2)/sqrtu du =2/3`

Answer 2

`I=int_1^4 (u-2)/sqrtudu=2/3`

Explanation:

Here,

`I=int_1^4 (u-2)/sqrtudu`

`=int_1^4 [u/sqrtu-2/sqrtu]du`

`=int_1^4 [sqrtu-2/sqrtu]du`

`=int_1^4 [u^(1/2)-2u^(-1/2)]du`

`=[u^(1/2+1)/(1/2+1)-2xx(u^(-1/2+1))/(-1/2+1)]_1^4`

`=[u^(3/2)/(3/2)-2xxu^(1/2)/(1/2)]_1^4`

`=[(4)^(3/2)/(3/2)-2xx(4)^(1/2)/(1/2)]- [(1)^(3/2)/(3/2)-2xx(1)^(1/2)/(1/2)]`

`=[(2)^3/(3/2)-2xx2/(1/2)]-[1/(3/2)-2xx1/(1/2)]`

`=[2/3xx8-8]-[2/3-4]`

`=16/3-8-2/3+4`

`=14/3-4`

`=2/3`