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

Question

3440 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-08T04:59:17

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

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 · 2018-06-08T05:01:17

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

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#