How do you evaluate the definite integral int (u-2)/sqrtuu2u from [1,4]?

2 Answers
Jun 8, 2018

phi= int_1^4(u-2)/sqrtu du =2/3ϕ=41u2udu=23

Explanation:

Let,
phi= int_1^4(u-2)/sqrtu duϕ=41u2udu

u=t^2u=t2 => du=2tdtdu=2tdt

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

phi= int_1^2(t^2-2)/t 2tdtϕ=21t22t2tdt

phi=2 int_1^2t^2-2dtϕ=221t22dt

phi=2 [t^3/3-2t]_1^2ϕ=2[t332t]21

phi=2 [8/3-4-1/3+2]ϕ=2[83413+2]

phi=2 [7/3-2]ϕ=2[732]

phi=2 [1/3]ϕ=2[13]

phi=2/3ϕ=23

phi= int_1^4(u-2)/sqrtu du =2/3ϕ=41u2udu=23

Jun 8, 2018

I=int_1^4 (u-2)/sqrtudu=2/3I=41u2udu=23

Explanation:

Here,

I=int_1^4 (u-2)/sqrtuduI=41u2udu

=int_1^4 [u/sqrtu-2/sqrtu]du=41[uu2u]du

=int_1^4 [sqrtu-2/sqrtu]du=41[u2u]du

=int_1^4 [u^(1/2)-2u^(-1/2)]du=41[u122u12]du

=[u^(1/2+1)/(1/2+1)-2xx(u^(-1/2+1))/(-1/2+1)]_1^4=[u12+112+12×u12+112+1]41

=[u^(3/2)/(3/2)-2xxu^(1/2)/(1/2)]_1^4=[u32322×u1212]41

=[(4)^(3/2)/(3/2)-2xx(4)^(1/2)/(1/2)]- [(1)^(3/2)/(3/2)-2xx(1)^(1/2)/(1/2)]=(4)32322×(4)1212(1)32322×(1)1212

=[(2)^3/(3/2)-2xx2/(1/2)]-[1/(3/2)-2xx1/(1/2)]=[(2)3322×212][1322×112]

=[2/3xx8-8]-[2/3-4]=[23×88][234]

=16/3-8-2/3+4=163823+4

=14/3-4=1434

=2/3=23