How do you evaluate the definite integral #int sqrtt ln(t)dt# from 2 to 1?

Question

3484 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-14T14:11:25

#4/9(2sqrt2-1)-((4sqrt2)/3)ln2.#

We use the Fundamental Theorem of Integral Calculus

#: intf(t)dt=F(t) rArr ""int_a^bf(t)dt=F(b)-F(a).#

So, we first find out the Indefinite Integral #I=intsqrttlntdt,# using the

Method of Integration by Parts,which states that,

#intuvdt=uintvdt-int{(du)/dtintvdt}dt.#

We apply this by taking #u=lnt, &, v=sqrtt,# so that,

#(du)/dt=1/t, &, intvdt=(t^(3/2))/(3/2)=(2/3)t^(3/2).#

#:. I=(2/3)t^(3/2)lnt-int{1/t*(2/3)t^(3/2)}dt,#

#=(2/3)t^(3/2)lnt-(2/3)intt^(1/2)dt,#

#=(2/3)t^(3/2)lnt-(2/3)(2/3)t^(3/2)#,

#:. I=(2/3)t^(3/2)lnt-(4/9)t^(3/2)+C.#

Hence, #int_2^1intsqrttlntdt=[(2/3)*1^(3/2)*ln1-(4/9)*1^(3/2)]-[(2/3)*2^(3/2)*ln2-(4/9)*2^(3/2)]#

#=[0-4/9]-[(2^(5/2)*ln2)/3-4/9*2^(3/2)]#

#=4/9(2^(3/2)-1)-2^(5/2)/3*(ln2)#

#=4/9(2sqrt2-1)-(4/3sqrt2)ln2.#

Answer 2 · 2016-07-14T14:23:05

#= 2/3( (4 sqrt2 )/3 - 2/3 - 2 sqrt2 * ln 2 ) #

#int_2^1 sqrtt ln(t)dt#

#= int_2^1 d/dt (2/3 t^(3/2) ) ln(t)dt#

and so by IBP: #int uv' = uv - int u' v#

#= [2/3 t^(3/2)*ln(t)]_2^1 - int_2^1 2/3 t^(3/2) \ d/dt ( ln(t)) \ dt#

#= [2/3 t^(3/2)*ln(t)]_2^1 - int_2^1 2/3 t^(3/2) \ 1/t \ dt#

#= [2/3 t^(3/2)*ln(t)]_2^1 - 2/3 int_2^1 t^(1/2) \ dt#

#= 2/3[ t^(3/2)*ln(t) - 2/3 t^(3/2) \]_2^1#

#= 2/3[ t^(3/2) (ln(t) - 2/3 ) \]_2^1#

#= 2/3([ 1 * (0 - 2/3 ) \] -[ 2 sqrt2 * (ln 2 - 2/3 ) \] ) #

#= 2/3( (4 sqrt2 )/3 - 2/3 - 2 sqrt2 * ln 2 ) #