Question

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

Answer 1

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

Explanation:

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

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

Explanation:

`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 )`