Show that int_1^e(lnx)^2dx = e - 2 ?
2 Answers
We want to show that:
int_1^e \ (lnx)^2 \ dx = e - 2
If we consider the indefinite integral:
I = int \ (lnx)^2 \ dx
Then, we can then apply Integration By Parts:
Let
{ (u,=ln^2x, => (du)/dx,=2lnx/x), ((dv)/dx,=1, => v,=x ) :}
Then plugging into the IBP formula:
int \ (u)((dv)/dx) \ dx = (u)(v) - int \ (v)((du)/dx) \ dx
We have:
int \ ln^2x \ 1 \ dx = (ln^2x)(x) - int \ (x)(2lnx/x) \ dx
:. I = xln^2x - 2 \ int \ lnx \ dx Let
{ (u,=lnx, => (du)/dx,=1/x), ((dv)/dx,=1, => v,=x ) :}
And, applying IBP to the second integral, we get:
int \ (lnx)(1) \ dx = (lnx)(x) - int \ (x)(1/x) \ dx
:. int \ lnx \ dx = xlnx - x
Combining these results we get:
I = xln^2x - 2 (xlnx-x) + C
So we can now evaluate the definite integral:
int_1^e \ (lnx)^2 \ dx = [ xln^2x - 2 (xlnx-x) ]_1^e
" " = (eln^2e - 2 (elne-e)) - (ln^2 1 - 2 (ln1-1))
" " = (e - 2 (e-e)) - (0 - 2 (0-1))
" " = (e) - (2)
" " = e-2 \ \ \ QED
See below
Explanation:
Lets say
Lets do the integral
Lets say
In our first case we had...
Applying Barrow's rule: