Show that int_1^e(lnx)^2dx = e - 2 ?

2 Answers
Mar 8, 2018

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

Mar 8, 2018

See below

Explanation:

int(lnx)^2dx= lets use the "Integration by parts method"

Lets say int(lnx)^2dx=intlnx·lnx·dx and make the change

u=lnx and dv=lnxdx, then du=1/xdx and v=intlnxdx

Lets do the integral intlnxdx by the same method.

Lets say u_1=lnx and dv_1=dx. Hence we have

intlnxdx=u_1v_1-intv_1du_1=xlnx-x+C=x(lnx-1)+C

In our first case we had...u=lnx and v=x(lnx-1)

int(lnx)^2dx=x·lnx·(lnx-1)-int1/x·x(lnx-1)dx= xlnx(lnx-1)-x(lnx-1)+x+c=x((lnx)^2-2lnx+2)+C=F(x)

Applying Barrow's rule: F(e)-F(1)=e(1^2-2·1+2)-1(0-0+2)=e-2