Question

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

Answer 1

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

Answer 2

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`