Question

How do you find the integral of x ln x from (1,0)?

Answer 1

I found: `int_1^0xln(x)dx=1/4`

Explanation:

You can integrat by Parts to get:
`intxln(x)dx=x^2/2ln(x)-intx^2/2*1/xdx=`
`=x^2/2ln(x)-intx/2dx=x^2/2ln(x)-x^2/4|_1^0` `=`
`=0-(-1/4)=1/4`

Answer 2

You can do this slightly differently as well.

`int_(1)^0 xlnxdx`

From the Fundamental Theorem of Calculus and basic properties of Integrals, where
`int_a^b f(x)dx = -int_b^a f(x)dx`

`= F(b) - F(a) = -(F(a) - F(b))`

`=> -int_(0)^1 xlnxdx`

Using Integration by Parts, let:
`u = lnx`
`du = 1/xdx`
`dv = xdx`
`v = x^2/2`

`uv - intvdu`

`= -[(x^2lnx)/2 - int x^2/2*1/xdx]|_(0)^(1`

`= -[(x^2lnx)/2 - 1/2int xdx]|_(0)^(1`

`= [x^2/4 - (x^2lnx)/2]|_(0)^(1)`

`= [1^2/4 - cancel((1^2ln1)/2)] - [0^2/4 - (0^2ln0)/2]`

`= 1/4 + 0*ln0`

`= 1/4 + 0*(-oo)`

Let's do the limit instead, on the right term, due to an indeterminate form, and then L'Hopital's Rule.

`lim_(x->0) (x^2lnx)/2 = lim_(x->0) lnx/(2x^(-2))`

`= lim_(x->0) (1/x)/(-4/x^3) = lim_(x->0) -x^3/(4x)`

`= lim_(x->0) -x^2/4 = 0`

Thus, the answer is `1/4`.