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

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Answer 1 · 2015-07-14T17:36:00

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

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 · 2015-07-16T06:52:53

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 $1/4$ .