Evaluate this Integral?

Question

∫xlnx dx

Please look at this picture
http://img529.imageshack.us/img529/6642/49348752qi6.jpg

b=1 and a=0

Answer 1

∫x ln x dx

First, integrate by parts -- let u=ln x, du=1/x dx, dv = x dx, v=x²/2. Then we have:

1/2 x² ln x - 1/2 ∫x²/x dx
1/2 x² ln x - 1/2 ∫x dx
1/2 x² ln x - x²/4

Now, evaluating from 0 to 1, we note that we must contend with limits, since the function is unbounded near 0. So we have:

[a → 0]lim 1/2 * 1 * ln 1 - 1/4 - (1/2 a² ln a - a²/4)
[a → 0]lim 0 - 1/4 + a²/4 - 1/2 a² ln a
-1/4 + [a→ 0]lim a²/4 - 1/2 a² ln a
-1/4 + [a→ 0]lim a²/4 - [a → 0]lim 1/2 a² ln a
-1/4 + 0 - [a → 0]lim 1/2 a² ln a
-1/4 - [a → 0]lim 1/2 ln a/(1/a²)

This is now ∞/∞ form, so using L'hopital's rule:

-1/4 - [a → 0]lim 1/2 1/a/(-2/a³)
-1/4 - [a → 0]lim -1/4 a²
-1/4 - 0
-1/4

And we are done.

Answer 2

is it a and b are the limits of that eq?

at that type of eq, we do a process which called "integration by parts". the main eq of that process is:

int( u dv) = uv - int(v du)

we let dv= x dx and integrating it gives a value of v = 1/2 x^2
we let u = ln x and differentiating it, du = 1/x dx

int(x ln x dx) = 1/2 x^2 ln x - int(1/2 x^2 *1/x dx)
int(x ln x dx) = 1/2 x^2 ln x - int(1/2 x dx)
int(x ln x dx) = 1/2 x^2 ln x - 1/4 x^2

substituting the value of limits,

int(x ln x dx) = 1/2 1^2 ln 1 - 1/4 1^2 - (1/2 0^2 ln 0 - 1/4 0^2)
int(x ln x dx) = -1/4 -1/2 0^2 ln 0

but this expression, 1/2 0^2 ln 0, is an indeterminate, so by using L'Hospital's rule, we first equate 1/2 x^2 ln x to (ln x) / (2x^(-2)), and differentiating it (derivative of numerator over the derivative of the denominator), gives a result of:

(1/x)/(-2/3 x^(-3))

or we can say

-3/2 x^2

but x is approaches to zero, so

the value is zero

we go back to our equation,

int(x ln x dx) = -1/4 - 0
int(x ln x dx) = -1/4 is the answer!

Answer 3

http://math2.org/math/integrals/more/ln.htm

Step by step^

Once you get to the answer, evaluate from 0 to 1. To do this, sub 1 for x in the answer, and get that value. Then sub 0 for x into the answer, and get a second value. Subtract this second value from the first to get the final final answer.

(Don't worry about the 'C' on the website... you only put C at the end when you're dealing with an indefinite integral (i.e. an integral that doesn't give you a and b)).

Answer 4

In other words, you want to evaluate this integral from 0 to 1. Use integration by parts, letting u = ln(x) and dv = x dx.

Answer 5

∫ x ln x dx

integrating by parts

let u = ln x

dv = x

du = 1/x

v = x^2/2

∫ x ln x dx = uv - ∫v du

=>x ln x- ∫ x^2/2 (1/x) dx

=> x ln x - 1/2∫ x dx

= > x ln x - 1/2 x^2/2

=> x lnx - 1/4 x^2
1
∫ x ln x =
0
=> 1 ln 1 - 1/4(1) - [0 - 0]

=> - 1/4