Question

i need help with this one
integrate x^x from 0 to 1

Answer 1

i think i did it before for one of the users let me search for it

Answer 2

\[\int\limits_{0}^{1} x^x dx\]
?

Answer 3

yes thats it

Answer 4

\[\int x^x dx\]
let u = x^x
du =

side solution:
a = x^x
ln a = x ln x
1/a a' = 1 + ln x
a' = x^x(1+ln x)

so du = x^x (1+ ln x)

dv = dx
v = x

\[\int x^x dx = x \cdot x^x - \int x \cdot x^x (1+ \ln x)dx\]
\[\int x^x dx = x^{x+1} - \int x^{x+1}(1+\ln x)dx\]
this is getting complicated

Answer 5

there is no anti derivative for x^x

Answer 6

here it is
plz let me know if u dont get it

Answer 7

no possible solution in terms of analytical form possible

Answer 8

in analytical way it will get complicated can be solved numerically .

Answer 9

oh...nice.... mukushala i think i should just follow your steps
anyway thank u

Answer 10

@moachi you know numerical methods ? like mid point rule simpson rule ? if you know i can solve it numerically.

Answer 11

welcome

Answer 12

@sami-21
yes but i wanted to know if there is any analytical solution for this or not!

Answer 13

http://www.numberempire.com/definiteintegralcalculator.php

and i got

\[1-\frac{1}{2^2}+\frac{1}{3^3}-\frac{1}{4^4}+... \approx 0.7813\]

Answer 14

using simpsons rule better accuracy can be achieved using larger n

Answer 15

and thats fine with numerical computation...