Question

Question from integral calculus

Answer 1

let integrand \(x^x=y\) and take log both sides to get exponent down. then proceed the leftover

Answer 2

Check the integrand . And how will we take logarithms ? If you have solution, please post

Answer 3

Taking logarithms doesnot help.

Answer 4

if y = x^x

log(y) = x log(x)

taking the derivative we get

y'/y = log(x) + x/x

y' = x^x (log(x) + 1)

what do you know of taylor series?

Answer 5

or, is this one of those problems which lets you know that integrating is not as cut and dry as taking derivatives?

Answer 6

are you able to work a reimann sum using simpsons, left/right/mid, traps ...

Answer 7

the integrand is undefined at the lower limit. you have to consider it.

Answer 8

yes i know about taylor, simpson, riemann, ...

Answer 9

\[\large e^x = \sum\limits_{n=0}^{\infty}\dfrac{x^n}{n!}\]

Answer 10

\[\large x^x = e^{x\ln x} = \sum\limits_{n=0}^{\infty}\dfrac{(x\ln x)^n}{n!}\]

Answer 11

\[\large \int\limits_0^1 x^x ~dx= \int\limits_0^1\sum\limits_{n=0}^{\infty}\dfrac{(x\ln x)^n}{n!}~dx\]

Answer 12

knw how to handle that beast ?

Answer 13

http://en.wikipedia.org/wiki/Sophomore's_dream

Answer 14

i would use right hand triangles, since its not defined at x=0 :)

\[\sum_{i=1}^{n} f(a+i\frac{b-a}{n})\frac{b-a}{n}\]

Answer 15

an + ib -ia

(n-i)a + ib

since a=0, b=1

sum f(i) 1/n, i = 1 to n

sum i^i 1/n, i = 1 to n

something like that if memory serves

Answer 16

so what should be the final answer? Will we keep in summation notation only?

Answer 17

from that wiki link :\[\large \int\limits_0^1 x^x ~dx = -\sum\limits_{n=1}^{\infty} (-n)^{-n} \approx 0.78343\]

which is some constant like e, however we don't know whether that constant is irrational like e

Answer 18

However, is there any way of solving this problem, using numerical analysis? If yes, please let me know... The solution is very nice .

Answer 19

if u just want to approximate, try riemann sum ?

Answer 20

or any other fancy numerical technique tha tu knw

Answer 21

try this in matlab :

```
fun = @(x) x.^x;

q = quadcc(fun,0,1)
```