# Thread: More Simpson Rule Question

1. ## More Simpson Rule Question

Hi

The following question i don't know why it is incorrect.
1) $\int_1^3 x^{x}dx$ n=4

$x_0 = 0$

$x_1 = \frac{1}{4}$

$x_2 = \frac{2}{4}$

$x_3 = \frac{3}{4}$

$x_4 = \frac{4}{4}$

$\int_0^1 x^{x} dx = \frac{1}{12}[0+4(\frac{1}{4}^{\frac{1}{4}})+2(\frac{1}{2}^{\fra c{1}{2}})+4(\frac{3}{4}^{\frac{3}{4}})+1]$

=0.70553

2. Originally Posted by Paymemoney
Hi

The following question i don't know why it is incorrect.
1) $\int_1^3 x^{x}dx$ n=4

$x_0 = 0$

$x_1 = \frac{1}{4}$

$x_2 = \frac{2}{4}$

$x_3 = \frac{3}{4}$

$x_4 = \frac{4}{4}$

$\int_0^1 x^{x} dx = \frac{1}{12}[0+4(\frac{1}{4}^{\frac{1}{4}})+2(\frac{1}{2}^{\fra c{1}{2}})+4(\frac{3}{4}^{\frac{3}{4}})+1]$

=0.70553

Why have you used $0^0=0$ in this case? You need $\lim_{x \to 0}x^x=1$

Next time you post a question try to make it clear what your question is, and don't change the detail part way through.

CB

3. Originally Posted by Paymemoney
Hi

The following question i don't know why it is incorrect.
1) $\int_0^1 x^{x}dx$ n=4

$x_0 = 0$

$x_1 = \frac{1}{4}$

$x_2 = \frac{2}{4}$

$x_3 = \frac{3}{4}$

$x_4 = \frac{4}{4}$

$\int_0^1 x^{x} dx = \frac{1}{12}[0+4(\frac{1}{4}^{\frac{1}{4}})+2(\frac{1}{2}^{\fra c{1}{2}})+4(\frac{3}{4}^{\frac{3}{4}})+1]$

=0.70553

srry about that, the above is the correct equation.

4. $\int_0^1 x^{x} dx = \frac{1}{12}[1+4(\frac{1}{4}^{\frac{1}{4}})+2(\frac{1}{2}^{\fra c{1}{2}})+4(\frac{3}{4}^{\frac{3}{4}})+1]$

=0.7889

5. Originally Posted by CaptainBlack
$\int_0^1 x^{x} dx = \frac{1}{12}[1+4(\frac{1}{4}^{\frac{1}{4}})+2(\frac{1}{2}^{\fra c{1}{2}})+4(\frac{3}{4}^{\frac{3}{4}})+1]$

=0.7889
but isn't $x_0$ equal to 0?

6. Originally Posted by Paymemoney
but isn't $x_0$ equal to 0?
No that's why I said it's not in post #2 in this thread. $0^0$ is undefined here you need to use $\lim_{x\to 0}x^x=1$ for the value of the integrand at $x=0$

CB