Calculate the integral $\int_1^{10}x^xdx$ $$\int_1^{10}x^xdx$$
I figured it with a relative error of 1% and I have a response $$≈0.3*10^{10}$$ 
But I don´t know how to accurately calculate it ...
 A: One way to solve this problem is to expand in terms of a power series as follows:
First notice that
$x^x = e^{\ln(x) x} = \sum_{n=0}^\infty \frac{(\ln(x) x)^n}{n!} $
Integrating on both sides from $1$ to $10$ gives
$\int_1^{10} x^x dx = \sum_{n=0}^\infty \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx $
The function $x^n ln(x)^n$ is monotonically increasing, so the integral inside the sum is bounded by:
$ \frac{1}{n!} \int_1^{10} x^n \ln(x)^n dx < \frac{9}{n!} \cdot \ln(10)^n \cdot 10^n $
So if you want the sum to whatever precision, you can use that expression to determine how small the remainder term is. Next, how on earth do you compute that integral? Well it looks ugly but there is a recursive formula. Let $k, l$ be positive integers. Then using integration by parts
$ \int_1^{10} x^k \ln(x)^l dx = \frac{x^{k+1} ln(x)^l}{k+1} \Bigr \vert_1^{10} - \frac{l}{k+1} \int_1^{10} x^k \ln(x)^{l-1} dx $
Since you chose the limits of integration to be from $1$ to $10$ you might have to take quite a few terms in this series in order to compute an accurate value, but that is to be expected as $x^x$ blows up so quickly.
