Consider $\int^{10}_0x^2g'(x)dx$ where g has the values in the following table.
\begin{array}{r|l|l|l|l|l} x& 0 & 2 &4 & 6 & 8 & 10\\\hline g(x) & 2.9 &3.7 & 5.2&6.5&6.6&7.5 \end{array}
Q: $\int^{10}_0x^2g'(x)dx$ ≈ ?
My Try: Using integration by-parts then I got
$x^2g(x)|^{10}_0-\int^{10}_02xg(x)dx$
And then what I am confusing is how to use Riemann Right Hand Sum to plug the number into the equation I got above?
Well for the first term, you just have to substitute in the values at $x = 0$ and $x = 10$. The second term, you'd then write the integral as a Riemann sum:
$$\int_0^{10} 2xg(x)\ dx \approx \sum_{x = 2}^{10}\left ( 2xg(x) \times 2\right )$$
since you're evaluating over rectangles of width 2 (given that's the spacing you have for values of $g(x)$ and height equal to the function within the integral, evaluated at the right-hand side of the rectangle).