Express the integral as a limit, integral from -5 to -1 of x sin (x) dx using left endpoints
so it would be i= 1 to n f(-5+ (i-1) 4/n * 4/n and then I don't know how to continue
looks like you've just about got it. It's easier here to use 0 as the sum starting index.
$$\sum_{i=0}^{n-1}f\left(-5+\frac{4i}{n}\right) \frac{4}{n}$$
now make use of the fact that
$$f(x)=x \sin(x)$$
and note that
$$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$$
you'll end up with an approximation formula for the integral in terms of n. Then take the limit as n goes to infinity.
Not quite.
in the sum
$$x_i=-5+\frac{4i}{n}$$
$$dx = \frac{4}{n}$$
$$f_i=x_i \sin(x_i) = \left(-5+\frac{4i}{n}\right) \sin\left(-5+\frac{4i}{n}\right)$$
so your sum becomes
$$\sum_{i=0}^{n-1}f_i dx = \sum_{i=0}^{n-1} \left(-5+\frac{4i}{n}\right) \sin\left(-5+\frac{4i}{n}\right) \frac{4}{n}$$
you can work the algebra to finish it up
a numerical check shows this does converge (very slowly) to the correct answer.