"With Riemann" means that you treat this as a Riemann sum- which is used to define the definite integral.
If we divide the x-axis, from [itex]x= a[/itex] to [itex]x= b[/itex], into [itex]n[/itex] intervals then each interval has length [itex]\Delta x= (b-a)/n[/itex]. And, if we take the value of the integrand, [tex]f(x)[/tex], in each interval at the left end of that interval, we have [tex]f\left(\frac{(b-a)k}{n}\right)[/tex] as the height of the rectangle we are forming on that interval so its area is [tex]f(\frac{(b-a)k}{n})\left(\frac{b-a}{n}\right)[/tex] and the whole area is
[tex]\displaystyle (b- a)\sum_{k=0}^n f\left( \frac{(b-a)k}{n}\right)\cdot\frac{1}{n}[/tex]
If we take the limit as the number of rectangles [itex]n \to \infty[/itex] this becomes [tex]\int^b_a f(x) dx[/tex].
So you need to indentify f(x), a, and b in this particular sum, and integrate.