How can I convert this Riemann sum to a definite integral?
$$\lim_\limits{n\to\infty}\sum_{i=1}^n\pi\biggl(1.6875+\frac{.75775i}{n}\biggl)^2\frac{1.625}{n}$$
I'm confused because the usual definition of
$$\int_a^b f(x)dx=\lim_\limits{n\to\infty}\sum_{i=1}^n f(x_i^*)\Delta x$$
doesn't work because $1.625 \ne .75775$. Help would be appreciated, thanks in advance.
Hint
$$1.6875 =\frac{27}{16} \qquad\qquad 0.75775=\frac{3031}{4000} \qquad \qquad1.625=\frac{13}{8}$$ $$\sum_{i=1}^n\pi\biggl(1.6875+\frac{.75775i}{n}\biggl)^2\frac{1.625}{n}=\frac{119430493 }{128000000}\pi\sum_{i=1}^n \left(\frac{i}{n}+\frac{6750}{3031}\right)^2\frac1 n$$