Q.From the definition of definite integral find the value of ∫₀¹ (2x + 1) dx.

Definite Integral as a Limit of a Sum

Many limits of large sums are evaluated by recognising them as Riemann sums. If a

sum can be written as $\dfrac1n\sum_{r=1}^{n} f\!\Big(\dfrac{r}{n}\Big)$, then

$\lim_{n\to\infty}\dfrac1n\sum_{r=1}^{n} f\!\Big(\dfrac rn\Big)=\int_0^1 f(x)\,dx,$

because $\dfrac rn\to x$ and $\dfrac1n\to dx$ as the partition is refined.

The method: factor out $\dfrac1n$, express the general term purely in $\dfrac rn$, let

$\dfrac rn\to x$, and read the limits of integration from the smallest and largest

values of $\dfrac rn$ (e.g. a sum from $r=n$ to $2n$ gives $x$ ranging over $[1,2]$).

The resulting definite integral is then evaluated by standard techniques. This

converts otherwise intractable sums such as

$\lim\dfrac1n\sum\sqrt{\dfrac{n+r}{n-r}}$ or $\lim\sum\dfrac{r}{r^2+n^2}$ into single …