In short

A definite integral is literally the limit of a sum: \int_0^1 f(x)\,dx = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n} f(r/n). So whenever a limit-of-sum problem has the right shape — a \frac{1}{n} outside and r/n sitting inside a function — you can rewrite the sum as an integral and evaluate it in one line. This is the single most elegant application of integration in JEE.

Evaluate

\lim_{n\to\infty}\left[\frac{1}{n+1} + \frac{1}{n+2} + \frac{1}{n+3} + \cdots + \frac{1}{2n}\right].

There are n terms here, and they are all close to \frac{1}{n}, so the sum is a bunch of \frac{1}{n}-sized numbers. If you replace each one with the crude lower bound \frac{1}{2n}, the sum is at least n\cdot\frac{1}{2n} = \frac{1}{2}. If you replace with the crude upper bound \frac{1}{n+1}, the sum is at most n\cdot\frac{1}{n+1}\approx 1. So the answer is somewhere between \frac{1}{2} and 1. But where, exactly?

There is no obvious telescoping trick or closed form for this sum. No factoring, no partial fractions on the term \frac{1}{n + r}, no magical identity. And yet the answer turns out to be \ln 2 \approx 0.693, a clean and specific number — and it comes out in two lines, not by cleverness, but by recognising the sum as a Riemann sum.

That recognition is the whole technique of this article.

The Riemann sum, rewritten

When you first meet the definite integral, one of the definitions you see is this. Split the interval [a, b] into n equal subintervals of width \frac{b - a}{n}, and pick any point in each subinterval. The height of the function at that point times the width \frac{b-a}{n} is the area of a rectangle; sum all n rectangles to get a Riemann sum; let n \to \infty to get the integral.

In symbols, one clean version is: let x_r^* = a + \frac{r}{n}(b-a) (the right endpoint of the r-th subinterval). Then

\int_a^b f(x)\,dx = \lim_{n\to\infty}\sum_{r=1}^n f(x_r^*)\cdot\frac{b-a}{n}.

Specialise to the interval [0, 1], where b - a = 1 and x_r^* = r/n. The formula simplifies dramatically:

\int_0^1 f(x)\,dx = \lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n f\!\left(\frac{r}{n}\right).

Look at that formula. On the right, the \frac{1}{n} is the width of each rectangle, and f(r/n) is the height. On the left, the integral is the exact area. The formula says: if you have a sum of the form \frac{1}{n}\sum f(r/n), you can compute its limit by doing the integral \int_0^1 f(x)\,dx.

That is the whole trick. Read from right to left: if a limit-of-sum problem has the shape \frac{1}{n} times a sum where the summand is a function of r/n, it is an integral in disguise.

Applying it to the opening problem

Back to the opening sum:

\lim_{n\to\infty}\left[\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n}\right] = \lim_{n\to\infty}\sum_{r=1}^{n}\frac{1}{n + r}

Pull a factor of \frac{1}{n} out of every term:

\frac{1}{n + r} = \frac{1}{n}\cdot\frac{1}{1 + r/n}

So the sum becomes

\sum_{r=1}^n\frac{1}{n + r} = \sum_{r=1}^n\frac{1}{n}\cdot\frac{1}{1 + r/n} = \frac{1}{n}\sum_{r=1}^n\frac{1}{1 + r/n}

That is exactly the shape \frac{1}{n}\sum f(r/n), with f(x) = \frac{1}{1 + x}. So its limit is

\int_0^1 \frac{dx}{1 + x} = \ln(1 + x)\Big|_0^1 = \ln 2

Total work: four lines. Compare with trying to evaluate that sum by elementary means.

The formal version, with limits

Before doing more examples, let me write down the recipe as a formal three-step procedure.

The limit-of-sum as integral rule

If f is continuous on [0, 1], then

\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{n} f\!\left(\frac{r}{n}\right) = \int_{0}^{1} f(x)\,dx

More generally, if f is continuous on [a, b] and you sample at x_r = a + \frac{r(b-a)}{n}, then

\lim_{n\to\infty}\frac{b-a}{n}\sum_{r=1}^{n} f(x_r) = \int_{a}^{b} f(x)\,dx

The three-step recipe.

  1. Factor out \frac{1}{n} from every term in the sum. Rewrite each term as \frac{1}{n} times some expression.
  2. Identify f(r/n) inside the sum. This tells you what f(x) is — the function you need to integrate.
  3. Integrate f(x) from 0 to 1 (or from a to b if the sum samples from a different interval).

The trickiest step is step 1. For the first example, factoring \frac{1}{n} out was easy. In harder problems, you sometimes have to do algebraic rewrites — like pulling n^2 out of a denominator, or rewriting n! as something you can manipulate — before the \frac{1}{n} appears. The pattern is always the same, though: a \frac{1}{n} outside, f(r/n) inside.

Which interval?

The default interval is [0, 1], and the sum runs from r = 1 to r = n (or from r = 0 to r = n - 1; it does not matter, because in the limit both give the same answer — the shifted-by-one version just corresponds to using left endpoints instead of right endpoints).

If the summand has r/n ranging up to 2 or higher — say, \frac{2r}{n} or \frac{r}{n} with r running up to 2n — then you adjust the interval. The general form

\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^{kn} f\!\left(\frac{r}{n}\right) = \int_0^k f(x)\,dx

handles the case where the sum has kn terms. Here r/n sweeps from 0 to k, so the integral runs from 0 to k.

For a sum that samples from [a, b] — like \frac{r}{n} shifted so that r/n runs over [1, 2] instead of [0, 1] — use the general form of the rule above.

The first full worked example

A classic JEE problem.

Example 1: $\displaystyle\lim_{n\to\infty}\sum_{r=1}^{n}\frac{r}{n^2 + r^2}$

Step 1. Rewrite the summand so n appears as a factor. The denominator n^2 + r^2 can be factored as n^2(1 + r^2/n^2), so

\frac{r}{n^2 + r^2} = \frac{r}{n^2(1 + (r/n)^2)} = \frac{r/n^2}{1 + (r/n)^2} = \frac{1}{n}\cdot\frac{r/n}{1 + (r/n)^2}

Why: the first move is always to get \frac{1}{n} factored out explicitly and to make sure everything else is written as a function of r/n. Here the numerator had to be split as r = n\cdot(r/n), producing an extra \frac{1}{n}.

Step 2. Identify f.

\frac{1}{n}\sum_{r=1}^n \frac{r/n}{1 + (r/n)^2} = \frac{1}{n}\sum_{r=1}^n f\!\left(\frac{r}{n}\right)\quad\text{where}\quad f(x) = \frac{x}{1 + x^2}

Why: reading off f is just pattern-matching. Whatever is multiplying \frac{1}{n} in the sum, with r/n replaced by x, is the function.

Step 3. Recognise the limit as an integral.

\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n f\!\left(\frac{r}{n}\right) = \int_0^1 \frac{x\,dx}{1 + x^2}

Why: the sum runs from r = 1 to r = n, so r/n sweeps across [0, 1] as n \to \infty. The integral uses the standard [0, 1] form of the rule.

Step 4. Evaluate the integral. Substitute u = 1 + x^2, du = 2x\,dx:

\int_0^1 \frac{x\,dx}{1 + x^2} = \frac{1}{2}\int_1^2 \frac{du}{u} = \frac{1}{2}\ln u\bigg|_1^2 = \frac{1}{2}\ln 2

Why: the integrand \frac{x}{1+x^2} is a derivative in disguise — the numerator is half the derivative of the denominator. Substitution is the right move.

Result: \displaystyle\lim_{n\to\infty}\sum_{r=1}^n\frac{r}{n^2 + r^2} = \dfrac{\ln 2}{2}.

The curve $f(x) = \frac{x}{1 + x^2}$ on $[0, 1]$. The area under it is the limit you were trying to compute. The function rises from $0$ to its peak of $\frac{1}{2}$ at $x = 1$, and its integral over $[0, 1]$ works out to $\frac{\ln 2}{2} \approx 0.347$ — just a little under half the area of the unit square.

The picture is worth a moment. The sum you started with was adding up values of f at n equally-spaced points and scaling by \frac{1}{n}. For n = 100, you are building a staircase of 100 narrow rectangles that approximates the shaded area. As n \to \infty, the staircase smooths out into the exact area — and the area is the integral.

A more algebraic example

The second example is harder on the algebra side but the same recipe.

Example 2: $\displaystyle\lim_{n\to\infty}\frac{1^p + 2^p + 3^p + \cdots + n^p}{n^{p+1}}$ for $p > 0$

Step 1. Recognise the sum in the numerator: \sum_{r=1}^n r^p. So the limit is

\lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{r=1}^n r^p

Why: the dividing by n^{p+1} looks odd, but there is a reason — you want the \frac{1}{n} out front and r/n inside the function.

Step 2. Rewrite each term.

\frac{r^p}{n^{p+1}} = \frac{1}{n}\cdot\frac{r^p}{n^p} = \frac{1}{n}\left(\frac{r}{n}\right)^p

Why: split the n^{p+1} in the denominator as n\cdot n^p. The n^p combines with r^p to form (r/n)^p. What is left is a clean \frac{1}{n} outside.

Step 3. Read off f and integrate.

\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\left(\frac{r}{n}\right)^p = \int_0^1 x^p\,dx = \frac{x^{p+1}}{p+1}\bigg|_0^1 = \frac{1}{p+1}

Result: \displaystyle\lim_{n\to\infty}\frac{1^p + 2^p + \cdots + n^p}{n^{p+1}} = \dfrac{1}{p+1} for any p > 0.

The functions $y = x$, $y = x^2$, and $y = x^3$ on $[0, 1]$. The areas under them are $\frac{1}{2}$, $\frac{1}{3}$, and $\frac{1}{4}$ respectively — matching the result $\frac{1}{p+1}$. These are the limits of $\frac{1 + 2 + \cdots + n}{n^2}$, $\frac{1 + 4 + 9 + \cdots + n^2}{n^3}$, and $\frac{1 + 8 + 27 + \cdots + n^3}{n^4}$.

This result is worth savouring. The sums 1 + 2 + 3 + \cdots + n, 1^2 + 2^2 + \cdots + n^2, and 1^3 + 2^3 + \cdots + n^3 have closed-form formulas — \frac{n(n+1)}{2}, \frac{n(n+1)(2n+1)}{6}, \frac{n^2(n+1)^2}{4} — but those formulas do not extend to non-integer p. The integral does. For p = 1/2 (summing \sqrt{r}), the integral approach gives \frac{1}{3/2} = \frac{2}{3} — a clean answer for a sum that has no elementary closed form.

A product-turned-into-sum example

The third flavour is products. If you have a product of terms, take the logarithm — a log of a product becomes a sum, and the sum may be a Riemann sum.

Example. Evaluate \displaystyle\lim_{n\to\infty}\left[\frac{(n+1)(n+2)\cdots(n+n)}{n^n}\right]^{1/n}.

Take the log of the expression inside the limit:

\ln\left[\frac{(n+1)(n+2)\cdots(2n)}{n^n}\right]^{1/n} = \frac{1}{n}\sum_{r=1}^n \ln\!\left(\frac{n + r}{n}\right) = \frac{1}{n}\sum_{r=1}^n\ln\!\left(1 + \frac{r}{n}\right)

Now it is a Riemann sum with f(x) = \ln(1 + x). The limit of the log is

\int_0^1 \ln(1 + x)\,dx = \left[(1 + x)\ln(1 + x) - (1 + x)\right]_0^1 = 2\ln 2 - 1

Exponentiate to undo the log:

\lim_{n\to\infty}\left[\frac{(n+1)(n+2)\cdots(2n)}{n^n}\right]^{1/n} = e^{2\ln 2 - 1} = \frac{4}{e}

\frac{4}{e} \approx 1.472. A clean answer for a limit that, attacked directly, would be impenetrable.

Common confusions

Going deeper

The core technique is what you have just seen. The rest of this section looks at why the Riemann-sum rule is really the fundamental theorem of calculus in disguise, how to handle sums whose dominant interval is not [0, 1], and a couple of spectacular applications.

The connection to the fundamental theorem

The definition of the integral as a limit of Riemann sums predates the fundamental theorem by several centuries. In its earliest form — Archimedes' computation of the area of a parabolic segment — it is exactly the three-step recipe above, except Archimedes did not have the notation. When Newton and Leibniz in the 1670s proved the fundamental theorem, they turned this clumsy limit-of-sum into a clean antiderivative calculation.

The miracle of the fundamental theorem is that it makes the Riemann-sum interpretation optional: if you want a definite integral, you can compute it by finding an antiderivative and subtracting. But in limit-of-sum problems, the interpretation runs in the other direction — from sum to integral — and that direction is where the power is. You are using the theorem to replace a sum that has no closed form with an integral that does.

Sampling from [a, b] instead of [0, 1]

For the general-interval rule, here is a concrete example.

\lim_{n\to\infty}\frac{1}{n}\sum_{r=1}^n\sqrt{1 + \frac{r}{n}}

Here f(x) = \sqrt{1 + x}, and r/n runs over [0, 1] — so in fact this is the [0, 1] rule applied to \sqrt{1 + x}:

\int_0^1\sqrt{1 + x}\,dx = \frac{2}{3}(1 + x)^{3/2}\bigg|_0^1 = \frac{2}{3}(2\sqrt{2} - 1) = \frac{4\sqrt{2} - 2}{3}

But if the sum had been \frac{1}{n}\sum_{r=n+1}^{2n}\sqrt{r/n}, with r running from n + 1 to 2n, then r/n sweeps [1, 2] and the integral is \int_1^2\sqrt{x}\,dx = \frac{2}{3}(2\sqrt{2} - 1). Same answer — which is not a coincidence; the substitution x \mapsto 1 + x makes them identical.

Stirling's approximation as a sum-to-integral result

One of the most famous applications of this technique is Stirling's approximation for n!. Take the log:

\ln(n!) = \ln 1 + \ln 2 + \cdots + \ln n = \sum_{r=1}^n \ln r

The sum is not directly a Riemann sum, but you can compare it to the integral \int_1^n \ln x\,dx = n\ln n - n + 1. More carefully (using the trapezoidal rule or Euler–Maclaurin), you get

\ln(n!) \approx n\ln n - n + \frac{1}{2}\ln(2\pi n)

which exponentiates to Stirling's formula:

n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n

This is the link between a discrete product and a smooth integral, and it is the same idea you have been using — a sum compared to an integral — pushed to its logical conclusion.

What about sums that aren't Riemann sums?

Not every sum submits to this trick. The classic non-example is

\sum_{r=1}^n\frac{1}{r^2} \to \frac{\pi^2}{6}

This sum does converge, but the limit is not a Riemann sum for any elementary function. The dominant contribution is from small r, not from r \sim n, so there is no f(r/n) structure to exploit. Evaluating it requires Fourier series or generating-function tricks.

The rule of thumb: if the summand depends on r/n in a smooth way, the Riemann-sum trick works. If the summand depends on r alone (like 1/r^2) and the role of n is only in the range, not in the expression, the trick does not apply.

Where this leads next