The Square Function and Sums of Squares

Why Squares Lead to an Exact Integral

A definite integral can be defined through limits of Riemann sums. For the square function, \(f(x)=x^2\), uniform partitions make those sums especially manageable: the sample points are equally spaced, and their squared values involve the familiar sum \(1^2+2^2+\cdots+n^2\).

The calculation has two important parts. First, we need an exact formula for that sum of squares. Then we use the formula inside a Riemann sum and let the number of subintervals grow. This gives an exact value for the integral without using antiderivatives.

The factor \(\Delta x\) represents the width of each subinterval. A Riemann sum adds the sampled function value multiplied by that width; the sum of sampled values by itself is not the Riemann sum.

The Formula for the First \(n\) Squares

The key finite-sum identity is

Here \(n\) is a positive integer. We prove the identity by induction, so the formula is justified for every positive integer rather than inferred from a few examples.

For completeness, the simplification in the last step is

The final expression is exactly \((n+1)((n+1)+1)(2(n+1)+1)/6\). Thus the identity holds at \(n+1\), completing the induction.

We will also use \(\sum_{k=1}^{n}k=n(n+1)/2\). This follows by pairing the first and last terms: each pair has sum \(n+1\), and the resulting formula applies whether \(n\) is even or odd.

First Calculation: The Integral on \([0,1]\)

A Second Interval

General Interval: The Riemann-Sum Derivation

Let \(a<b\) be real numbers. Divide \([a,b]\) into \(n\) equal pieces, so that \(\Delta x=(b-a)/n\). To keep the algebra readable, set \(L=b-a\). The right endpoints are \(x_k=a+kL/n\), and the right-endpoint sum is

Expand the square and separate the sums. The constant term occurs \(n\) times, the terms proportional to \(k\) use the sum of the first \(n\) integers, and the terms proportional to \(k^2\) use the sum-of-squares formula:

As \(n\to\infty\), the terms containing \(1/n\) or \(1/n^2\) vanish. Consequently,

Since \(L=b-a\), this limit can be written in terms of the endpoints. Indeed,

The middle equality follows by expanding \((a+L)^3=a^3+3a^2L+3aL^2+L^3\). This computation explains why the endpoint expression emerges: the finite Riemann sum has terms that approach the expansion of the difference of two cubes.

Using the General Calculation

Common Mistake: Leaving Out the Width

A frequent error is to compute the function values at the sample points and add them without multiplying by \(\Delta x\). On \([0,1]\), the right endpoints are \(k/n\). The incorrect calculation would be

This expression is approximately \(n/3\), so it grows rather than approaching the integral. The mistake is that it adds heights, not rectangle areas. Each height must be multiplied by the subinterval width \(1/n\):

Another error is to use the sum-of-squares formula with an incorrect power of \(n\). For \([0,1]\), the sampled square contributes \(k^2/n^2\), and the width contributes a further factor \(1/n\). Together these give \(k^2/n^3\), not \(k^2/n^2\). Keeping the sample value and the width visible in the first line helps prevent this error.

When the Result Applies

The derivation requires a finite interval with \(a<b\), uniform subintervals, and right endpoints \(a+k(b-a)/n\). The function is specifically \(x^2\), which is continuous on every closed bounded interval \([a,b]\). By the earlier result that continuous functions are Riemann integrable, the limit of these tagged sums equals the definite integral. The finite-sum formulas apply for every positive integer \(n\), and the integral value follows by taking \(n\to\infty\).

The procedure is useful because it turns the definition of the integral into a concrete calculation: identify the width and sample points, expand the sampled function, apply finite-sum identities, and take the limit. The sum-of-squares identity is not merely a shortcut; it is the exact algebraic step that makes this Riemann-sum evaluation possible.