The Cubic Function and Sums of Cubes

From Cubic Values to an Integral

The function \(f(x)=x^3\) changes across an interval, so its integral is not simply the function value at one point multiplied by the interval length. The Riemann-sum definition instead divides the interval into smaller pieces, samples the function on each piece, and takes a limit as the pieces become narrow.

For the cubic function, a uniform partition makes each sample point an integer multiple of a single step size, after accounting for the left endpoint. Expanding the cube then reduces the Riemann sum to familiar finite sums. The sum-of-cubes formula is especially useful: its exact expression makes the limiting value visible.

Uniform Right-Endpoint Sums

Fix real numbers \(a<b\), and let \(n\in\mathbb{N}\) be the number of subintervals. The common width of the subintervals is

The partition points are \(x_k=a+k\Delta x\), for \(k=0,1,\ldots,n\). On the \(k\)-th subinterval \([x_{k-1},x_k]\), choose the right endpoint \(x_k\) as the sample point. The corresponding Riemann sum for \(x^3\) is

The mesh size, meaning the largest subinterval width, is \((b-a)/n\), which tends to zero as \(n\) increases. The function \(x^3\) is continuous on the closed bounded interval \([a,b]\), so it is Riemann integrable there. By the definition of the definite integral as a limit of tagged sums, these right-endpoint sums converge to the integral.

The Finite-Sum Formulas We Need

Expanding \((a+k\Delta x)^3\) produces sums of \(1\), \(k\), \(k^2\), and \(k^3\). For every positive integer \(n\), the needed identities are

The sum-of-cubes identity can be justified directly by induction. For \(n=1\), the sum is \(1\), which equals \((1\cdot2/2)^2\). Suppose the identity holds for \(n\). Adding the next cube gives

This is the required formula with \(n+1\) in place of \(n\), so induction proves it for every positive integer \(n\). The formulas for the first and second power sums are the standard finite-sum identities; we will use them alongside the cubic identity.

Deriving the Cubic Integral

Write \(h=b-a\), so \(\Delta x=h/n\) and \(x_k=a+kh/n\). Substitute this into the right-endpoint sum and expand the cube using \((u+v)^3=u^3+3u^2v+3uv^2+v^3\):

Now apply the four finite-sum formulas. Keeping each contribution separate makes the powers of \(n\) and the limit easier to check:

Cancel the common factors in each term. For the term involving squares, use \((n+1)(2n+1)=2n^2+3n+1\). This gives an exact expression for every \(n\):

As \(n\to\infty\), both \(1/n\) and \(1/n^2\) tend to zero. Therefore the sums converge to

Since \(h=b-a\), the binomial expansion of \((a+h)^4\) shows that this expression is \(((a+h)^4-a^4)/4\). Also \(a+h=b\). Thus the integral is

This derivation does not depend on whether the interval lies above, below, or across zero. The endpoints enter through the partition points, and the same finite-sum calculation applies in all three situations.

Worked Examples

A Common Indexing and Scaling Mistake

A frequent error is to confuse the right endpoint \(x_k=a+kh/n\) with \(kh/n\), which is only correct when \(a=0\). On \([1,3]\), for instance, \(h=2\), so the right endpoint is \(1+2k/n\), not \(2k/n\). Omitting the \(1\) changes every sampled function value.

Another common error is to use the cube-sum identity without preserving its square. For example, the incorrect calculation \(\sum_{k=1}^n k^3=n(n+1)/2\) gives the first-power sum, not the sum of cubes. At \(n=2\), the actual sum is \(1^3+2^3=9\), while that incorrect expression gives \(2\cdot3/2=3\). The correct formula gives \((2\cdot3/2)^2=9\).

Why the Sum-of-Cubes Formula Matters

The cubic function is a simple example of a broader Riemann-sum strategy. The partition gives sample points in a regular pattern; algebra turns function values into powers of the index \(k\); finite-sum identities then produce an exact expression in \(n\). The integral is obtained only after examining the limit of that expression.

For this function, the terms containing \(1/n\) or \(1/n^2\) record finite-partition corrections. They affect the approximating sums but disappear as the mesh shrinks. The remaining terms combine into the difference of fourth powers of the endpoints. This is a direct demonstration of how an accumulation limit can produce a compact endpoint formula.