Matrix of Integration with a Fixed Constant Convention

Integration as a Linear Map

Every polynomial has many antiderivatives: adding any constant to one antiderivative gives another. To represent integration as a map, we must choose one antiderivative for each input. Requiring the result to be zero at the origin makes that choice precise. It also determines the first row of the matrix: every output has constant coefficient zero.

Work over \(\mathbb{R}\). For \(n\geq 0\), let \(P_n\) denote the vector space of real polynomials of degree at most \(n\), together with the zero polynomial. Its ordered monomial basis is \(\mathcal{B}_n=(1,x,\ldots,x^n)\). Thus if \(p\in P_n\), there are unique real coefficients \(a_0,\ldots,a_n\) such that \(p(x)=a_0+a_1x+\cdots+a_nx^n\). We use \(\mathcal{B}_{n+1}=(1,x,\ldots,x^{n+1})\) for the codomain.

Integrating each monomial gives an explicit formula for the output. If \(p(x)=\sum_{k=0}^{n}a_kx^k\), then

In particular, the degree of \(T(p)\) is at most \(n+1\), as required for the codomain. The formula also shows that \(T\) is linear: for \(p,q\in P_n\) and \(a,b\in\mathbb{R}\), integration of the coefficientwise sum gives \(T(ap+bq)=aT(p)+bT(q)\). The zero-at-origin condition is essential to this formula; an additional constant term would change the map.

Building the Matrix from Monomials

The matrix of \(T\) relative to \(\mathcal{B}_n\) in the domain and \(\mathcal{B}_{n+1}\) in the codomain has \(n+2\) rows and \(n+1\) columns. Its \(j\)th column is the coordinate vector of \(T(x^{j-1})\) in \(\mathcal{B}_{n+1}\). Since \(T(x^{j-1})=x^j/j\), that column has \(1/j\) in row \(j+1\) and zero in every other row.

Proof. Assume \(p\in P_n\), and write \(p(x)=\sum_{k=0}^{n}a_kx^k\). The \(j\)th domain basis vector is \(x^{j-1}\), and its image is \(T(x^{j-1})=\int_0^x t^{j-1}\,dt=x^j/j\). In the ordered codomain basis \((1,x,\ldots,x^{n+1})\), the coordinate of \(x^j/j\) is \(1/j\) in position \(j+1\), with all other coordinates zero. Therefore these image coordinate vectors are exactly the columns specified in the theorem. Multiplying the matrix by the input coordinate vector forms the same linear combination of those columns as \(p\) forms of its basis vectors: \(\sum_{j=1}^{n+1}a_{j-1}[T(x^{j-1})]_{\mathcal{B}_{n+1}}\). By linearity this is \([T(p)]_{\mathcal{B}_{n+1}}\), proving the stated coordinate equation. \(\square\)

Written out, the matrix has a zero first row, followed by the scaled entries for \(x,x^2,\ldots,x^{n+1}\):

The first row is not an omission. It records that the constant coefficient of every output is zero. The denominators come from integrating the input monomials; the entries are not all \(1\), because \(x^k\) integrates to \(x^{k+1}/(k+1)\).

Worked Examples

Reading the Matrix and Avoiding a Common Error

Matrix multiplication makes the coefficient rule visible. If the input coefficients are \(a_0,\ldots,a_n\), then the output coefficient of \(1\) is \(0\), the coefficient of \(x\) is \(a_0\), the coefficient of \(x^2\) is \(a_1/2\), and, in general, the coefficient of \(x^{k+1}\) is \(a_k/(k+1)\). Integration shifts each input coefficient up by one power and divides it by the new exponent.

Common mistake: putting the input constant coefficient into the output constant position. A tempting but incorrect first column for the matrix is \((1,1,0,\ldots)^\top\), on the mistaken idea that integrating \(1\) leaves a constant \(1\) and adds \(x\). That would claim \(\int_0^x 1\,dt=1+x\). But \(\int_0^x 1\,dt=x\): the lower limit forces the value at the origin to be zero. The correct first column has \(0\) in the constant row and \(1\) in the \(x\) row.

The distinction between the domain and codomain bases is important when reading coordinates. The input coefficient of \(x^k\) occupies position \(k+1\) in \(\mathcal{B}_n\); after integration, its contribution occupies position \(k+2\) in \(\mathcal{B}_{n+1}\), scaled by \(1/(k+1)\). The matrix is rectangular because the domain has \(n+1\) basis vectors while the codomain has \(n+2\). The extra codomain coordinate is the highest possible power \(x^{n+1}\); the constant coordinate, although part of the codomain basis, is always zero for outputs of this particular map.

This representation is useful whenever polynomial inputs and outputs are handled by their coefficient lists. It replaces an integration rule with an ordinary matrix-vector multiplication while preserving the chosen constant convention. Changing that convention can change not just the matrix entries but whether the rule is linear at all.