Fixing the Integration Constant
An indefinite integral is not a single polynomial: an antiderivative may be changed by adding any constant. Fixing the lower limit at \(0\) removes that ambiguity. The resulting map sends each polynomial \(p\) to a particular polynomial in \(x\), namely the accumulated integral from \(0\) to \(x\). Its outputs always vanish at \(x=0\), a fact that will determine both its image and whether it is onto a chosen codomain.
For an integer \(n\geq 0\), let \(P_n(\mathbb{R})\) denote the vector space of real polynomials of degree at most \(n\), including the zero polynomial. We define integration coefficient by coefficient: if \(p(t)=\sum_{k=0}^{n}a_k t^k\), then \(\int_0^x p(t)\,dt=\sum_{k=0}^{n}\frac{a_k}{k+1}x^{k+1}\). This follows from integrating each monomial and evaluating the resulting antiderivative at the endpoints \(0\) and \(x\).
The coefficient formula shows why this codomain works: integrating a polynomial of degree at most \(n\) produces one of degree at most \(n+1\). It also shows that the constant term of every output is zero. The zero constant term is not a claim that every polynomial in the codomain can occur; it is a restriction on the outputs within that codomain.
Linearity and the Coefficient Formula
A map between vector spaces is linear when it preserves addition and scalar multiplication. Here both properties can be checked directly from coefficients. If \(p(t)=\sum_{k=0}^n a_k t^k\) and \(r(t)=\sum_{k=0}^n c_k t^k\), and if \(\alpha,\beta\in\mathbb{R}\), then \(\alpha p+\beta r\) has coefficient \(\alpha a_k+\beta c_k\) on \(t^k\). Integrating that coefficient gives \(\frac{\alpha a_k+\beta c_k}{k+1}x^{k+1}\), which is the corresponding coefficient in \(\alpha T_n(p)+\beta T_n(r)\).
Proof. For each \(k\) from \(0\) to \(n\), the antiderivative of \(a_k t^k\) that is zero at \(0\) is \(\frac{a_k}{k+1}x^{k+1}\), since its value at \(x=0\) is \(0\). Adding these antiderivatives gives the displayed formula. For \(p,r\in P_n(\mathbb{R})\) and \(\alpha,\beta\in\mathbb{R}\), write their coefficients as \(a_k,c_k\). The coefficient of \(x^{k+1}\) in \(T_n(\alpha p+\beta r)\) is \(\frac{\alpha a_k+\beta c_k}{k+1}=\alpha\frac{a_k}{k+1}+\beta\frac{c_k}{k+1}\), which is the coefficient of \(x^{k+1}\) in \(\alpha T_n(p)+\beta T_n(r)\). Both polynomials have zero constant term, so their coefficients agree at every power. Therefore \(T_n(\alpha p+\beta r)=\alpha T_n(p)+\beta T_n(r)\), proving linearity. \(\square\)
Worked Example: Integrating a Quadratic
Let \(n=2\) and \(p(t)=3-2t+6t^2\). Apply \(T_2\) term by term:
The output is in \(P_3(\mathbb{R})\), as expected. To check the computation, differentiate \(3x-x^2+2x^3\): its derivative is \(3-2x+6x^2\), the original polynomial with its variable written as \(x\). The output also has value \(0\) at \(x=0\), as required by the lower limit.
Kernel, Image, and the Exact Output Space
For a linear map \(T:V\to W\), the kernel is the set of domain vectors mapped to the zero vector, and the image is the set of all outputs \(T(\mathbf{v})\) as \(\mathbf{v}\) ranges over the domain. In this polynomial setting, the zero vector is the zero polynomial. We can find both sets by using the coefficient formula, rather than trying to infer them from a few examples.
Proof. Write an arbitrary domain polynomial as \(p(t)=\sum_{k=0}^n a_k t^k\). If \(T_n(p)\) is the zero polynomial, then \(\sum_{k=0}^n\frac{a_k}{k+1}x^{k+1}=0\). Uniqueness of polynomial coefficients implies \(\frac{a_k}{k+1}=0\) for every \(k\). Since each \(k+1\) is nonzero, every \(a_k=0\), so \(p=0\). Conversely, \(T_n(0)=0\), so the kernel is exactly \(\{0\}\).
Every output has the displayed form because the coefficient formula contains only powers \(x^1,\ldots,x^{n+1}\). For the reverse inclusion, take any \(q(x)=\sum_{j=1}^{n+1}b_jx^j\) and define \(p(t)=\sum_{j=1}^{n+1}j b_jt^{j-1}\), which lies in \(P_n(\mathbb{R})\). Applying the coefficient formula gives \(T_n(p)(x)=\sum_{j=1}^{n+1}\frac{j b_j}{j}x^j=q(x)\). Thus every polynomial of the displayed form is an output. Finally, a polynomial in \(P_{n+1}(\mathbb{R})\) has value zero at \(0\) exactly when its constant coefficient is zero, so this description is equivalent to the condition \(q(0)=0\). \(\square\)
The set described in the theorem is also a polynomial space in its own right: it is \(\operatorname{span}\{x,x^2,\ldots,x^{n+1}\}\). This gives a sharper possible codomain than all of \(P_{n+1}(\mathbb{R})\). Choosing a codomain is part of specifying a map, so the codomain choice matters when asking whether the map is onto.
Injectivity and Surjectivity Depend on the Codomain
A map is injective if distinct domain elements always have distinct outputs. It is surjective onto its stated codomain if every element of that codomain is an output. The zero kernel proves injectivity here directly: if \(T_n(p)=T_n(r)\), linearity gives \(T_n(p-r)=0\); the kernel result gives \(p-r=0\), so \(p=r\). Conversely, \(T_n\) cannot identify two different polynomials.
With codomain \(P_{n+1}(\mathbb{R})\), the map is not surjective: the constant polynomial \(1\) belongs to that codomain but cannot be an output, because every output has value zero at \(0\). If instead we declare the codomain to be \(W_n=\operatorname{span}\{x,x^2,\ldots,x^{n+1}\}\), the theorem proves that every element of \(W_n\) is an output, so the same formula defines a surjective map \(T_n:P_n(\mathbb{R})\to W_n\). Its injectivity is unchanged.
Worked Example: Finding a Preimage in the Image
For \(n=2\), find a polynomial \(p\in P_2(\mathbb{R})\) that maps to \(q(x)=4x+3x^2-2x^3\).
The coefficient formula says the coefficient of \(x^{k+1}\) in the output is the coefficient of \(t^k\) in the input divided by \(k+1\). Therefore the input coefficients must be \(4\), \(2\cdot3=6\), and \(3\cdot(-2)=-6\). Set \(p(t)=4+6t-6t^2\). Then
As a check, the derivative of \(4x+3x^2-2x^3\) is \(4+6x-6x^2\), and the output is zero at \(x=0\). The theorem also says this preimage is unique: if another polynomial had the same output, injectivity would force it to equal \(p\).
Worked Example: A Polynomial the Map Cannot Reach
For \(n=2\), test whether the constant polynomial \(q(x)=1\) is an output when the codomain is \(P_3(\mathbb{R})\). Write a general input \(p(t)=a+bt+ct^2\). The coefficient formula gives
To equal \(1\), the constant coefficient of this expression would need to be \(1\). But that coefficient is \(0\) for every choice of \(a,b,c\). Hence no input maps to \(1\), and \(T_2:P_2(\mathbb{R})\to P_3(\mathbb{R})\) is not surjective. This agrees with the general fact that every output vanishes at zero, whereas \(q(0)=1\).
Reading the Map in Monomial Coordinates
The coefficient rule can also be written as a matrix action. The ordered monomial basis for \(P_2(\mathbb{R})\) is \((1,t,t^2)\); for \(P_3(\mathbb{R})\), use \((1,x,x^2,x^3)\). The coefficient column \((a,b,c)^\top\) represents \(a+bt+ct^2\), and its output has coefficient column \((0,a,b/2,c/3)^\top\). Thus the matrix of \(T_2\) in these bases is
Worked Example: Checking the Coordinate Matrix
Use the preceding matrix for \(p(t)=3-2t+6t^2\), whose domain coordinate column is \(\begin{bmatrix}3&-2&6\end{bmatrix}^{\top}\). Multiplying gives
The result is the coefficient column of \(3x-x^2+2x^3\), matching the direct integration. In these monomial bases, the first row is zero because no output has a constant term. If the codomain is instead \(W_2\), use the basis \((x,x^2,x^3)\); the coordinate column of the output is then \((a,b/2,c/3)^\top\), and the matrix is \(\begin{bmatrix}1&0&0\\0&1/2&0\\0&0&1/3\end{bmatrix}\). That matrix also displays why every vector in the chosen codomain has a preimage.
Common Mistake: Confusing an Antiderivative with This Map
A common incorrect calculation is to say \(T_n(1)=1\), on the grounds that an antiderivative of \(1\) is \(1\). That is false: the derivative of \(1\) is \(0\), not \(1\). The antiderivative of \(1\) is \(t+C\), and the condition that the integral from \(0\) to \(x\) is zero at \(x=0\) selects \(x\). Indeed, \(T_n(1)(x)=\int_0^x1\,dt=x\). Keeping the lower-limit condition in view prevents the arbitrary constant from being mistaken for part of the output.
A related mistake is to say the map is simply “onto” without naming its codomain. It is not onto \(P_{n+1}(\mathbb{R})\), since constants other than zero cannot occur. It is onto \(W_n=\operatorname{span}\{x,\ldots,x^{n+1}\}\), since every polynomial in that space has a preimage. In both versions it is injective, and its kernel remains \(\{0\}\).
Check Your Understanding
Use the coefficient formula and the declared codomain to justify each answer.
- For \(n=3\), compute \(T_3(2-t+3t^2)\) and check it by differentiation and evaluation at zero.
- Describe \(\operatorname{im}T_1\) as a span and state whether \(T_1:P_1(\mathbb{R})\to P_2(\mathbb{R})\) is surjective.
- Find the unique input in \(P_2(\mathbb{R})\) that maps to \(2x-x^2+5x^3\).
- Explain why the only polynomial in the kernel of \(T_n\) is the zero polynomial.
- For which codomain is the integration map surjective: \(P_{n+1}(\mathbb{R})\), or \(\operatorname{span}\{x,\ldots,x^{n+1}\}\)? Explain the role of the constant term.
