Why does the integration function map the polynomial vector space to the real vector space?

I'm reading a linear algebra book and it defines the integration operation as $T \epsilon L(P(\Bbb R), \Bbb R)$. However, it defines the differentiation operation as T $\epsilon$ $L(P(\Bbb R), P(\Bbb R))$. Don't they both map to the vector space that contains all polynomials? Why is integration as a linear function defined this way?

• Is the integration definite, i.e. with fixed bounds? Then, in contrast to indefinite integration, which indeed gives you a polynomial again, you would get a real number for every definite integral of a polynomial. – DominikS Jul 28 '17 at 16:03
• Oh, I see. That's why since it's, indeed, defined with fixed bounds. Would it be defined the same way as differentiation if it was boundless? – Shocked Jul 28 '17 at 16:04
• I saw your edited comment. Thanks. – Shocked Jul 28 '17 at 16:05

Because integration gives you a real number value

Integration is considered a transformation T: $P(\mathbb R) \to \mathbb R$ because a definite integral of the form $\int_a^b p(x) dx$ will typically give you a real number value: i.e. the area under the curve of p(x) on [a,b]. For example, consider: $$T(p(x))= \int_0^1 p(x) dx$$

Let's take this transformation for $p(x)=x^2$: $$\int_0^1 x^2 dx = \frac{1}{3}$$

The transformation T has taken our polynomial, $x^2$, which is in the set $P(\mathbb R)$, and produced a fraction, which is in the set $\mathbb R$, so for any polynomial in the set, the transformation (the definite integral on [0,1] in this case) will produce a real number. That's why we consider the integral to be a linear transformation between these two spaces.

A derivative applied to a first order polynomial $(ax+C)$ will give you a real number in the same way.

However, for the indefinite integral $\int p(x) dx$, you are correct that most of the time you will get a polynomial back as an answer, and the derivative of a higher-order polynomial will also give you a polynomial.

TL;DR: the indefinite integral will map to $\mathbb R$ but the definite integral and most derivatives tend to remain in $P(\mathbb R)$.