Does a definite integral define a linear functional? Would $\displaystyle\int_0^1 t^2x(t)\,dt$ be a linear functional? 
For each $x$ in $P$ the function $y$ is defined by $\displaystyle\int_0^1 t^2x(t)\,dt$. I have to show that $y(ax+bz) = ay(x)+by(z)$.  I am confused as to whether or not the $t^2$ term messes it up or not.
 A: I'm just going to apologize in advance. If there is any lesson to be learned here, it's that

*

*clearly stating your goals and cast of characters, as well as

*good notation

both help immensely.

SCENE I
Enter Vector Space $P$, the space of polynomials with real coefficients and variable $t$, with his servant the Mapmaker.
Vector Space $P$:
 Mapmaker! I had commissioned you to craft a map which, when applied to my vector inhabitants, shall send them linearly to my base field, $\Bbb R$. You have advised the use of a map you call $T$, which takes a vector $f$,  henceforth known as a polynomial, and sends it to the number $$T(f) = \int_0^1 t^2 f(t)~dt.$$
Mapmaker
 I have painstakingly crafted the map, sire; so much of my vigor has been spent on its creation that I lie now, infirm. I am quite certain that the map $T$ is linear, but a splendid Space of your nobility deserves proof. Alas, I am unable to provide this with my remaining strength.
Vector Space $P$:
 Fret not, for you have served me well. Any student would be lucky to apply himself to your work. I shall send for one now.
Vector Space $P$'s guards bring a STUDENT from the local university.
Vector Space $P$:
 Ah, a Student! Young man, I require your assistance. My faithful servant Mapmaker has given me a map $T\colon P \to \Bbb R$ that transforms vectors $f$ into numbers $T(f) = \int_0^1 t^2 f(t)~dt$. While I would that he himself were able to, he has not provided me assurance that this map is linear. I myself was never much for cartography; the late Queen was always quick to remind me that I could not tell a linear map from a quadratic form. But enough dithering about.
STUDENT:
 Yes my liege, at once.
I shall take your two plainest subjects, the vectors $f$ and $g$, as well as the brutish and arbitrary scalar inhabitants $\alpha$ and $\beta$ of the base field $\Bbb R$, to verify that indeed $$T(\alpha f + \beta g) = \alpha T(f) + \beta T(g).$$
The task is easy at first; the map clearly dictates that
$$T(\alpha f + \beta g) = \int_0^1 t^2\big(\alpha f(t) + \beta g(t)\big)~dt,$$
to which a quick distribution yields that we have in fact possessed
$$\int_0^1 \big(\alpha t^2 f(t) + \beta t^2 g(t)\big)~dt$$ all along. I remember dimly that integrals are themselves linear and may be split across sums
$$\int_0^1 \alpha t^2 f(t)~dt + \int_0^1 \beta t^2 g(t)~dt$$ and have their multipliers forced in front
$$\alpha \int_0^1 t^2 f(t)~dt + \beta\int_0^1 t^2 g(t)~dt.$$
This was of course the key insight, for now it is clear to me that this is $\alpha T(f) + \beta T(g)$, that which was to be demonstrated. $\square$
STUDENT is sent way, while the Court Chaplain arrives to administer the Last Rites to Mapmaker.
A: The "functional"
$$\phi(x):=\int_0^1 t^2\>x(t)\>dt$$
takes as input a function $x:\>t\mapsto x(t)$ and produces as output a number. According to the rules about integrals you have learned in Calculus 101 (and have applied a hundred times ever since) it is obvious that
$$\phi(x+y)=\phi(x)+\phi(y), \qquad \phi(\lambda x)=\lambda\>\phi(x)$$
for all admissible functions $x$, $y$, and all $\lambda\in{\mathbb R}$.
A: Let $F(x) = \int_0^1 t^2 x(t) dt$
$$F(\alpha x(t) + \beta y(t)) = \int_0^1t^2(\alpha x(t) + \beta y(t))dt = \int_0^1\alpha t^2x(t) dt + \int_0^1\beta t^2y(t) dt = \alpha \int_0^1t^2x(t)dt + \beta \int_0^1 t^2y(t) dt = \alpha F(x) + \beta F(y)$$
A: A linear functional is a linear map from a vector space to its scalar field. As runaround has already shown, the integral is linear (which can be rigorously proven in, say, the Riemann or Lebesgue sense), and thus is a linear functional. The $t^2$ is the integrand and does not have to be linear - you can replace it with any integrable function and your integral will still be a linear functional.
