Answer
The preimage of $1$ is $\displaystyle a{x^2} + bx + \left( {1 - \frac{a}{3} - \frac{b}{2}} \right)$
Work Step by Step
The domain is $P_2$, which consists of all polynomials of degree at most $2$.
So, let the preimage of $1$ be $f(x)=ax^2+bx+c$.
Then $T(f(x))=1$
$\displaystyle \begin{array}{l}
\displaystyle\int\limits_0^1 {(a{x^2} + bx + c)} dx = 1\\
\displaystyle \Rightarrow \left. {\left[ {a\frac{{{x^3}}}{3} + b\frac{{{x^2}}}{2} + cx} \right]} \right|_{x = 0}^{x = 1} = 1\\
\displaystyle \Rightarrow \left[ {a\frac{{{1^3}}}{3} + b\frac{{{1^2}}}{2} + c} \right] - 0 = 1\\
\displaystyle \Rightarrow \frac{a}{3} + \frac{b}{2} + c = 1\\
\displaystyle \Rightarrow c = 1 - \frac{a}{3} - \frac{b}{2}
\end{array}$
Hence, the preimage of $1$ is $\displaystyle a{x^2} + bx + \left( {1 - \frac{a}{3} - \frac{b}{2}} \right)$
