= Solution
{c}
The monic degree-zero polynomial is $\phi_0=1$. Orthogonality of $x-a$ to one gives $a=\int_0^1x\,dx=1/2$, so $\phi_1=x-1/2$.
The relevant integrals are
$$
\|\phi_0\|_2^2=1,\qquad \|\phi_1\|_2^2=\frac1{12},\qquad
(\phi_1,x\phi_1)=\frac1{24}.
$$
Insert these into the <three-term recurrence for monic orthogonal polynomials>:
$$
\phi_2=(x-\tfrac12)(x-\tfrac12)-\frac1{12}
=x^2-x+\frac16.
$$
Thus
$$
\boxed{\phi_0=1,\qquad\phi_1=x-\tfrac12,\qquad
\phi_2=x^2-x+\tfrac16.}
$$
Their mutual orthogonality can also be checked directly by integrating on $[0,1]$.
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