Using the facts that $\phi_{0}$ and $\phi_{1}$ are orthogonal to $\phi_{n}$ for $n>1$, we have
$$\begin{aligned}
\int_{a}^{b}(\alpha x+\beta) \phi_{n}(x) d x &=\alpha \int_{a}^{b} x \phi_{n}(x) d x+\beta \int_{a}^{b} 1 \cdot \phi_{n}(x) d x \\
&=\alpha \int_{a}^{b} \phi_{1}(x) \phi_{n}(x) d x+\beta \int_{a}^{b} \phi_{0}(x) \phi_{n}(x) d x \\
&=\alpha \cdot 0+\beta \cdot 0=0
\end{aligned}$$
for $n=2,3,4, \dots$