This problem is linked to a physical problem of linear system (structural analysis of a beam with $n+2$ supports), so this matrix $A$ should be invertible. The matrix is $n\times n$ and has elements $a_{i,j}$ given by $$a_{i,j}=\int_0^L\phi_i(x)\phi_j(x)dx$$ for $i,j=${$1,2,...,n$}, $L$ is the length of the beam and $$\phi_i(x)=\frac{x(L-x_i)}{L}$$ if $x<x_i$ and $$\phi_i(x)=\frac{x_i(L-x)}{L}$$ if $x>x_i$.
$x_i$ are the positions of the "internal supports", with $0<x_1<x_2<...<x_n<L$. I already see that $A$ is symmetric, which means that it is diagonalizable. So, if I could prove that $A$ is positive definite then it would have all eingenvalues positive, and since $A$ is diagonalizable then its determinant is the product of its eingenvalues, which means it is invertible.
Yes, $A$ is positive-definite for "reasonable" basis functions $\phi$.
Hint: for any vector $v$,
\begin{align*} v^TAv &= \sum_{i,j} v_i \left(\int_0^L \phi_i(x)\phi_j(x)\,dx \right)v_j \\ &= \int_0^L \sum_{i,j} v_i\phi_i(x)v_j\phi_j(x)\,dx\\ &= \int_0^L \left(\sum_i v_i\phi_i(x)\right)^2\,dx \end{align*}
and can you take it from here? You should now also see the conditions necessary on $\phi$ for $A$ to be positive-definite.
Stepping back, you should recognize that the $A$ matrix represents an $L^2$ inner product on the discrete function space spanned by the basis elements $\phi$, and so must be positive-definite.