Determinant of a matrix and chech whether it is non negative definite or not 
Let $V = \{ f : [0,1] \to \mathbb R | f$ is a polynomial of degree less than or equal to n $\}$. Let $f_j(x)  = x^j$ for $0\leq j \leq n$ and let $A$ be the $(n+1) \times (n+1)$  given by $a_{ij} = \int_0^1 f_i(x)f_j(x) dx$. Then which of the followings are true.
  
  
*
  
*$\dim V = n$
  
*$\dim V >n$
  
*$A$ is non negative definite.
  
*$\det A >0$.

Clearly the option 1. is false and 2. is correct
I knnow the matrix $A$ is 
$$
\begin{bmatrix}
a_{00} & a_{01} & a_{02} & \cdots & a_{0n}\\
a_{10} & a_{11} & a_{12} & \cdots & a_{1n}\\
a_{20} & a_{21} & a_{22} & \cdots & a_{2n}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
a_{n0} & a_{n1} & a_{n2} & \cdots & a_{nn}\\
\end{bmatrix}
$$
$$A = 
\begin{bmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n+1}\\
\frac{1}{2} & \frac{1}{3} &\frac{1}{4} & \cdots & \frac{1}{n+2}\\
\frac{1}{3}& \frac{1}{4} & \frac{1}{5} & \cdots & \frac{1}{n+3}\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
\frac{1}{n+1} & \frac{1}{n+2} & \frac{1}{n+3} & \cdots & \frac{1}{2n+1}\\
\end{bmatrix}
$$
I am unable to find the determinant of the matrix $A$, so please help to find the determinant of the matrix and the matrix $A$ is non negative definnite or not. Thank you

Hint: for any real vector $v$, express $v^T A v$ as an integral.
