This result seems to apply to vector spaces over both $\Bbb R$ and $\Bbb C$, so I'll resort to the complex case in the following; the real case is then implied. Thus I take $\langle \cdot, \cdot \rangle$ to be a hermitian inner product. Furthermore, I'll simply assume $A(t)$ has the necessary integrability properties to make things fly; continuity should be sufficient.
Let
$A(t) = [A_{ij}(t)]; \tag{1}$
then by definition
$\int_0^1 A(t) dt = [\int_0^1 A_{ij}(t) dt]. \tag{2}$
Let $\vec x \ne 0$ be any constant vector:
$\vec x = \begin{pmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{pmatrix}; \tag{3}$
for any fixed $t \in [0, 1]$ we have
$0 < \langle \vec x, A(t) \vec x \rangle = \sum_{i, j} A_{ij}(t) \bar x_i x_j, \tag{4}$
since the $A(t)$ are positive definite. Integrating (4) over $[0, 1]$ we see that
$0 < \int_0^1 \langle \vec x, A(t) \vec x \rangle dt = \int_0^1 (\sum_{i, j} A_{ij}(t) \bar x_i x_j) dt; \tag{5}$
by linearity of the integral we have
$\int_0^1 (\sum_{i, j} A_{ij}(t) \bar x_i x_j) dt = \sum_{i, j} \bar x_i x_j \int_0^1 A_{ij}(t) dt = \langle x, (\int_0^1 A(t) dt) x \rangle. \tag{6}$
Combining (5) and (6) yields
$\langle x, (\int_0^1 A(t) dt) x \rangle > 0 \tag{7}$
for every non-zero vector $x$; $\int_0^1 A(t) dt$ is thus positive definite.
Note: I don't see exactly how looking at $\text{Tr}(A)$ and $\det(A)$ can help much with this one, since $\int \det A(t) dt$ and $\det \int A(t) dt$ are not obviously related. In any event, it appears that $\text{Tr}(A) > 0$ and $\det (A) > 0$ are determinative for positive definiteness in only the $2 \times 2$ case; the above argument binds for matrices of any size $n$. End of Note.
Hope this helps. Cheers,
and as ever,
Fiat Lux!!!
For a Hermitian matrix $A$ to be positive semi-definite, it is necessary for its leading principal minors to be non-negative, but it is not sufficient as the following example shows:
Consider the matrix
$$
A =
\begin{bmatrix}
0 & 0 \\
0 & -1 \\
\end{bmatrix}
$$
with both leading principle minors
$$
M_{0} = \det(A) = 0*(-1) - 0*0 = 0,
$$
$$
M_{1} =
\det(A_{2,2}) =
\det(\begin{bmatrix} 0 & \square \\ \square & \square \\ \end{bmatrix}) =
\det(\begin{bmatrix} 0 \end{bmatrix}) = 0,
$$
non-negative, but $A$ is not positive semi-definite as it has a negative eigenvalue $-1$.
To check if a Hermitian matrix $A$ is positive semi-definite one has to test if all principal minors (not only the leading principal minors) are non-negative. (proof)
If we look at the example above, the principal minors are
$$
M_{0,0} = \det(A) = M_0 = 0,
$$
$$
M_{1,1} =
\det(A_{1,1}) =
\det(\begin{bmatrix} \square & \square \\ \square & -1 \\ \end{bmatrix}) =
\det(\begin{bmatrix} -1 \end{bmatrix}) = -1,
$$
$$
M_{2,2} = \det(A_{2,2}) = M_1 = 0.
$$
We see that $M_{1,1}$ is negative, so the matrix is not positive semi-definite.
Best Answer
Let $H_k$ be the $k$th Hankel matrix of the sequence.
For the forward direction of the equivalence, we use the fact that positive definiteness of each $H_k$ individually implies positive determinant of $H_k$.
In more detail: Akhiezer defines a positive sequence as one in which every quadratic (Hankel) form is positive definite. This translates to $c^T H_k c > 0$ for all nonzero $c$. For example in the case $k=2$, the quadratic form is $s_0c_0^2 + 2s_1c_0c_1 + s_2c_1^2 = c^T H_2 c$.
Now let $c$ be any eigenvector of $H_k$ with eigenvalue $\lambda$ ($c$ is nonzero by definition of eigenvector). Then $0 < c^T H_k c = c^T \lambda c = \lambda |c|^2,$ which shows $\lambda > 0.$ so all eigenvalues of $H_k$ are greater than 0, and the determinant of $H_k$, which is the product of its eigenvalues, is positive. Note that this argument is not specific to Hankel or even symmetric matrices, in general, positive definiteness implies positive determinant.
The other direction follows immediately from Sylvester's Criterion (https://en.wikipedia.org/wiki/Sylvester%27s_criterion) , which characterizes positive definite matrices as having all their leading minors positive (necessarily and sufficiently).
For $n \geq k$ we find $H_k$ embedded in $H_n$ as the $k$th leading minor of $H_n$. Assuming that $H_i$ for $i \leq n$ has positive determinant, Sylvester's Criterion implies that $H_n$ is positive definite. Since $H_k$ by assumption has positive determinant for all $k$, this shows that all Hankel matrices of the sequence are positive-definite.