Answer
$(-1,1)$
Work Step by Step
Given
\begin{aligned}F(x)&=\int_1^xf(t)dt \end{aligned}
To find the interval of concave downward for $F''(x)$, since
\begin{aligned}
F'(x)& = \frac{d}{dx}\int_1^xf(t)dt \\
&=f(x)
\end{aligned}
Then
\begin{aligned}
F''(x)&= f'(x)
\end{aligned}
Hence the interval of concave downward for $F''(x)$ is the decreasing interval for $f(x)$, from the given graph, we know that f(x) is decreasing on $(-1,1)$. It follows that $F(x)$ is concave downward on $(-1,1)$