Answer
the interval is the curve
$$
y=\int_{0}^{x} \frac{t^{2}}{t^{2}+t+2} d t
$$
concave downward is $(-4, 0)$,
Work Step by Step
$$
y=\int_{0}^{x} \frac{t^{2}}{t^{2}+t+2} d t
$$
the derivatives of the function $y$ are given by the following:
$$
y^{\prime}=\frac{x^{2}}{x^{2}+x+2}
$$
and
$$
\begin{aligned}
y^{\prime \prime}&=\frac{\left(x^{2}+x+2\right)(2 x)-x^{2}(2 x+1)}{\left(x^{2}+x+2\right)^{2}}\\
&=\frac{2 x^{3}+2 x^{2}+4 x-2 x^{3}-x^{2}}{\left(x^{2}+x+2\right)^{2}}\\
&=\frac{x^{2}+4 x}{\left(x^{2}+x+2\right)^{2}} \\
&=\frac{x(x+4)}{\left(x^{2}+x+2\right)^{2}}
\end{aligned}
$$
The curve $y$ is concave downward when $y^{\prime \prime} \lt 0$ that is, on the interval $(-4, 0)$,
So, the interval is the curve $y$ concave downward is $(-4, 0)$,
