Answer
The curve is concave downward on the interval $(-4,0)$
Work Step by Step
$y = \int_{0}^{x}\frac{t^2}{t^2+t+2}~dt$
Then:
$y' = \frac{x^2}{x^2+x+2}$
$y$ is concave down when $y'' \lt 0$
We can find $y''$:
$y'' = \frac{d}{dx}(\frac{x^2}{x^2+x+2}) = \frac{(2x)(x^2+x+2)-(x^2)(2x+1)}{(x^2+x+2)^2} = \frac{x^2+4x}{(x^2+x+2)^2}$
We can find the interval when $y'' \lt 0$:
$\frac{x^2+4x}{(x^2+x+2)^2} \lt 0$
$x^2+4x \lt 0$
$x(x+4) \lt 0$
$-4 \lt x \lt 0$
Therefore:
The curve is concave downward on the interval $(-4,0)$
