Let $h$ be the function with the first derivative given by $h'(x)=\int\limits_0^x e^{-t^2}  dt$. Which of the following must be true on the interval $0 \lt x \lt 3$ ?

Summary Submit Skip Question

The graph of $e^{-t^2}$ is always positive. The area under the curve therefore will be positive for the interval $0 \lt x\lt 3$.

$h'(x)$ is therefore always positive, and the graph of $h$ is increasing on the interval.

By the second fundamental theorem of calculus,

$$h'(x)=\int\limits_0^x e^{-t^2}   dt$$ $$h''(x)=e^{-x^2}$$

Similar to the previous case, the exponential function is positive for all values. Therefore, $h$ is also concave up on the interval.