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 \(0 \lt x \lt 3 \) ?

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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.