Let $f(x)=\int_{-2}^{x^2-3x} e^{t^2}\,dt$

Key takeaway: This is an optimization problem for a function defined by an integral with a variable upper limit. The sign of the integrand is crucial: since $e^{t^2}>0$ for all $t$, the value of $f(x)$ depends only on the direction and size of the interval from $-2$ to $x^2-3x$.

Let

$f(x)=\int_{-2}^{x^2-3x} e^{t^2}\,dt.$

Since

$e^{t^2}>0 \quad \text{for all } t,$

the integral is increasing as its upper limit increases. So $f(x)$ is minimized when the upper limit

$u(x)=x^2-3x$

is as small as possible.

Complete the square:

$x^2-3x=\left(x-\frac{3}{2}\right)^2-\frac{9}{4}.$

This expression is smallest when

$x=\frac{3}{2}.$

So $f(x)$ attains its minimum at

$\boxed{\frac{3}{2}}$

and the correct choice is C.

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Pitfalls the pros know 👇 Do not differentiate first and then look only at critical points without checking what the integral itself is doing. Because the integrand is always positive, the minimum comes from making the upper limit as small as possible. Also, do not confuse the minimum of $u(x)=x^2-3x$ with the minimum of the integral over a fixed interval.

What if the problem changes? If the integrand were negative everywhere, such as $-e^{t^2}$, then the same upper limit $x^2-3x$ would produce the opposite behavior: the function would be minimized where the upper limit is largest, not smallest. The sign of the integrand determines the direction of monotonicity.

Tags: variable upper limit, optimization, positive integrand