If $F(x)=\int_0^x f(t)\,dt$, and $F(a)=-2$ and $F(b)=-2$

Expert intro: This question links the Fundamental Theorem of Calculus with a value comparison. Since $F(a)=F(b)$ and $F$ is differentiable, the Mean Value Theorem applies to $F$ on $[a,b]$. That forces a point where the derivative of $F$ is zero.

Detailed walkthrough

Because $F$ is differentiable, we can apply the Mean Value Theorem to $F$ on $[a,b]$.

Since

$F(a)=F(b)=-2,$

we have

$\frac{F(b)-F(a)}{b-a}=0.$

By the Mean Value Theorem, there exists some $c$ with

$a<c<b$

such that

$F'(c)=0.$

From the Fundamental Theorem of Calculus,

$F'(x)=f(x).$

So there must be some $c$ in $(a,b)$ such that

$f(c)=0.$

Therefore the statement that must be true is

$\boxed{\text{A. } f(x)=0 \text{ for some } x \text{ with } a<x<b}.$

💡 Pitfall guide

A frequent mistake is to think equal endpoint values mean the function must stay above or below them throughout the interval. That is not guaranteed. The only conclusion forced by differentiability is the existence of at least one point where the derivative is zero.

🔄 Real-world variant

If the problem had said $F(a)<F(b)$ instead of $F(a)=F(b)$, then the Mean Value Theorem would only guarantee that $F'(x)$ is positive at some point if the average slope is positive. With equal endpoint values, the average slope is exactly zero, so a zero derivative is guaranteed.

🔍 Related terms

Mean Value Theorem, Fundamental Theorem of Calculus, derivative zero