If F(x) = ∫_{a}^{x} f(t) dt, what is F'(x)?

If F(x) = ∫_{a}^{x} f(t) dt, what is F'(x)?

• A. f(x)
• B. f'(x)
• C. ∫ f(t) dt
• D. 0

Answer: (A)

The rate at which the accumulated area changes when the upper limit moves is given by the integrand at that upper limit. This is a direct consequence of the Fundamental Theorem of Calculus: for F(x) = ∫ from a to x of f(t) dt, the derivative F'(x) equals f(x), as long as f is continuous on [a, x]. Intuitively, increasing x by a tiny amount adds a thin strip of height f(x) and width dx, so the change in F is about f(x)·dx. So F'(x) = f(x). The other forms don’t match the situation: differentiating the integrand would give f'(x), not the rate of change of the accumulated area; an indefinite integral ∫ f(t) dt is an antiderivative with respect to t, not a function of x in this context; and 0 would imply no change, which isn’t true since the upper limit x is changing.

The rate at which the accumulated area changes when the upper limit moves is given by the integrand at that upper limit. This is a direct consequence of the Fundamental Theorem of Calculus: for F(x) = ∫ from a to x of f(t) dt, the derivative F'(x) equals f(x), as long as f is continuous on [a, x]. Intuitively, increasing x by a tiny amount adds a thin strip of height f(x) and width dx, so the change in F is about f(x)·dx.

So F'(x) = f(x). The other forms don’t match the situation: differentiating the integrand would give f'(x), not the rate of change of the accumulated area; an indefinite integral ∫ f(t) dt is an antiderivative with respect to t, not a function of x in this context; and 0 would imply no change, which isn’t true since the upper limit x is changing.