1. Which formula correctly extends FTC Part 1 when the upper limit of integration is a function $g(x)$ rather than just $x$?
- A.$\frac{d}{dx}\int_a^{g(x)} f(t)\, dt = f(g(x))$
- B.$\frac{d}{dx}\int_a^{g(x)} f(t)\, dt = f(g(x)) \cdot g'(x)$
- C.$\frac{d}{dx}\int_a^{g(x)} f(t)\, dt = f'(g(x)) \cdot g'(x)$
- D.$\frac{d}{dx}\int_a^{g(x)} f(t)\, dt = f(x) \cdot g'(x)$
View Answer
Answer: $\frac{d}{dx}\int_a^{g(x)} f(t)\, dt = f(g(x)) \cdot g'(x)$
Apply FTC Part 1 with the Chain Rule: Let $F(u) = \int_a^u f(t)\,dt$. Then $F'(u) = f(u)$ by FTC Part 1. With $u = g(x)$, the chain rule gives $\frac{d}{dx}F(g(x)) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$. Why Option B is correct: It correctly combines FTC Part 1 with the chain rule by evaluating the integrand at $g(x)$ and multiplying by $g'(x)$. Why distractors fail: Option A omits the chain rule factor $g'(x)$. Option C incorrectly differentiates $f$ to get $f'(g(x))$ — FTC Part 1 returns the integrand, not its derivative. Option D evaluates $f$ at $x$ instead of at $g(x)$.