Question

If \[ \int f(x)\,dx=F(x)+C, \] then \[ \frac{d}{dt}\int_{g(t)}^{h(t)}f(x)\,dx= \]

• A. \(f(h(t))-f(t)\)
• B. \(F(h(t))-F(g(t))\)
• C. \(F(h(t))h'(t)-F(g(t))g'(t)\)
• D. \(f(h(t))h'(t)-f(g(t))g'(t)\)

Hint:

For variable limits, use Leibniz rule: derivative is value of integrand at upper limit times derivative of upper limit minus value of integrand at lower limit times derivative of lower limit.

Answer 1

The Correct Option is D

Step 1: Use the given anti-derivative.
Given, \[ \int f(x)\,dx=F(x)+C \] Therefore, \[ F'(x)=f(x) \]

Step 2: Apply Fundamental Theorem of Calculus.
We have \[ \int_{g(t)}^{h(t)}f(x)\,dx \] Using the anti-derivative \(F(x)\), \[ \int_{g(t)}^{h(t)}f(x)\,dx = F(h(t))-F(g(t)) \]

Step 3: Differentiate with respect to \(t\).
Now, \[ \frac{d}{dt}\int_{g(t)}^{h(t)}f(x)\,dx = \frac{d}{dt}\left[F(h(t))-F(g(t))\right] \] Using chain rule, \[ \frac{d}{dt}F(h(t))=F'(h(t))h'(t) \] and \[ \frac{d}{dt}F(g(t))=F'(g(t))g'(t) \] Therefore, \[ \frac{d}{dt}\left[F(h(t))-F(g(t))\right] = F'(h(t))h'(t)-F'(g(t))g'(t) \]

Step 4: Substitute \(F'(x)=f(x)\).
Thus, \[ F'(h(t))=f(h(t)) \] and \[ F'(g(t))=f(g(t)) \] Hence, \[ \frac{d}{dt}\int_{g(t)}^{h(t)}f(x)\,dx = f(h(t))h'(t)-f(g(t))g'(t) \]

Step 5: Final conclusion.
Therefore, \[ \boxed{f(h(t))h'(t)-f(g(t))g'(t)} \]