For $-4 \leq x \leq 4$ , define $F(x)=\int_{-4}^{x} \sqrt{16-t^{2}} d t$ .Find F'(x).
A) $\sqrt{2 x}$
B) $-\sqrt{2 x}$
C) $-\sqrt{16-x^{2}}$
D) $\sqrt{16-x^{2}}$

Name: For $-4 \leq x \leq 4$ , define $F(x)=\int_{-4}^{x} \sqrt{16-t^{2}} d t$ .Find F'(x). A) $\sqrt{2 x}$ B) $-\sqrt{2 x}$ C) $-\sqrt{16-x^{2}}$ D) $\sqrt{16-x^{2}}$
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Question

Quizplus · Accepted Answer

The derivative of $F(x)$ with respect to $x$ is the integrand evaluated at $x$, which is $\sqrt{16-x^2}$. This follows from the Fundamental Theorem of Calculus.

For −4≤x≤4-4 \leq x \leq 4−4≤x≤4 , Define F(x)=∫−4x16−t2dtF(x)=\int_{-4}^{x} \sqrt{16-t^{2}} d tF(x)=∫−4x​16−t2​dt

For $-4 \leq x \leq 4$ , Define $F(x)=\int_{-4}^{x} \sqrt{16-t^{2}} d t$