g(x)=∫₁ˣ√(19 - t)dt\n g(3)=□
g(x)=∫₁ˣ√(19 - t)dt\n g(3)=□
Answer
Explanation:
Step1: Apply the fundamental theorem of calculus
If $g(x)=\int_{a}^{x}f(t)dt$, then $g'(x) = f(x)$. Here $a = 1$, $f(t)=\sqrt{19 - t}$, so $g'(x)=\sqrt{19 - x}$.
Step2: Evaluate $g'(3)$
Substitute $x = 3$ into $g'(x)$. We get $g'(3)=\sqrt{19-3}$. $g'(3)=\sqrt{16}=4$.
Answer:
$4$