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g(x)=int_(1)^(x)sqrt(19-t)dt

g^(')(3)=

$g(x)=\int_{1}^{x} \sqrt{19-t} d t$ $\newline$ $g^{\prime}(3)=$

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Evaluate definite integrals using the chain rule

Full solution

Q.

g(x)=\int_{1}^{x} \sqrt{19-t} d t

\newline

g^{\prime}(3)=

Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus which states that if $g(x) = \int_{a}^{x}f(t)\,dt$ , then $g'(x) = f(x)$ .
Determine $f(t)$ : Here, $f(t) = \sqrt{19-t}$ . So, $g'(x) = \sqrt{19-x}$ .
Find $g'(x)$ : Evaluate $g'(3) = \sqrt{19-3}$ .
Substitute $x=3$ : Calculate $g'(3) = \sqrt{16}$ .
Simplify the Result: Simplify $\sqrt{16}$ to get $4$ .

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