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g(x)=int_(1)^(x)sqrt(2t+7)dt

g^(')(9)=

$g(x)=\int_{1}^{x} \sqrt{2 t+7} d t$ $\newline$ $g^{\prime}(9)=$

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Evaluate definite integrals using the chain rule

Full solution

Q.

g(x)=\int_{1}^{x} \sqrt{2 t+7} d t

\newline

g^{\prime}(9)=

Apply Fundamental Theorem of Calculus: First, we need to use the Fundamental Theorem of Calculus which says that if $g(x)$ is the integral from $a$ to $x$ of $f(t)$ dt, then $g'(x)$ is $f(x)$ .
Calculate $g'(x)$ : So, $g'(x) = \sqrt{2x+7}$ .
Find $g'(9)$ : Now we just plug in $x=9$ into $g'(x)$ to find $g'(9)$ .
Evaluate $g'(9)$ : $g'(9) = \sqrt{2\cdot 9 + 7} = \sqrt{18 + 7} = \sqrt{25}$ .
Final Result: And we know that $\sqrt{25}$ is $5$ .

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