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g(x)=int_(0)^(x)sqrt(10 t+1)dt

g^(')(8)=

$g(x)=\int_{0}^{x} \sqrt{10 t+1} d t$ $\newline$ $g^{\prime}(8)=$

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Evaluate definite integrals using the chain rule

Full solution

Q.

g(x)=\int_{0}^{x} \sqrt{10 t+1} d t

\newline

g^{\prime}(8)=

Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus which states that if $g(x)$ is defined as an integral from $a$ to $x$ of $f(t)$ dt, then $g'(x)$ is $f(x)$ . $\newline$ $g'(x) = \sqrt{10x+1}$
Evaluate at $x=8$ : Evaluate $g'(x)$ at $x=8$ . $g'(8) = \sqrt{10\cdot 8+1}$
Calculate inside square root: Calculate the value inside the square root. $\sqrt{10 \times 8 + 1} = \sqrt{80 + 1}$
Simplify square root: Simplify the square root. $\sqrt{81} = 9$

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