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g(x)=int_(0)^(x)sqrt(5+4cos t)dt

g^(')(pi)=

$g(x)=\int_{0}^{x} \sqrt{5+4 \cos t} d t$ $\newline$ $g^{\prime}(\pi)=$

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Evaluate definite integrals using the power rule

Full solution

Q.

g(x)=\int_{0}^{x} \sqrt{5+4 \cos t} d t

\newline

g^{\prime}(\pi)=

Apply Fundamental Theorem: We need to use the Fundamental Theorem of Calculus to find $g'(x)$ . Since $g(x)$ is an integral from $0$ to $x$ , $g'(x)$ is just the integrand evaluated at $x$ .
Find $g'(x)$ : So, $g'(x) = \sqrt{5 + 4\cos(x)}$ .
Evaluate at $x = \pi$ : Now we evaluate $g'(x)$ at $x = \pi$ : $g'(\pi) = \sqrt{5 + 4\cos(\pi)}$ .
Substitute $\cos(\pi)$ : We know that $\cos(\pi) = -1$ , so plug that in: $g'(\pi) = \sqrt{5 + 4(-1)}$ .
Simplify expression: Simplify the expression: $g'(\pi) = \sqrt{5 - 4}.$
Calculate value: Calculate the value: $g'(\pi) = \sqrt{1}.$
Final result: The square root of $1$ is $1$ , so $g'(\pi) = 1$ .

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