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g(x)=int_(-10)^(x)sqrt(t^(2)+11)dt

g^(')(-5)=

$g(x)=\int_{-10}^{x} \sqrt{t^{2}+11} d t$ $\newline$ $g^{\prime}(-5)=$

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Evaluate definite integrals using the power rule

Full solution

Q.

g(x)=\int_{-10}^{x} \sqrt{t^{2}+11} d t

\newline

g^{\prime}(-5)=

Apply Fundamental Theorem: We know that the derivative of an integral with respect to its upper limit is the integrand evaluated at that limit, according to the Fundamental Theorem of Calculus.
Find $g'(x)$ : So, $g'(x) = \sqrt{x^2 + 11}$ .
Evaluate $g'(-5)$ : Now we need to evaluate $g'(-5)$ , which means we plug in $x = -5$ into the derivative we found.
Substitute $x=-5$ : $g'(-5) = \sqrt{(-5)^2 + 11} = \sqrt{25 + 11} = \sqrt{36}$ .
Calculate $g'(-5)$ : $\sqrt{36}$ is $6$ , so $g'(-5) = 6$ .

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