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${:[g(x)=int_(-10)^(x)(10-3t)dt],[g^(')(-4)=]:}$

$\begin{array}{l}g(x)=\int_{-10}^{x}(10-3 t) d t \\ g^{\prime}(-4)=\end{array}$

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Evaluate definite integrals using the chain rule

Full solution

Q.

\begin{array}{l}g(x)=\int_{-10}^{x}(10-3 t) d t \\ g^{\prime}(-4)=\end{array}

Find Derivative of $\newline$ $g(x)$ : First, we need to find the derivative of $\newline$ $g(x)$ using the Fundamental Theorem of Calculus, which tells us that if $\newline$ $g(x)$ is defined as an integral from a constant to $\newline$ $x$ , then $\newline$ $g'(x)$ is the integrand evaluated at $\newline$ $x$ .
Calculate $g'(x)$ : So, $g'(x) = 10 - 3x$ .
Evaluate $g'(-4)$ : Now we just plug in $x = -4$ into $g'(x)$ to find $g'(-4)$ . $\newline$ $g'(-4) = 10 - 3(-4) = 10 + 12 = 22$ .

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