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g(x)=int_(-1)^(x)(8-t)dt

g^(')(1)=

$g(x)=\int_{-1}^{x}(8-t) d t$ $\newline$ $g^{\prime}(1)=$

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Evaluate definite integrals using the chain rule

Full solution

Q.

g(x)=\int_{-1}^{x}(8-t) d t

\newline

g^{\prime}(1)=

Find derivative of $g(x)$ : First, let's find the derivative of $g(x)$ using the Fundamental Theorem of Calculus. $\newline$ $g'(x) = \frac{d}{dx} \left[\int_{-1}^{x}(8-t)dt\right] = 8 - x$
Evaluate $g'(x)$ at $x=1$ : Now, we need to evaluate $g'(x)$ at $x=1$ .
$g'(1) = 8 - 1 = 7$

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