g(x)=int_(-pi)^(x)sin(t)dt g^(')(pi)=

Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus which states that if g(x) = int(a to x) f(t)dt, then g'(x) = f(x).Calculate g'(x): Since g(x) = int(-pi to x) sin(t)dt, then g'(x) = sin(x).Evaluate g'(pi): Evaluate g'(x) at x = pi. So, g'(pi) = sin(pi).Final Result: Since sin(pi) = 0, g'(pi) = 0.

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

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Evaluate definite integrals using the power rule

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g(x)=\int_{-\pi}^{x} \sin (t) d t

\newline

g^{\prime}(\pi)=

Apply Fundamental Theorem of Calculus: Use the Fundamental Theorem of Calculus which states that if $g(x) = \int_{a}^{x} f(t)\,dt$ , then $g'(x) = f(x)$ .
Calculate $g'(x)$ : Since $g(x) = \int_{-\pi}^{x} \sin(t)\,dt$ , then $g'(x) = \sin(x)$ .
Evaluate $g'(\pi):</b> Evaluate \$g'(x)$ at $x = \pi$ . So, $g'(\pi) = \sin(\pi)$ .
Final Result: Since $\sin(\pi) = 0$ , $g'(\pi) = 0$ .

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g(x)=∫−πxsin⁡(t)dt g(x)=\int_{-\pi}^{x} \sin (t) d t g(x)=∫−πx​sin(t)dt\newlineg′(π)= g^{\prime}(\pi)= g′(π)=

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