G(x)=cos(3x) g(x)=G^(')(x) int_(0)^(pi)g(x)dx=

Find g(x): First, we need to find g(x) by differentiating G(x). G(x) = cos(3x), so g(x) = G'(x) = -3sin(3x).Set up integral: Now, let's set up the integral. int_(0)^(pi)g(x)dx = int_(0)^(pi)(-3sin(3x))dx.Integrate -3sin(3x): Integrate -3sin(3x) with respect to x. The antiderivative of sin(ax) is -1/a * cos(ax), so the antiderivative of -3sin(3x) is -3 * (-1/3) * cos(3x) = cos(3x).Evaluate integral: Evaluate the integral from 0 to pi. int_(0)^(pi)(-3sin(3x))dx = [cos(3x)]_(0)^(pi) = cos(3pi) - cos(0).Calculate values: Calculate the values of cos(3pi) and cos(0). cos(3pi) = -1 and cos(0) = 1.Substitute values: Substitute these values into the equation. cos(3pi) - cos(0) = -1 - 1 = -2.Final answer: The final answer is -2.

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G(x)=cos(3x)

g(x)=G^(')(x)

int_(0)^(pi)g(x)dx=

$G(x)=\cos (3 x)$ $\newline$ $g(x)=G^{\prime}(x)$ $\newline$ $\int_{0}^{\pi} g(x) d x=$

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Calculus

Evaluate definite integrals using the power rule

Full solution

G(x)=\cos (3 x)

\newline

g(x)=G^{\prime}(x)

\newline

\int_{0}^{\pi} g(x) d x=

Find $g(x)$ : First, we need to find $g(x)$ by differentiating $G(x)$ . $\newline$ $G(x) = \cos(3x)$ , so $g(x) = G'(x) = -3\sin(3x)$ .
Set up integral: Now, let's set up the integral. $\int_{0}^{\pi}g(x)\,dx = \int_{0}^{\pi}(-3\sin(3x))\,dx$ .
Integrate $-3\sin(3x)$ : Integrate $-3\sin(3x)$ with respect to $x$ . The antiderivative of $\sin(ax)$ is $-\frac{1}{a} \cos(ax)$ , so the antiderivative of $-3\sin(3x)$ is $-3 \cdot (-\frac{1}{3}) \cdot \cos(3x) = \cos(3x)$ .
Evaluate integral: Evaluate the integral from $0$ to $\pi$ . $\newline$ $\int_{0}^{\pi}(-3\sin(3x))\,dx = [\cos(3x)]_{0}^{\pi} = \cos(3\pi) - \cos(0).$
Calculate values: Calculate the values of $\cos(3\pi)$ and $\cos(0)$ . $\cos(3\pi) = -1$ and $\cos(0) = 1$ .
Substitute values: Substitute these values into the equation. $\newline$ $\cos(3\pi) - \cos(0) = -1 - 1 = -2$ .
Final answer: The final answer is $-2$ .