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${:[G(x)=(x+3)^(2)],[g(x)=G^(')(x)],[int_(-1)^(7)g(x)dx=]:}$

$\begin{array}{l}G(x)=(x+3)^{2} \\ g(x)=G^{\prime}(x) \\ \int_{-1}^{7} g(x) d x=\end{array}$

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Find higher derivatives of rational and radical functions

Full solution

Q.

\begin{array}{l}G(x)=(x+3)^{2} \\ g(x)=G^{\prime}(x) \\ \int_{-1}^{7} g(x) d x=\end{array}

Find Derivative of G(x): First, we need to find the derivative of G(x), which is $g(x)$ . $g(x) = G'(x) = \frac{d}{dx}[(x+3)^2]$ Using the power rule, the derivative of $(x+3)^2$ is $2*(x+3)$ .So, $g(x) = 2*(x+3)$
Calculate Integral of $g(x)$ : Now, we need to calculate the integral of $g(x)$ from $-1$ to $7$ .
$\int_{-1}^{7} 2\cdot(x+3) \, dx$
This is a simple polynomial integration problem.

Integrate $2\cdot(x+3)$ with respect to $x$ .
The antiderivative of $2\cdot(x+3)$ is $2\cdot(\frac{1}{2})\cdot x^2 + 2\cdot3\cdot x$ , which simplifies to $x^2 + 6x$ .
So, $g(x)$ $0$
Evaluate Antiderivative: Evaluate the antiderivative from $-1$ to $7$ . $\newline$ Plug in the upper limit: $(7)^2 + 6*(7) = 49 + 42 = 91$ $\newline$ Plug in the lower limit: $(-1)^2 + 6*(-1) = 1 - 6 = -5$ $\newline$ Now subtract the lower limit result from the upper limit result. $\newline$ $91 - (-5) = 91 + 5 = 96$

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