$\begin{array}{l}H(x)=-x-5 \ h(x)=H^{\prime}(x) \ \int_{-3}^{6} h(x) d x=\end{array}$

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∻Solution by Steps∻ ‖ step 1 ⋮ Identify the function to be integrated. ‖ ‖ step 2 ⋮ Set up the definite integral for the function $H(x) = -x - 5$ from $-3$ to $6$. ‖ ‖ step 3 ⋮ Compute the integral $\int_{-3}^{6} (-x - 5) \, dx$. ‖ ‖ step 4 ⋮ Apply the fundamental theorem of calculus to evaluate the integral. ‖ ‖ step 5 ⋮ The result of the integral is the difference $H(6) - H(-3)$. ‖ ‖ step 6 ⋮ Calculate $H(6) = -(6) - 5 = -11$ and $H(-3) = -(-3) - 5 = 3 - 5 = -2$. ‖ ‖ step 7 ⋮ Find the difference $H(6) - H(-3) = -11 - (-2) = -11 + 2 = -9$. ‖ ‖ step 8 ⋮ The definite integral $\int_{-3}^{6} h(x) \, dx$ is equal to $-9$. ‖ ∻Answer∻ ⚹ $\int_{-3}^{6} h(x) \, dx = -9$ ⚹ ∻Key Concept∻ ⚹Definite Integral Evaluation⚹ ∻Explanation∻ ⚹The definite integral of a function over an interval is the net area under the curve, which can be found by evaluating the antiderivative at the upper limit and subtracting the antiderivative evaluated at the lower limit.⚹◊What is the integral of the function $h(x)$ defined as the derivative of $H(x)$, given $H(x) = -x - 5$?⍭ Generate me a similar question◊

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