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H(x)=-x-5

h(x)=H^(')(x)

int_(-3)^(6)h(x)dx=

$H(x)=-x-5$ $\newline$ $h(x)=H^{\prime}(x)$ $\newline$ $\int_{-3}^{6} h(x) d x=$

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Evaluate definite integrals using the power rule

Full solution

Q.

H(x)=-x-5

\newline

h(x)=H^{\prime}(x)

\newline

\int_{-3}^{6} h(x) d x=

Integrate $h(x)$ : Now we integrate $h(x)$ from $-3$ to $6$ . $\int_{-3}^{6} h(x) \, dx = \int_{-3}^{6} (-1) \, dx$ This is a simple integral, the antiderivative of $-1$ is $-x$ .
Evaluate antiderivative: We evaluate the antiderivative at the bounds $6$ and $-3$ . $-(-6) - (-(-3)) = 6 - 3 = 3$
Correct evaluation: But wait, there's a mistake in the previous step. The correct evaluation should be: $\newline$ $-(-6) - (-(-3)) = 6 + 3 = 9$

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