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Question
Math
Posted 3 months ago
H(x)=x5h(x)=H(x)36h(x)dx=\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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Answer from Sia
Posted 3 months ago
Solution by Steps
step 1
Identify the function to be integrated
step 2
Set up the definite integral for the function H(x)=x5H(x) = -x - 5 from 3-3 to 66
step 3
Compute the integral 36(x5)dx\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)H(6) - H(-3)
step 6
Calculate H(6)=(6)5=11H(6) = -(6) - 5 = -11 and H(3)=(3)5=35=2H(-3) = -(-3) - 5 = 3 - 5 = -2
step 7
Find the difference H(6)H(3)=11(2)=11+2=9H(6) - H(-3) = -11 - (-2) = -11 + 2 = -9
step 8
The definite integral 36h(x)dx\int_{-3}^{6} h(x) \, dx is equal to 9-9
Answer
36h(x)dx=9\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.

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