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Wed 17Apr Applications of integration: Unit test What is the average value of h(x)=(x^(3)+7x^(2)+3)/(2x^(2)-5x+4) on the interval [-8,-2] ? Use a graphing calculator and round your answer to three decimal places. ◻

Wed 17Apr 17 \mathrm{Apr} \newlineApplications of integration: Unit test\newlineWhat is the average value of h(x)=x3+7x2+32x2−5x+4 h(x)=\frac{x^{3}+7 x^{2}+3}{2 x^{2}-5 x+4} on the interval [−8,−2] [-8,-2] ?\newlineUse a graphing calculator and round your answer to three decimal places.\newline□ \square

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Q. Wed 17Apr 17 \mathrm{Apr} \newlineApplications of integration: Unit test\newlineWhat is the average value of h(x)=x3+7x2+32x2−5x+4 h(x)=\frac{x^{3}+7 x^{2}+3}{2 x^{2}-5 x+4} on the interval [−8,−2] [-8,-2] ?\newlineUse a graphing calculator and round your answer to three decimal places.\newline□ \square
  1. Calculate Interval Width: To find the average value of a function h(x)h(x) on the interval [a,b][a, b], use the formula: Average value = 1(b−a)⋅∫abh(x) dx\frac{1}{(b-a)} \cdot \int_{a}^{b} h(x) \, dx.
  2. Set Up Integral: First, calculate the width of the interval: b−a=−2−(−8)=6b - a = -2 - (-8) = 6.
  3. Compute Definite Integral: Now, set up the integral to find the average value: Average value = (1/6)×∫−8−2x3+7x2+32x2−5x+4dx(1/6) \times \int_{-8}^{-2} \frac{x^3 + 7x^2 + 3}{2x^2 - 5x + 4} dx.
  4. Calculate Average Value: Use a graphing calculator to compute the definite integral from −8-8 to −2-2 of the function h(x)h(x).
  5. Calculate Average Value: Use a graphing calculator to compute the definite integral from −8-8 to −2-2 of the function h(x)h(x).After calculating the integral on the calculator, suppose we get a value of II (we will not actually compute this value here as it requires a calculator).
  6. Calculate Average Value: Use a graphing calculator to compute the definite integral from −8-8 to −2-2 of the function h(x)h(x).After calculating the integral on the calculator, suppose we get a value of II (we will not actually compute this value here as it requires a calculator).The average value is then (1/6)×I(1/6) \times I. Round this to three decimal places as instructed.

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