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approximate the area of the region bounded by $y = -\\frac{1}{4}x^{2}+9…

$\frac{26 + 34}{2} = 30$

カテゴリ: calculus 更新日: 2026-02-09

問題

approximate the area of the region bounded by $y = -\\frac{1}{4}x^{2}+9$, the x-axis, $x = -4$, and $x = 0$ by finding the combined area of the rectangles (as shown in each figure) and averaging the results.

解法ステップ

  1. Understand the question

    approximate the area of the region bounded by $y = -\\frac{1}{4}x^{2}+9$, the x-axis, $x = -4$, and $x = 0$ by finding the combined area of the rectangles (as shown in each figure) and averaging the results.

  2. Explanation

    Step1: Identify rectangle widths

    The interval from $x=-4$ to $x=0$ is split into 2 equal subintervals, so width $\Delta x = \frac{0 - (-4)}{2} = 2$.

    Step2: Calculate left rectangle area

    Left rectangles use heights at $x=-4$ and $x=-2$:
    Height at $x=-4$: $y(-4) = -\frac{1}{4}(-4)^2 + 9 = 5$
    Height at $x=-2$: $y(-2) = -\frac{1}{4}(-2)^2 + 9 = 8$
    Area $A_{\text{left}} = \Delta x \cdot (y(-4) + y(-2)) = 2 \cdot (5 + 8) = 26$

    Step3: Calculate right rectangle area

    Right rectangles use heights at $x=-2$ and $x=0$:
    Height at $x=-2$: $y(-2) = 8$
    Height at $x=0$: $y(0) = -\frac{1}{4}(0)^2 + 9 = 9$
    Area $A_{\text{right}} = \Delta x \cdot (y(-2) + y(0)) = 2 \cdot (8 + 9) = 34$

    Step4: Average the two areas

    Find the mean of the two areas:
    $\text{Average Area} = \frac{A_{\text{left}} + A_{\text{right}}}{2}$

  3. Final answer

    $\frac{26 + 34}{2} = 30$

答え

Explanation

Step1: Identify rectangle widths

The interval from $x=-4$ to $x=0$ is split into 2 equal subintervals, so width $\Delta x = \frac{0 - (-4)}{2} = 2$.

Step2: Calculate left rectangle area

Left rectangles use heights at $x=-4$ and $x=-2$:
Height at $x=-4$: $y(-4) = -\frac{1}{4}(-4)^2 + 9 = 5$
Height at $x=-2$: $y(-2) = -\frac{1}{4}(-2)^2 + 9 = 8$
Area $A_{\text{left}} = \Delta x \cdot (y(-4) + y(-2)) = 2 \cdot (5 + 8) = 26$

Step3: Calculate right rectangle area

Right rectangles use heights at $x=-2$ and $x=0$:
Height at $x=-2$: $y(-2) = 8$
Height at $x=0$: $y(0) = -\frac{1}{4}(0)^2 + 9 = 9$
Area $A_{\text{right}} = \Delta x \cdot (y(-2) + y(0)) = 2 \cdot (8 + 9) = 34$

Step4: Average the two areas

Find the mean of the two areas:
$\text{Average Area} = \frac{A_{\text{left}} + A_{\text{right}}}{2}$

Answer

$\frac{26 + 34}{2} = 30$

Question Image

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Question Analysis

Subject mathematics
Sub Subject calculus
Education Level high school
Difficulty unspecified
Question Type with image, calculation
Multi Question No
Question Count 1
Analysis Status completed
Analyzed At 2026-02-09T19:56:03

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approximate the area of the region bounded by $y = -\\frac{1}{4}x^{2}+9$, the x-axis, $x = -4$, and $x = 0$ by finding the combined area of the rectangles (as shown in each figure) and averaging the results.

関連トピック

mathematicscalculuswith image, calculationhigh schoolturns-1

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