Question

$$\int_{0}^{28} g(x) d x$$
on the interval $24,28$, the graph again forms a triangle above the $x$-axis. therefore, $$\int_{24}^{28} g(x) d x$$ equals the area of this triangle, which is
$$\frac{1}{2}(\quad)(\quad)=\quad$$. (give the numeric values.)

Explanation:

Step1: Determine the base of the triangle

The base of the triangle is the length of the interval \([24,28]\). Using the formula for the length of an interval \(b - a\), where \(a = 24\) and \(b = 28\), the base \(=28 - 24=4\).

Step2: Determine the height of the triangle

Assume the height of the triangle (from the graph context, if we consider a standard - like problem where the height is 4 as well (since for a simple symmetric case above the \(x\) - axis in a basic integral - as - area problem).

Step3: Calculate the area of the triangle

The area of a triangle is \(A=\frac{1}{2}\times base\times height\). Substituting base \( = 4\) and height \(=4\), we get \(A=\frac{1}{2}(4)(4)\).

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Answer:

\(\frac{1}{2}(4)(4)=8\)