An approximation for the area under the graph [math: y=12x] and the x-axis over the interval x = 1 to x = 4 using the trapezoidal rule and interval widths of 1 unit is:

Question

BlackTom AI · Accepted Answer

The trapezoidal rule uses the average of function values at interval endpoints for each subinterval. For n subintervals of equal width h on [a,b] the formula is:
T = (h/2)[f(x0) + 2f(x1) + 2f(x2) + ... + 2f(x_{n-1}) + f(xn)].

Identify key points:
- Function: f(x) = 12/x.
- Interval: a = 1, b = 4.
- Interval width given: h = 1, so number of subintervals n = (b−a)/h = (4−1)/1 = 3.
- Nodes: x0 = 1, x1 = 2, x2 = 3, x3 = 4.

Compute function values (showing intermediate steps):
f(1) = 12/1 = 12
f(2) = 12/2 = 6
f(3) = 12/3 = 4
f(4) = 12/4 = 3

Apply the trapezoidal formula:
Sum endpoints: f(1) + f(4) = 12 + 3 = 15
Sum doubled middle terms: 2[f(2) + f(3)] = 2(6 + 4) = 2(10) = 20
Bracket total: 15 + 20 = 35
Multiply by h/2: T = (1/2) * 35 = 17.5

Why this is correct: the trapezoidal rule calculation above follows the standard formula with the correct h, node values, and weightings (endpoints counted once, interior nodes doubled). For reference, the exact integral is ∫_1^4 12/x dx = 12 ln 4 ≈ 16.635, so the trapezoidal approximation 17.5 is reasonably close.

Why other choices are incorrect:
- 35 would be the bracket sum before multiplying by h/2 (i.e., forgetting the 1/2 factor).
- 70 is twice 35 and corresponds to doubling rather than halving — incorrect scaling.
- 8.75 is half of 17.5 and would result from mistakenly using h/4 instead of h/2; it underestimates here.

Conclusion: The correct trapezoidal approximation is 17.5 square units, matching the provided answer (d).

An approximation for the area under the graph [math: y=12x] and the x-axis over the interval x = 1 to x = 4 using the trapezoidal rule and interval widths of 1 unit is:单项选择题

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