What is the approximate value of the following integral using two applications of the composite trapezoidal method? [math: f(x)∫−4.54(x3cos⁡(x))dx]

Question

BlackTom AI · Accepted Answer

The composite trapezoidal rule with two subintervals (n = 2) is applied. Relevant formula: for interval [a,b] with step h = (b − a)/n,
T = (h/2)[f(x0) + 2 f(x1) + f(x2)].

Here a = −4.5, b = 4, so h = (4 − (−4.5))/2 = 8.5/2 = 4.25. The nodes are:
x0 = −4.5, x1 = −4.5 + h = −0.25, x2 = 4.

Evaluate the integrand f(x) = x^3 cos x at each node (keep a few extra digits for accuracy):
f(x0) = (−4.5)^3 cos(−4.5) = −91.125 × (−0.2107957994) ≈ +19.202012
f(x1) = (−0.25)^3 cos(−0.25) = −0.015625 × 0.968912421 ≈ −0.0151405
f(x2) = 4^3 cos(4) = 64 × (−0.6536436209) ≈ −41.83111174

Form the trapezoidal sum:
S = f(x0) + 2 f(x1) + f(x2)
  = 19.202012 + 2(−0.0151405) + (−41.83111174)
  = 19.202012 − 0.0302810 − 41.83111174
  = −22.65938074 (rounded)

Multiply by h/2 = 4.25/2 = 2.125:
T ≈ 2.125 × (−22.65938074) ≈ −48.1412526

Rounded to the given option precision gives −48.1412, which is option a.

Why the other choices are incorrect: they are numerically far from the computed composite trapezoid result (for instance, −32.0564 and −57.4414 differ by many units, and −96.2825 is roughly double the correct magnitude), so they do not match the correct trapezoidal computation shown above.

Therefore the composite trapezoidal approximation with two subintervals yields −48.1412, confirming option a.

What is the approximate value of the following integral using two applications of the composite trapezoidal method? [math: f(x)∫−4.54(x3cos⁡(x))dx]单项选择题

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