A function $f(x)$, that is smooth and convex-shaped (concave downward) on the interval $(x_l,x_u)$ is shown. The function is observed at an odd number of regularly spaced points. If the area under the function is computed numerically, then
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Sign of $f''(x)$ tells the bias: if $f''>0$ (convex up), trapezoidal overestimates; if $f''<0$ (concave down), trapezoidal underestimates.
the numerical value of the area obtained using the trapezoidal rule will be \textbf{less} than the actual
the numerical value of the area obtained using the trapezoidal rule will be \textbf{more} than the actual
the numerical value of the area obtained using the trapezoidal rule will be \textbf{exactly equal} to the actual
with the given details, the numerical value of area cannot be obtained using trapezoidal rule
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The Correct Option isA
Solution and Explanation
For a smooth function with negative second derivative on the interval ($f''(x)<0$; concave downward), the straight-line segment between two data points (the basis of the trapezoidal rule) lies below the curve.
Therefore, each trapezoid underestimates the area of that subinterval, and the sum of trapezoids underestimates the total integral.
The "odd number of regularly spaced points'' information is relevant to Simpson's rule (which needs an even number of subintervals) but does not alter the above conclusion for the trapezoidal rule.
Hence, the trapezoidal-rule area is less than the exact area.
\[
\boxed{\text{Trapezoidal approximation }<\ \text{true area for concave-down functions.}}
\]
