Question

Multiple Choice The trapezoidal approximation of \(\int_{0}^{\pi} \sin x d x\) using 4 equal subdivisions of the interval of integration is $$ \begin{array}{l}{\text { (A) } \frac{\pi}{2}} \\ {\text { (B) } \pi} \\\ {\text { (C) } \frac{\pi}{4}(1+\sqrt{2})} \\ {\text { (D) } \frac{\pi}{2}(1+\sqrt{2})} \\ {\text { (E) } \frac{\pi}{4}(2+\sqrt{2})}\end{array} $$

Short answer

The trapezoidal approximation of \(\int_{0}^{\pi} \sin x d x\) using 4 equal subdivisions of the interval of integration is \(\frac{\pi}{4}(1+\sqrt{2})\). So, the correct answer choice is (C) \(\frac{\pi}{4}(1+\sqrt{2})\).

Step-by-step solution

Breaking Into Sub-intervals

Divide the range of integration \[0, \pi] into 4 equal subintervals. As a result, each subinterval will have length \(\pi/4\). Thus, the intervals are \[0, \pi/4], \[\pi/4, \pi/2], \[\pi/2, 3\pi/4], and \[3\pi/4, \pi].

Apply Trapezoidal Rule

The trapezoidal rule formula is \(\Delta x/2 [f(x_0) + 2f(x_1) + ... + 2f(x_{n-1}) + f(x_n)]\). So, first calculate the function values at these points: \(f(x) = \sin(x)\) at points \(x = 0, \pi/4, \pi/2, 3\pi/4, \pi\). You will find these values are \(0, \sqrt{2}/2, 1, \sqrt{2}/2, 0\). Then compute the trapezoidal approximation: \((\pi/4)/2 * (0 + 2 * \sqrt{2}/2 + 2*1 + 2* \sqrt{2}/2 + 0)\)

Simplify the Result

This simplifies to \(\frac{\pi}{4}(1+\sqrt{2})\)