1 Multiple Choice 1 point Find the area of the region bounded by the graphs of the functions \( f(x)=\sin 2 x, g(x)=\cos x, p i / 6 \leq x \leq \mathrm{pi} / 2 \) None of these \( 1 / 8 \) \( 1 / 2 \) \( 1 / 4 \)

To find the area of the region bounded by the graphs of the functions \( f(x) = \sin 2x \) and \( g(x) = \cos x \) within the interval \( \frac{\pi}{6} \leq x \leq \frac{\pi}{2} \), we need to calculate the definite integral of the difference between the two functions over the given interval. The area of the region bounded by the graphs of the functions can be calculated using the formula: \[ \text{Area} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (f(x) - g(x)) \, dx \] Substitute the functions \( f(x) = \sin 2x \) and \( g(x) = \cos x \) into the formula: \[ \text{Area} = \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin 2x - \cos x) \, dx \] Now, we can calculate the definite integral to find the area of the region bounded by the graphs of the functions. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{\frac{\pi }{6}}^{\frac{\pi }{2}} \left(\sin\left(2x\right)-\cos\left(x\right)\right) dx\) - step1: Evaluate the integral: \(\int \left(\sin\left(2x\right)-\cos\left(x\right)\right) dx\) - step2: Use properties of integrals: \(\int \sin\left(2x\right) dx+\int -\cos\left(x\right) dx\) - step3: Evaluate the integral: \(-\frac{1}{2}\cos\left(2x\right)+\int -\cos\left(x\right) dx\) - step4: Evaluate the integral: \(-\frac{1}{2}\cos\left(2x\right)-\sin\left(x\right)\) - step5: Return the limits: \(\left(-\frac{1}{2}\cos\left(2x\right)-\sin\left(x\right)\right)\bigg |_{\frac{\pi }{6}}^{\frac{\pi }{2}}\) - step6: Calculate the value: \(\frac{1}{4}\) The area of the region bounded by the graphs of the functions \( f(x) = \sin 2x \) and \( g(x) = \cos x \) within the interval \( \frac{\pi}{6} \leq x \leq \frac{\pi}{2} \) is \( \frac{1}{4} \) or 0.25.