Solve the problem.
-The length of one arch of the curve $y = 3 \sin 2 x$ is given by
$$L = \int _ { 0 } ^ { \pi / 2 } \sqrt { 1 + 36 \cos ^ { 2 } 2 x } d x$$
Estimate $\mathrm { L }$ by the Trapezoidal Rule with $\mathrm { n } = 6$.
A) 6.2807
B) 6.4115
C) 5.8169
D) 6.1995

Question

Quizplus · Accepted Answer

The Trapezoidal Rule with $n = 6$ gives$$L \approx \frac{\pi/2 - 0}{6} \left[ \sqrt{1 + 36 \cos^2 0} + 2\sqrt{1 + 36 \cos^2 \frac{\pi}{12}} + 2\sqrt{1 + 36 \cos^2 \frac{\pi}{6}} + 2\sqrt{1 + 36 \cos^2 \frac{\pi}{4}} + 2\sqrt{1 + 36 \cos^2 \frac{5\pi}{12}} + \sqrt{1 + 36 \cos^2 \frac{\pi}{2}} \right] \approx 6.2807$$

Solve the Problem y=3sin⁡2xy = 3 \sin 2 xy=3sin2x Is Given By

Solve the Problem $y = 3 \sin 2 x$ Is Given By