Set up an integral for the length of the curve.
-$$y = 6 \cos x , 0 \leq x \leq \pi$$
A) $\int _ { 0 } ^ { \pi } \sqrt { 1 + 6 \sin x } d x$
B) $\int _ { 0 } ^ { \pi } \sqrt { 1 - 6 \sin x } d x$
C) $\int _ { 0 } ^ { \pi } \sqrt { 1 + 36 \cos ^ { 2 } x } d x$
D) $\int _ { 0 } ^ { \pi } \sqrt { 1 + 36 \sin ^ { 2 } x } d x$

Question

Quizplus · Accepted Answer

The formula for the arc length of a curve $ y = f(x) $ is $\int \sqrt{1 + (f'(x))^2} \, dx$. Here, $ f(x) = 6 \cos x $, so $ f'(x) = -6 \sin x $. Thus, the integral becomes $\int \sqrt{1 + 36 \sin^2 x} \, dx$.

Set Up an Integral for the Length of the Curve y=6cos⁡x,0≤x≤πy = 6 \cos x , 0 \leq x \leq \piy=6cosx,0≤x≤π

Set Up an Integral for the Length of the Curve $y = 6 \cos x , 0 \leq x \leq \pi$