Set up, but do not evaluate, an integral for the length of the curve.
$y = 6 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 9 \pi } { 2 }$
A) $L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 - 36 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
B) $L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
C)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 - 36 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
D)
$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
E)
$$L = \int _ { 0 } ^ { 9 x / 2 } \sqrt { 1 + 36 e ^ { 2 x } ( 1 - \sin x ) } d x$$

Question

Quizplus · Accepted Answer

The formula for the length of a curve $ y = f(x) $ from $ x = a $ to $ x = b $ is $ L = \int_a^b \sqrt{1 + (f'(x))^2} \, dx $. Here, $ f(x) = 6e^x \sin x $, so $ f'(x) = 6e^x (\sin x + \cos x) $. Thus, $ (f'(x))^2 = 36e^{2x} (1 + \sin 2x) $.

Set Up, but Do Not Evaluate, an Integral for the Length