Set up, but do not evaluate, an integral for the length of the curve.
$$y = 2 e ^ { x } \sin x , \quad 0 \leq x \leq \frac { 3 \pi } { 2 }$$
A)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 - 4 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$
B)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 + 4 e ^ { 2 x } ( 1 - \sin x ) } d x$
C)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 + 4 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
D)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 - 4 e ^ { 2 x } ( 1 - \sin 2 x ) } d x$
E)
$L = \int _ { 0 } ^ { 3 x / 2 } \sqrt { 1 + 4 e ^ { 2 x } ( 1 + \sin 2 x ) } d x$

Question

Quizplus · Accepted Answer

The length of a curve $y = f(x)$ from $x = a$ to $x = b$ is given by the integral $\int_a^b \sqrt{1 + (f'(x))^2} dx$. For $y = 2e^x \sin x$, we first find the derivative $y' = 2e^x \sin x + 2e^x \cos x = 2e^x(\sin x + \cos x)$. Squaring $y'$ gives $(2e^x(\sin x + \cos x))^2 = 4e^{2x}(\sin^2 x + 2\sin x \cos x + \cos^2 x) = 4e^{2x}(1 + \sin 2x)$ because $\sin^2 x + \cos^2 x = 1$ and $2\sin x \cos x = \sin 2x$. Substituting into the formula for the length of a curve gives the integral in option E.

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