7. if $\\frac{dy}{dx}=16\\sin^{3}x\\cos x$ and $y = 16$ when $x=\\frac{\\pi}{2}$, what is the value of $y$…

Explanation:

Step1: Integrate the derivative

We know that (y=\int 16\sin^{3}x\cos xdx). Let (u = \sin x), then (du=\cos xdx). The integral becomes (y = 16\int u^{3}du). Using the power - rule (\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n\neq - 1)), we have (y=16\times\frac{u^{4}}{4}+C = 4u^{4}+C). Substituting back (u=\sin x), we get (y = 4\sin^{4}x+C).

Step2: Find the constant (C)

Since (y = 16) when (x=\frac{\pi}{2}), and (\sin\frac{\pi}{2}=1). Substitute into (y = 4\sin^{4}x+C): (16=4\times(1)^{4}+C). Solving for (C), we get (C = 12). So (y=4\sin^{4}x + 12).

Step3: Calculate (y) when (x=\frac{\pi}{6})

When (x=\frac{\pi}{6}), (\sin\frac{\pi}{6}=\frac{1}{2}). Substitute into (y=4\sin^{4}x + 12): (y=4\times(\frac{1}{2})^{4}+12). First, ((\frac{1}{2})^{4}=\frac{1}{16}), then (4\times\frac{1}{16}=\frac{1}{4}). So (y=\frac{1}{4}+12=\frac{1 + 48}{4}=\frac{49}{4}).

Answer:

A. (\frac{49}{4})