Question

1. if $\frac{dy}{dx}=16\sin^{3}x\cos x$ and $y = 16$ when $x=\frac{\pi}{2}$, what is the value of $y$ when $x=\frac{\pi}{6}$? (a) $\frac{49}{4}$ (b) $\frac{1}{4}$ (c) 10 (d) 13

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 \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), 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. \(\frac{49}{4}\)