Question

The value of $\frac{16 S _{1}}{\pi}$ is_____

Answer 1

Correct Answer: 2

Step 1: Understanding the given functions
We are given two functions $g_1(x)$ and $g_2(x)$ defined as follows:
- $g_1(x) = 1$
- $g_2(x) = |4x - \pi|$
Also, we are given the function $f(x) = \sin^2 x$. The task is to evaluate $S_1$ and then find the value of $\frac{16S_1}{\pi}$.
Step 2: Define $S_1$
We are asked to compute the integral for $S_1$ defined as:
$$ S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_1(x) \, dx. $$
Since $g_1(x) = 1$, this simplifies to:
$$ S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \, dx. $$
Step 3: Simplify the integral using trigonometric identity
To simplify the integral, we use the trigonometric identity for $\sin^2 x$:
$$ \sin^2 x = \frac{1}{2} \left( 1 - \cos 2x \right). $$
Thus, the integral becomes:
$$ S_1 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \frac{1}{2} \left( 1 - \cos 2x \right) \, dx. $$
Step 4: Compute the integral
We can now compute the integral of each term separately:
$$ S_1 = \frac{1}{2} \left[ \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} 1 \, dx - \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos 2x \, dx \right]. $$
The first integral is straightforward:
$$ \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} 1 \, dx = \frac{3\pi}{8} - \frac{\pi}{8} = \frac{\pi}{4}. $$
Now, we compute the second integral:
$$ \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \cos 2x \, dx. $$
The integral of $\cos 2x$ is:
$$ \int \cos 2x \, dx = \frac{\sin 2x}{2}. $$
Evaluating this from $\frac{\pi}{8}$ to $\frac{3\pi}{8}$:
$$ \left[ \frac{\sin 2x}{2} \right]_{\frac{\pi}{8}}^{\frac{3\pi}{8}} = \frac{\sin \left( \frac{3\pi}{4} \right)}{2} - \frac{\sin \left( \frac{\pi}{4} \right)}{2} = \frac{\frac{\sqrt{2}}{2}}{2} - \frac{\frac{\sqrt{2}}{2}}{2} = 0. $$
Step 5: Combine the results
Now, we can combine the results:
$$ S_1 = \frac{1}{2} \left( \frac{\pi}{4} - 0 \right) = \frac{\pi}{8}. $$
Step 6: Find $\frac{16S_1}{\pi}$
Finally, we compute:
$$ \frac{16S_1}{\pi} = \frac{16 \times \frac{\pi}{8}}{\pi} = 2. $$
Thus, the value of $\frac{16S_1}{\pi}$ is 2.