Question

The value of $\frac{48 \,S _{2}}{\pi^{2}}$ is_____

Answer 1

Correct Answer: 1.5

Step 1: Understand the given functions
We are given the 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 $f(x) = \sin^2 x$. We are tasked with finding the value of $\frac{48S_2}{\pi^2}$, where $S_2$ is defined as:
$$ S_2 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} f(x) \cdot g_2(x) \, dx = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \left|4x - \pi\right| \, dx. $$
Step 2: Simplify the expression for $S_2$
We need to evaluate the integral for $S_2$. First, express $|4x - \pi|$ as $4|x - \frac{\pi}{4}|$.
Thus, the integral becomes:
$$ S_2 = \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \cdot 4 \left|x - \frac{\pi}{4}\right| \, dx. $$
We can factor out the constant 4:
$$ S_2 = 4 \int_{\frac{\pi}{8}}^{\frac{3\pi}{8}} \sin^2 x \left|x - \frac{\pi}{4}\right| \, dx. $$
Step 3: Change of variables
Now, let’s make the substitution $x - \frac{\pi}{4} = t$, so that $dx = dt$. When $x = \frac{\pi}{8}$, $t = -\frac{\pi}{8}$, and when $x = \frac{3\pi}{8}$, $t = \frac{\pi}{8}$.
Thus, the integral becomes:
$$ S_2 = 4 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \sin^2 \left( \frac{\pi}{4} + t \right) |t| \, dt. $$
Step 4: Simplify the expression for $\sin^2 \left( \frac{\pi}{4} + t \right)$
We use the trigonometric identity for $\sin^2 \left( \frac{\pi}{4} + t \right)$:
$$ \sin^2 \left( \frac{\pi}{4} + t \right) = \frac{1}{2} \left( 1 - \cos 2 \left( \frac{\pi}{4} + t \right) \right). $$
Substituting this back into the integral, we get:
$$ S_2 = 4 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \frac{1}{2} \left( 1 - \cos \left( 2 \times \left( \frac{\pi}{4} + t \right) \right) \right) |t| \, dt. $$
Simplifying the cosine term:
$$ S_2 = 2 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \left( 1 - \cos \left( \frac{\pi}{2} + 2t \right) \right) |t| \, dt. $$
Step 5: Expand and simplify
The cosine term simplifies as:
$$ \cos \left( \frac{\pi}{2} + 2t \right) = -\sin(2t). $$
Thus, the integral becomes:
$$ S_2 = 2 \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} \left( 1 + \sin(2t) \right) |t| \, dt. $$
We now split the integral into two parts:
$$ S_2 = 2 \left( \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \, dt + \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \sin(2t) \, dt \right). $$
Step 6: Evaluate the integrals
The first integral is straightforward:
$$ \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \, dt = 2 \int_{0}^{\frac{\pi}{8}} t \, dt = 2 \left[ \frac{t^2}{2} \right]_0^{\frac{\pi}{8}} = 2 \times \frac{\pi^2}{128} = \frac{\pi^2}{64}. $$
The second integral is 0 because $\sin(2t)$ is an odd function and $|t|$ is even, so their product integrated over the symmetric interval $[-\frac{\pi}{8}, \frac{\pi}{8}]$ gives zero:
$$ \int_{-\frac{\pi}{8}}^{\frac{\pi}{8}} |t| \sin(2t) \, dt = 0. $$
Step 7: Combine the results
Thus, we have:
$$ S_2 = 2 \times \frac{\pi^2}{64} = \frac{\pi^2}{32}. $$
Step 8: Compute $\frac{48S_2}{\pi^2}$
Finally, we compute:
$$ \frac{48S_2}{\pi^2} = \frac{48 \times \frac{\pi^2}{32}}{\pi^2} = \frac{48}{32} = 1.5. $$
Thus, the value of $\frac{48S_2}{\pi^2}$ is 1.5.