Question

The maximum value of $\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x. f(x) dx$, subject to the condition $|f(x)| \leq 5$ is $M$, then $\frac{M}{10}$ is equal to

Answer 1

2

Answer 2

To find the maximum value of the integral $\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} \sin x \cdot f(x) dx$ subject to the condition $|f(x)| \leq 5$, we need to choose $f(x)$ such that the integrand $\sin x \cdot f(x)$ is maximized at every point $x$ in the interval.

The condition $|f(x)| \leq 5$ implies $-5 \leq f(x) \leq 5$. To maximize the product $\sin x \cdot f(x)$:

1. If $\sin x > 0$, we should choose $f(x)$ to be as large positive as possible, i.e., $f(x) = 5$. In this case, $\sin x \cdot f(x) = 5 \sin x$.
2. If $\sin x < 0$, we should choose $f(x)$ to be as large negative as possible, i.e., $f(x) = -5$. In this case, $\sin x \cdot f(x) = -5 \sin x$.

Combining these two cases, we can write the optimal choice for $f(x)$ as $f(x) = 5 \cdot \text{sgn}(\sin x)$. Substituting this into the integral, the integrand becomes: $\sin x \cdot f(x) = \sin x \cdot (5 \cdot \text{sgn}(\sin x)) = 5 \cdot (\sin x \cdot \text{sgn}(\sin x)) = 5 |\sin x|$.

So, the maximum value $M$ is given by: $M = \int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} 5 |\sin x| dx = 5 \int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} |\sin x| dx$.

Now, we need to evaluate the integral $\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} |\sin x| dx$. We split the integral based on the sign of $\sin x$ in the given interval:

• For $x \in [-\frac{\pi}{2}, 0]$, $\sin x \leq 0$, so $|\sin x| = -\sin x$.
• For $x \in [0, \pi]$, $\sin x \geq 0$, so $|\sin x| = \sin x$.
• For $x \in [\pi, \frac{3\pi}{2}]$, $\sin x \leq 0$, so $|\sin x| = -\sin x$.

Therefore, $M = 5 \left[ \int_{-\frac{\pi}{2}}^{0} (-\sin x) dx + \int_{0}^{\pi} \sin x dx + \int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx \right]$

Let's evaluate each part:

1. $\int_{-\frac{\pi}{2}}^{0} (-\sin x) dx = [\cos x]_{-\frac{\pi}{2}}^{0} = \cos 0 - \cos(-\frac{\pi}{2}) = 1 - 0 = 1$.
2. $\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = (-\cos \pi) - (-\cos 0) = (-(-1)) - (-1) = 1 + 1 = 2$.
3. $\int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx = [\cos x]_{\pi}^{\frac{3\pi}{2}} = \cos \frac{3\pi}{2} - \cos \pi = 0 - (-1) = 1$.

Summing these values: $\int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} |\sin x| dx = 1 + 2 + 1 = 4$.

Now, substitute this back into the expression for $M$: $M = 5 \times 4 = 20$.

The question asks for the value of $\frac{M}{10}$. $\frac{M}{10} = \frac{20}{10} = 2$.

The final answer is $\boxed{2}$.

Explanation of the solution: To maximize $\int g(x) f(x) dx$ with $|f(x)| \leq C$, choose $f(x) = C \cdot \text{sgn}(g(x))$. Here $g(x) = \sin x$ and $C=5$. So $f(x) = 5 \cdot \text{sgn}(\sin x)$. The integral becomes $M = \int_{-\frac{\pi}{2}}^{\frac{3\pi}{2}} 5 |\sin x| dx = 5 \left( \int_{-\frac{\pi}{2}}^{0} (-\sin x) dx + \int_{0}^{\pi} \sin x dx + \int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx \right)$. Evaluating the integrals: $\int_{-\frac{\pi}{2}}^{0} (-\sin x) dx = [\cos x]_{-\frac{\pi}{2}}^{0} = 1 - 0 = 1$. $\int_{0}^{\pi} \sin x dx = [-\cos x]_{0}^{\pi} = 1 - (-1) = 2$. $\int_{\pi}^{\frac{3\pi}{2}} (-\sin x) dx = [\cos x]_{\pi}^{\frac{3\pi}{2}} = 0 - (-1) = 1$. Thus, $M = 5 (1 + 2 + 1) = 5 \times 4 = 20$. Finally, $\frac{M}{10} = \frac{20}{10} = 2.