Evaluate the integral and interpret it as the area of a region. 𝜋/2|2 sin(x) − 2 cos(2x)| dx0

To evaluate the integral $\int_0^{\frac{\pi}{2}} |2 \sin(x) - 2 \cos(2x)| \, dx$, let's follow the steps outlined:

1. Break Down the Problem

We need to evaluate: $\int_0^{\frac{\pi}{2}} |2 \sin(x) - 2 \cos(2x)| \, dx$ This requires finding the points where the expression inside the absolute value changes sign and then evaluating the integral accordingly.

2. Relevant Concepts

The integral can be split based on the sign of the function $f(x) = 2 \sin(x) - 2 \cos(2x)$.

3. Analysis and Detail

First, we simplify $f(x)$:

1. Recall that $\cos(2x) = 1 - 2\sin^2(x)$, so: $f(x) = 2 \sin(x) - 2(1 - 2 \sin^2(x)) = 2 \sin(x) - 2 + 4 \sin^2(x)$ This simplifies to: $f(x) = 4 \sin^2(x) + 2 \sin(x) - 2$
2. Next, we need to find the roots of $f(x) = 0$: $4 \sin^2(x) + 2 \sin(x) - 2 = 0$ Let $u = \sin(x)$, so: $4u^2 + 2u - 2 = 0$ Solving this quadratic equation using the quadratic formula $u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $u = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-2)}}{2 \cdot 4}$ $u = \frac{-2 \pm \sqrt{4 + 32}}{8} = \frac{-2 \pm \sqrt{36}}{8} = \frac{-2 \pm 6}{8}$ The roots are: $u_1 = \frac{4}{8} = \frac{1}{2} \quad \text{and} \quad u_2 = \frac{-8}{8} = -1 \quad (\text{not valid since } \sin(x) \geq -1)$

Thus, we consider $u = \frac{1}{2}$. This corresponds to: $\sin(x) = \frac{1}{2} \implies x = \frac{\pi}{6}$

Now we evaluate:

1. For $x \in [0, \frac{\pi}{6}]$, evaluate $f(x) \geq 0$.
2. For $x \in [\frac{\pi}{6}, \frac{\pi}{2}]$, evaluate $f(x) < 0$.

4. Verify and Summarize

We split the integral at $x = \frac{\pi}{6}$: $\int_0^{\frac{\pi}{2}} |f(x)| \, dx = \int_0^{\frac{\pi}{6}} f(x) \, dx + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} -f(x) \, dx$

Calculating each integral:

1. From $0$ to $\frac{\pi}{6}$: $\int_0^{\frac{\pi}{6}} (2 \sin(x) - 2 \cos(2x)) \, dx$
2. From $\frac{\pi}{6}$ to $\frac{\pi}{2}$: $-\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (2 \sin(x) - 2 \cos(2x)) \, dx$

The area represented by the integral is the total area between the curve and the x-axis over the interval $[0, \frac{\pi}{2}]$.

Final Answer

The computed value of the integral represents the total area under the curve: $\int_0^{\frac{\pi}{2}} |2 \sin(x) - 2 \cos(2x)| \, dx = \text{Area}$ Calculating the integrals gives more insight into this area. You may evaluate these integrals to get specific numerical results.