**1. Visualize the Region:** The region is bounded by the line \(y = mx\), the circle \(x^2 + y^2 = 4\) (which has a radius of 2), and the x-axis in the first quadrant.

**2. Find the Intersection Point:** To find where the line and circle intersect, substitute \(y = mx\) into the circle's equation: \(x^2 + (mx)^2 = 4\) \(x^2 + m^2x^2 = 4\) \(x^2(1 + m^2) = 4\) \(x^2 = \frac{4}{1 + m^2}\) \(x = \frac{2}{\sqrt{1 + m^2}}\) (since we are in the first quadrant, x is positive) Now find the corresponding y value: \(y = m \cdot \frac{2}{\sqrt{1 + m^2}} = \frac{2m}{\sqrt{1 + m^2}}\) So, the intersection point is \(\left(\frac{2}{\sqrt{1 + m^2}}, \frac{2m}{\sqrt{1 + m^2}}\right)\).

**3. Set up the Integral:** The area of the region can be found by integrating the area under the circle from x = 0 to \(x = \frac{2}{\sqrt{1 + m^2}}\) and subtracting the area under the line \(y = mx\) from x = 0 to \(x = \frac{2}{\sqrt{1 + m^2}}\). First, express the circle as a function of x: \(y = \sqrt{4 - x^2}\). The area under the circle is \(\int_{0}^{\frac{2}{\sqrt{1 + m^2}}} \sqrt{4 - x^2} \, dx\). The area under the line is \(\int_{0}^{\frac{2}{\sqrt{1 + m^2}}} mx \, dx\). The area of the region is given by: Area = \(\int_{0}^{\frac{2}{\sqrt{1 + m^2}}} (\sqrt{4 - x^2} - mx) \, dx = \frac{\pi}{2}\)

**4. Evaluate the Integrals:** Let's evaluate the two integrals separately. * **Integral 1 (Circle):** \(\int_{0}^{\frac{2}{\sqrt{1 + m^2}}} \sqrt{4 - x^2} \, dx\) Use the substitution \(x = 2\sin\theta\), so \(dx = 2\cos\theta \, d\theta\). When \(x = 0\), \(\theta = 0\). When \(x = \frac{2}{\sqrt{1 + m^2}}\), \(\sin\theta = \frac{1}{\sqrt{1 + m^2}}\), so \(\theta = \arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)\). The integral becomes: \(\int_{0}^{\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)} \sqrt{4 - 4\sin^2\theta} \cdot 2\cos\theta \, d\theta = 4\int_{0}^{\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)} \cos^2\theta \, d\theta\) \( = 4\int_{0}^{\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)} \frac{1 + \cos(2\theta)}{2} \, d\theta = 2\left[\theta + \frac{1}{2}\sin(2\theta)\right]_{0}^{\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)}\) \( = 2\left[\theta + \sin\theta\cos\theta\right]_{0}^{\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)}\) Let \(\alpha = \arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right)\). Then \(\sin\alpha = \frac{1}{\sqrt{1 + m^2}}\) and \(\cos\alpha = \sqrt{1 - \frac{1}{1 + m^2}} = \sqrt{\frac{m^2}{1 + m^2}} = \frac{m}{\sqrt{1 + m^2}}\). So, the integral evaluates to: \(2\left[\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right) + \frac{1}{\sqrt{1 + m^2}} \cdot \frac{m}{\sqrt{1 + m^2}}\right] = 2\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right) + \frac{2m}{1 + m^2}\) * **Integral 2 (Line):** \(\int_{0}^{\frac{2}{\sqrt{1 + m^2}}} mx \, dx = m\left[\frac{1}{2}x^2\right]_{0}^{\frac{2}{\sqrt{1 + m^2}}} = \frac{m}{2} \cdot \frac{4}{1 + m^2} = \frac{2m}{1 + m^2}\)

**5. Solve for m:** Now, subtract the second integral from the first and set it equal to \(\frac{\pi}{2}\): \(2\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right) + \frac{2m}{1 + m^2} - \frac{2m}{1 + m^2} = \frac{\pi}{2}\) \(2\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right) = \frac{\pi}{2}\) \(\arcsin\left(\frac{1}{\sqrt{1 + m^2}}\right) = \frac{\pi}{4}\) \(\frac{1}{\sqrt{1 + m^2}} = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\) \(\sqrt{1 + m^2} = \sqrt{2}\) \(1 + m^2 = 2\) \(m^2 = 1\) \(m = 1\) (since \(m > 0\))