Definite integration

\[ \text{Area} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \, dx \end{aligned} \] By calculating the integral: \begin{aligned} \int \sin x \, dx & = -\cos x + C \\ \int_0^{\frac{\pi}{2}} \sin x \, dx & = -\cos\left(\frac{\pi}{2}\right) - (-\cos(0)) \\ & = 0 + 1 = 1 \end{aligned} \] Thus, area from \( \frac{\pi}{4} \) to \( \frac{\pi}{2} \): \begin{aligned} \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \, dx & = 1 - k \end{aligned} \] Since \( k = \int_0^{\frac{\pi}{4}} \sin x \, dx \), we substitute: \begin{aligned} 1 - k & = \text{Area between } \frac{\pi}{4} \text{ and } \frac{\pi}{2} \end{aligned} \] Hence, the required area is: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \sin x \, dx = 1 - k \] Thus, the area between the curve \( y = \sin x \) and the x-axis in the interval \([\frac{\pi}{4}, \frac{\pi}{2}]\) is \(1 - k\).