Okay, let's evaluate the integral shown in the image (Equation 4.3). This integral represents the total area under the current curve for a half-cycle of a half-wave rectified signal. Evaluate the integral for output current over half a cycle: The integral given is: $$ \int_{0}^{\pi} \frac{V_m \sin(\omega t)}{R_D + R_L} d(\omega t) $$ Step 1: Identify constants. The terms $V_m$, $R_D$, and $R_L$ are constants with respect to the integration variable $\omega t$. We can pull them out of the integral. $$ \frac{V_m}{R_D + R_L} \int_{0}^{\pi} \sin(\omega t) d(\omega t) $$ Step 2: Integrate $\sin(\omega t)$. The integral of $\sin(x)$ with respect to $x$ is $-\cos(x)$. $$ \int_{0}^{\pi} \sin(\omega t) d(\omega t) = [-\cos(\omega t)]_{0}^{\pi} $$ Step 3: Apply the limits of integration. Substitute the upper limit ($\pi$) and the lower limit ($0$) into the result of the integration. $$ [-\cos(\omega t)]_{0}^{\pi} = (-\cos(\pi)) - (-\cos(0)) $$ We know that $\cos(\pi) = -1$ and $\cos(0) = 1$. $$ = (-(-1)) - (-1) $$ $$ = 1 + 1 $$ $$ = 2 $$ Step 4: Combine the results. Multiply the constant term by the result of the definite integral. $$ \frac{V_m}{R_D + R_L} \times 2 $$ $$ = \frac{2 V_m}{R_D + R_L} $$ The value of the integral representing the output current over half a cycle is $\boxed{\frac{2 V_m}{R_D + R_L}}$. Last free one today — make it count tomorrow, or type /upgrade for unlimited.