## Problem Restatement Evaluate the definite integral: \[ \int_{\sqrt{5}}^{3} \frac{4z}{\sqrt{z^2-4}} \, dz \] --- ## Step 1: **Substitute and Find the Indefinite Integral** Let’s use substitution. Let \( u = z^2 - 4 \implies du = 2z\,dz \). So, \( 4z\,dz = 2 \cdot 2z\,dz = 2\,du \). Thus, the integral becomes: \[ \int \frac{4z}{\sqrt{z^2-4}}\,dz = \int \frac{2\,du}{\sqrt{u}} = 2\int u^{-1/2} du \] Integrate: \[ 2\int u^{-1/2} du = 2 \cdot 2u^{1/2} = 4u^{1/2} \] Substituting back \( u = z^2 - 4 \): \[ 4\sqrt{z^2 - 4} \] **Explanation:** We used substitution to simplify the integral, and then integrated using the power rule. --- ## Step 2: **Evaluate the Definite Integral** Now, evaluate from \( z = \sqrt{5} \) to \( z = 3 \): \[ 4\sqrt{z^2 - 4} \Big|_{\sqrt{5}}^{3} \] Calculate at \( z = 3 \): \[ 4\sqrt{3^2 - 4} = 4\sqrt{9 - 4} = 4\sqrt{5} \] Calculate at \( z = \sqrt{5} \): \[ 4\sqrt{(\sqrt{5})^2 - 4} = 4\sqrt{5 - 4} = 4\sqrt{1} = 4 \] Subtract: \[ 4\sqrt{5} - 4 \] **Explanation:** We plugged the limits into the antiderivative and subtracted to find the definite value. --- ## Final Answer \[ \boxed{4(\sqrt{5} - 1)} \]