1. **State the problem:** Find the exact area under the curve of the function $f_1(x) = \sqrt{4x - 7}$ from $x=4$ to $x=8$. 2. **Formula used:** The area under a curve $y=f(x)$ from $x=a$ to $x=b$ is given by the definite integral: $$\text{Area} = \int_a^b f(x) \, dx$$ 3. **Set up the integral:** $$\int_4^8 \sqrt{4x - 7} \, dx$$ 4. **Substitution:** Let $u = 4x - 7$, then $\frac{du}{dx} = 4$ or $dx = \frac{du}{4}$. When $x=4$, $u = 4(4) - 7 = 16 - 7 = 9$. When $x=8$, $u = 4(8) - 7 = 32 - 7 = 25$. 5. **Rewrite the integral in terms of $u$:** $$\int_9^{25} \sqrt{u} \cdot \frac{1}{4} \, du = \frac{1}{4} \int_9^{25} u^{\frac{1}{2}} \, du$$ 6. **Integrate:** $$\frac{1}{4} \cdot \left[ \frac{u^{\frac{3}{2}}}{\frac{3}{2}} \right]_9^{25} = \frac{1}{4} \cdot \left[ \frac{2}{3} u^{\frac{3}{2}} \right]_9^{25} = \frac{1}{4} \cdot \frac{2}{3} \left(25^{\frac{3}{2}} - 9^{\frac{3}{2}} \right)$$ 7. **Simplify constants:** $$\frac{1}{4} \cdot \frac{2}{3} = \frac{2}{12} = \frac{1}{6}$$ 8. **Evaluate powers:** $$25^{\frac{3}{2}} = (\sqrt{25})^3 = 5^3 = 125$$ $$9^{\frac{3}{2}} = (\sqrt{9})^3 = 3^3 = 27$$ 9. **Calculate the final area:** $$\frac{1}{6} (125 - 27) = \frac{1}{6} \times 98 = \frac{98}{6} = \frac{49}{3}$$ **Final answer:** $$\boxed{\frac{49}{3}}$$