1. **State the problem:** Calculate the definite integral $$\int_1^4 \frac{dx}{\sqrt{5 - x}}$$ using the change of variable method. 2. **Choose a substitution:** Let $$u = 5 - x$$. Then, $$du = -dx$$ or $$dx = -du$$. 3. **Change the limits of integration:** When $$x = 1$$, $$u = 5 - 1 = 4$$. When $$x = 4$$, $$u = 5 - 4 = 1$$. 4. **Rewrite the integral in terms of $$u$$:** $$\int_1^4 \frac{dx}{\sqrt{5 - x}} = \int_4^1 \frac{-du}{\sqrt{u}} = \int_1^4 \frac{du}{\sqrt{u}}$$ 5. **Simplify the integral:** $$\int_1^4 u^{-\frac{1}{2}} du$$ 6. **Integrate:** Use the formula $$\int u^n du = \frac{u^{n+1}}{n+1} + C$$ for $$n \neq -1$$. Here, $$n = -\frac{1}{2}$$, so $$\int u^{-\frac{1}{2}} du = \frac{u^{\frac{1}{2}}}{\frac{1}{2}} = 2u^{\frac{1}{2}} = 2\sqrt{u}$$. 7. **Evaluate the definite integral:** $$2\sqrt{u} \Big|_1^4 = 2(\sqrt{4} - \sqrt{1}) = 2(2 - 1) = 2$$. **Final answer:** $$\int_1^4 \frac{dx}{\sqrt{5 - x}} = 2$$.