To integrate the function $\sqrt{5 - 2x}$ from $-2$ to $0$, we can use the method of substitution.

Step-by-Step Solution:

1. Substitution: Let $u = 5 - 2x$. Then, differentiate $u$ with respect to $x$:

   $\frac{du}{dx} = -2 \implies dx = -\frac{1}{2} du$

2. Change of limits: When $x = -2$, $u = 5 - 2(-2) = 9$.

   When $x = 0$, $u = 5 - 2(0) = 5$.

   So the integral in terms of $u$ is now from $u = 9$ to $u = 5$.

3. Rewrite the integral: Substitute $u$ and $dx$ into the integral:

   $\int_{-2}^{0} \sqrt{5 - 2x} \, dx = \int_{9}^{5} \sqrt{u} \left(-\frac{1}{2} du\right)$

   $= -\frac{1}{2} \int_{9}^{5} u^{1/2} \, du$

4. Integrate: Now we integrate $u^{1/2}$:

   $= -\frac{1}{2} \left[ \frac{2}{3} u^{3/2} \right]_{9}^{5}$

   $= -\frac{1}{3} \left[ u^{3/2} \right]_{9}^{5}$

5. Evaluate at the limits:

   $= -\frac{1}{3} \left( 5^{3/2} - 9^{3/2} \right)$

   Calculate the values:

   • $5^{3/2} = 5 \cdot \sqrt{5}$
   • $9^{3/2} = 9 \cdot \sqrt{9} = 27$

   $= -\frac{1}{3} \left( 5\sqrt{5} - 27 \right)$

6. Final answer:

   $= \frac{27 - 5\sqrt{5}}{3}$

This is the evaluated result of the integral $\int_{-2}^{0} \sqrt{5 - 2x} \, dx$.

Do you want further details or have any questions?

Here are some related questions you might find interesting:

1. How do you use substitution for more complex integrals?
2. What are the different methods for solving definite integrals?
3. How do you handle improper integrals with square root functions?
4. How can you confirm the result of an integral using numerical methods?
5. What are the applications of definite integrals in real-world problems?

Tip: When substituting variables, always remember to change the limits of integration accordingly.