The image you provided contains the integral:

$\int_0^4 \frac{dx}{\sqrt{4 - x}}$

This is a standard integral that can be evaluated using substitution or known integral formulas.

Let's walk through the solution step by step:

1. Substitution: Let $u = 4 - x$. Therefore, $du = -dx$, and the limits of integration change accordingly:

   • When $x = 0$, $u = 4$.
   • When $x = 4$, $u = 0$.

   The integral now becomes: $\int_4^0 \frac{-du}{\sqrt{u}} = \int_0^4 \frac{du}{\sqrt{u}}.$

2. Simplification: We now have a standard integral: $\int_0^4 u^{-\frac{1}{2}} du.$

3. Integration: The integral of $u^{-\frac{1}{2}}$ is $2u^{\frac{1}{2}}$. Applying the limits, we get: $2\left[u^{\frac{1}{2}}\right]_0^4 = 2\left(4^{\frac{1}{2}} - 0^{\frac{1}{2}}\right) = 2(2 - 0) = 4.$

Thus, the value of the integral is:

$\boxed{4}.$

Would you like more details or have any questions? Here are some related questions to explore further:

1. What are other common integrals that involve square roots?
2. How can trigonometric substitution be applied to similar integrals?
3. What is the geometric interpretation of integrals involving square roots like this one?
4. How do you handle definite integrals with more complex limits?
5. Can you generalize the process for integrals of the form $\int \frac{dx}{\sqrt{a - bx}}$?

Tip: Whenever you encounter an integral with a square root expression like $\sqrt{a - x}$, consider substitution to simplify the integral into a standard form.