To evaluate the integral $\int_{0}^{9} e^{2\sqrt{x}} \, dx$, we will perform a substitution to simplify the expression.

Step-by-Step Solution

1. Substitute: Let $u = 2\sqrt{x}$. Then, $x = \frac{u^2}{4}$.

2. Differentiate: Compute $dx$ in terms of $du$.

   $\frac{du}{dx} = \frac{d}{dx}(2\sqrt{x}) = \frac{2}{2\sqrt{x}} = \frac{1}{\sqrt{x}}$

   Therefore,

   $dx = \frac{du}{2\sqrt{x}}$

   Since $\sqrt{x} = \frac{u}{2}$, we have:

   $dx = \frac{du}{2 \cdot \frac{u}{2}} = \frac{du}{u}$

   Hence,

   $dx = \frac{du}{2u}$

3. Change of limits: When $x = 0$, $u = 2\sqrt{0} = 0$. When $x = 9$, $u = 2\sqrt{9} = 6$.

4. Rewrite the integral: The integral in terms of $u$ is:

   $\int_{0}^{9} e^{2\sqrt{x}} \, dx = \int_{0}^{6} e^{u} \cdot \frac{u}{4} \, du$

   Simplify:

   $\int_{0}^{6} \frac{e^u}{u} \cdot \frac{u}{4} \, du = \frac{1}{4} \int_{0}^{6} ue^u \, du$

5. Integrate: Integrate $e^u$ with respect to $u$.

   $\int ue^u \, du = e^u(u - 1) + C$

6. Evaluate definite integral: Apply the limits from 0 to 6.

   $\left[\frac{1}{4} \left(e^u(u - 1)\right)\right]_{0}^{6} = \frac{1}{4} \left( (6 - 1)e^6 - (0 - 1)e^0 \right)$

   $= \frac{1}{4} \left(5e^6 + 1\right)$

7. Simplify:

   $= \frac{5e^6 + 1}{4}$

Final Answer

$\int_{0}^{9} e^{2\sqrt{x}} \, dx = \frac{5e^6 + 1}{4}$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How would you solve $\int_{0}^{9} e^{3\sqrt{x}} \, dx$?
2. What happens if the limits of integration change, for example, from 1 to 4?
3. How does integration by parts work, and when is it useful?
4. Can you derive the formula for $\int e^{ax} \, dx$ from scratch?
5. What is the importance of substitution in integration, and how does it work?

Tip: Always check the limits of integration when substituting variables to ensure they match the new variable's range.