To solve the integral $\int_{0}^{9} e^{2\sqrt{x}} \, dx$, we can use the substitution method.

Step-by-Step Solution:

1. Substitution: Let's make the substitution: $u = 2\sqrt{x}$ Then, differentiate $u$ with respect to $x$: $du = \frac{d}{dx} (2\sqrt{x}) = \frac{2}{2\sqrt{x}} \cdot \frac{d}{dx}(x) = \frac{1}{\sqrt{x}} \, dx$ So, we have: $du = \frac{1}{\sqrt{x}} \, dx$ Multiplying both sides by $\sqrt{x}$, we get: $\sqrt{x} \, du = dx$ Also, express $\sqrt{x}$ in terms of $u$: $\sqrt{x} = \frac{u}{2}$ Thus, the differential $dx$ becomes: $dx = \frac{2}{u} \, du$

2. Adjust the limits of integration: When $x = 0$, $u = 2\sqrt{0} = 0$. When $x = 9$, $u = 2\sqrt{9} = 6$.

3. Rewrite the integral: Substitute $u$ and $dx$ into the integral: $\int_{0}^{9} e^{2\sqrt{x}} \, dx = \int_{0}^{6} e^{u} \cdot \frac{2}{u} \, du$ Simplifying further: $\int_{0}^{6} 2e^{u} \, du$

4. Integrate: The integral of $e^{u}$ with respect to $u$ is simply $e^{u}$, so we have: $2 \int_{0}^{6} e^{u} \, du = 2 \left[ e^{u} \right]_{0}^{6}$ Evaluate at the bounds: $2 \left( e^{6} - e^{0} \right) = 2 \left( e^{6} - 1 \right)$

So, the final result is: $\int_{0}^{9} e^{2\sqrt{x}} \, dx = 2 \left( e^{6} - 1 \right)$

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Related Questions:

1. How can the substitution method be used in other types of integrals?
2. What is the significance of $e$ (Euler's number) in calculus?
3. How do you handle integrals with square roots in the integrand?
4. Can substitution be used for definite integrals in all cases?
5. How does integration by parts differ from substitution?

Tip: When performing substitution in integrals, always remember to adjust the limits of integration if you're dealing with a definite integral.