Evaluating the Integral of (e^{sqrt{x}}): Step-by-step Guide

Understanding and evaluating the integral of #8203;(e^{sqrt{x}}) can be a challenging task, but with the right techniques and approach, it becomes much easier. In this article, we will guide you through the process, using various methods such as substitution and integration by parts. Let's dive in!

1. Introduction and Basic Concepts

The integral we are considering is (int e^{sqrt{x}}, dx). At first glance, this integral may seem daunting, but we can break it down into simpler steps. The key is to use substitution and integration by parts effectively.

2. Approach using Substitution

Step 1: Substitution
Let (t sqrt{x}). This substitution simplifies the integral significantly.

2.1 Calculating (dt)

First, we need to express (dt) in terms of (dx).

dt frac{1}{2sqrt{x}}dx frac{1}{2t}dx

Therefore, (frac{dx}{dt} 2t). This means (dx 2t , dt).

2.2 Substituting into the Integral

Substituting (t sqrt{x}) and (dx 2t , dt) into the integral, we get:

int e^{sqrt{x}}, dx int e^t cdot 2t , dt

Now our integral looks much simpler: (int 2t e^t , dt).

2.3 Integration by Parts

Next, we apply the integration by parts formula (int u , dv uv - int v , du). Let's choose:

u 2t

dv e^t dt

Then:

du 2 dt

v (int e^t , dt e^t)

Applying integration by parts:

int 2t e^t , dt 2t e^t - int 2 e^t , dt

This simplifies to:

int 2t e^t , dt 2t e^t - 2 int e^t , dt 2t e^t - 2e^t C

Substituting back (t sqrt{x}), we get:

int e^{sqrt{x}}, dx 2sqrt{x} e^{sqrt{x}} - 2e^{sqrt{x}} C 2e^{sqrt{x}}(sqrt{x} - 1) C

3. Verification and Additional Insights

To verify, let's differentiate the result:

frac{d}{dx} [2e^{sqrt{x}}(sqrt{x} - 1) C] 2[1 cdot e^{sqrt{x}}(sqrt{x} - 1) sqrt{x} e^{sqrt{x}} cdot frac{1}{2sqrt{x}}]

Simplifying the derivative:

2[e^{sqrt{x}}(sqrt{x} - 1) frac{1}{2}e^{sqrt{x}}] 2e^{sqrt{x}}(sqrt{x} - frac{1}{2} frac{1}{2}) 2e^{sqrt{x}} e^{sqrt{x}}

This confirms that the integral is correct!

4. Conclusion

Evaluating integrals of complex functions like (e^{sqrt{x}}) requires a combination of substitution and integration by parts. By breaking down the problem into smaller, manageable parts, we can find a solution systematically. This approach not only helps in solving such problems but also enhances your understanding of integral calculus.