Chapter 5: Problem 50
Evaluate the integrals using appropriate substitutions. $$ \int \sqrt{e^{x}} d x $$
Short Answer
Expert verified
The integral of \( \sqrt{e^{x}} \) is \( 2\sqrt{e^{x}} + C \).
Step by step solution
01
Identify the Inner Function
Look at the integrand, which is \( \sqrt{e^{x}} \). Notice that the expression inside the square root, \( e^{x} \), can be used as the basis for substitution.
02
Choose the Substitution
Set \( u = \sqrt{e^{x}} \). Then we have \( u^{2} = e^{x} \). Differentiate both sides to find \( du \). We get \( 2u \frac{du}{dx} = e^{x} \), or \( du = \frac{1}{2}xe^{x/2}dx \) when expressed in terms of \( x \) and \( du \) for this substitution.
03
Express dx in Terms of u
To find \( dx \) in terms of \( du \), note that \( x = 2\ln(u) \) and differentiate with respect to \( u \) to get \( dx = \frac{2}{u} du \). Substitute this expression for \( dx \) in the integral.
04
Change the Integral to u-terms
Substitute into the integral: \( \int \sqrt{e^{x}} dx = \int u \cdot \frac{2}{u} du = 2 \int du \). This simplifies to \( 2 \int du \).
05
Integrate
Integrate the expression: \( 2 \int du = 2u + C \).
06
Substitute Back to x
Since \( u = \sqrt{e^{x}} \), substitute back: \( 2u + C = 2\sqrt{e^{x}} + C \). Now, the integral in terms of \( x \) is \( 2\sqrt{e^{x}} + C \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
The Substitution Method is a powerful technique for solving integrals, particularly when dealing with complex expressions. It simplifies integration by changing variables. Imagine you have a complicated function to integrate, like \( \sqrt{e^{x}} \). By using substitution, you find a new, simpler variable to express the problem.
In this example, choose \( u = \sqrt{e^{x}} \), which transforms the integral into a more manageable form. This step is crucial, as identifying a fitting substitution is often the key to solving the integral.
In this example, choose \( u = \sqrt{e^{x}} \), which transforms the integral into a more manageable form. This step is crucial, as identifying a fitting substitution is often the key to solving the integral.
- Select a part of the function as the new variable \( u \). Here, \( u = \sqrt{e^{x}} \).
- Find the derivative of this new variable, \( du \), in terms of the original variable \( x \).
- Substitute \( du \) and \( u \) back into the integral to simplify it.
Definite Integral
Definite integrals evaluate the area under a curve between two specific points, providing a number instead of a function. While this exercise didn't involve finding a definite integral, it's helpful to understand how it differs from indefinite integrals.
In a definite integral, you compute the integral from one number to another. The limits of integration are these specific numbers, like \( \int_{a}^{b} f(x) \). The result gives the total accumulated area under the curve between the limits.
In practice, you solve a definite integral by:
In a definite integral, you compute the integral from one number to another. The limits of integration are these specific numbers, like \( \int_{a}^{b} f(x) \). The result gives the total accumulated area under the curve between the limits.
In practice, you solve a definite integral by:
- Finding the antiderivative or integral of the function.
- Evaluating this antiderivative at the upper limit.
- Evaluating at the lower limit and subtracting the results.
Indefinite Integral
Indefinite integrals represent a family of functions, providing the general form of antiderivatives. It's the reverse process of differentiation. Unlike definite integrals, indefinite integrals include an arbitrary constant \( C \). This is because when you take the derivative, any constant term disappears, so the integral's constant remains unknown.
Consider the indefinite integral for \( \int \sqrt{e^{x}} \, dx \) performed in the exercise. The steps involve:
Consider the indefinite integral for \( \int \sqrt{e^{x}} \, dx \) performed in the exercise. The steps involve:
- Selecting \( u \) for substitution, simplifying the expression.
- Integrating the simplified expression \( 2 \int du \).
- Substituting back to the original variable.