Problem 88 - Thomas Calculus in SI Units

Question

Evaluate the integrals in Exercises \(83-92.\) $$\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x$$

Step-by-Step Solution

Answer

\( \frac{4}{\ln(2)} \)

Step 1: Substitution

To evaluate the integral \( \int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} dx \), we start with substitution. Let \( u = \sqrt{x} \) which means \( x = u^2 \), and \( \frac{dx}{du} = 2u \). This gives \( dx = 2u \, du \).

Step 2: Change the Limits of Integration

Convert the limits of integration. When \( x = 1 \), \( u = \sqrt{1} = 1 \). When \( x = 4 \), \( u = \sqrt{4} = 2 \). So the integral becomes \( \int_{1}^{2} \frac{2^u}{u} \, 2u \, du \).

Step 3: Simplify the Integral

Simplify the integral: \( \int_{1}^{2} 2^{u} \cdot 2 \, du = 2 \int_{1}^{2} 2^{u} \, du \).

Step 4: Integrate with Respect to u

The integral \( \int 2^u \, du \) can be solved using the formula \( \int a^u \, du = \frac{a^u}{\ln(a)} + C \), where \( a = 2 \). Therefore, this integral becomes \( \frac{2^u}{\ln(2)} \) evaluated from 1 to 2.

Step 5: Evaluate the Definite Integral

Now calculate \( 2 \left[ \frac{2^u}{\ln(2)} \right]_{1}^{2} = \frac{2}{\ln(2)} \left( 2^2 - 2^1 \right) = \frac{2}{\ln(2)} (4 - 2) = \frac{4}{\ln(2)} \).

Key Concepts

Substitution Method

The substitution method is a helpful technique in calculus used to simplify the integration process. In this method, we choose a new variable to replace an existing expression, making it easier to compute the integral.

This change of variable aims to reduce the complexity of the original function. In the example provided, the integral \( \int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} \, dx \) is simplified by letting \( u = \sqrt{x} \). This implies that \( x = u^2 \) and therefore, \( dx = 2u \, du \). The initial expression \( 2^{\sqrt{x}} \) becomes \( 2^u \) with this substitution.

By making this transformation, we convert the integral into a form that is more manageable to calculate. The substitution method is particularly useful when dealing with composite functions or functions that involve roots and exponents.

Integration Limits

When performing an integral using substitution, it is essential to change the integration limits along with the variable. The integration limits determine the range over which we integrate. If the limits are not correctly adjusted, the solution will be incorrect.

In the exercise, the original integral had limits from 1 to 4, corresponding to the variable \( x \). When we substitute \( u = \sqrt{x} \), these limits must also change. For \( x = 1 \), \( u = \sqrt{1} = 1 \). For \( x = 4 \), \( u = \sqrt{4} = 2 \).

Thus, the new limits for \( u \) range from 1 to 2. This adjustment ensures that the integration is performed over the correct interval, reflecting the transformation of the variable. Remembering to change the limits is as crucial as the substitution itself, as it preserves the integrity and correctness of the integral.

Exponential Functions

Exponential functions have the form \( a^x \), where the base \( a \) is positive and constant, and \( x \) is the variable exponent. In integrals, dealing with exponential functions requires specific techniques for evaluation.

In the context of the definite integral \( \int 2^u \, du \), we use the formula \( \int a^u \, du = \frac{a^u}{\ln(a)} + C \). This formula is key when integrating exponential functions, and is derived from the rules of differential calculus.

With \( a = 2 \), this transforms our problem into \( \frac{2^u}{\ln(2)} \). Finally, evaluating this from the new limits of 1 to 2 gives the solution to the integral. Working with exponential functions often involves recognizing patterns and the correct application of integration rules. Understanding these fundamentals allows for efficient and accurate solutions to integrals involving exponential expressions.