Problem 52 - Thomas Calculus

Question

Evaluate the integrals. \(\int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{4}{\ln 2} \).

Step 1: Define the Problem

We need to evaluate the integral \( \int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} \, dx \). This is a definite integral from 1 to 4.

Step 2: Perform a Substitution

To simplify the integral, use the substitution \( u = \sqrt{x} \). Then \( x = u^2 \) and \( dx = 2u \, du \). Adjust the limits too: when \( x = 1 \), \( u = 1 \); when \( x = 4 \), \( u = 2 \).

Step 3: Rewrite the Integral

The integral becomes \( \int_{1}^{2} \frac{2^{u}}{u} \cdot 2u \, du \), which simplifies to \( 2 \int_{1}^{2} 2^{u} \, du \).

Step 4: Integrate the Function

Now, we need to integrate \( 2^{u} \). The integral of \( 2^{u} \) is \( \frac{2^{u}}{\ln 2} \), so the integral \( 2 \int 2^{u} \, du = \frac{2^{u + 1}}{\ln 2} + C \), where C is the constant of integration.

Step 5: Apply the Limits

After finding the indefinite integral, plug in the limits: \( \left[ \frac{2^{u + 1}}{\ln 2} \right]_{1}^{2} = \frac{2^{3}}{\ln 2} - \frac{2^{2}}{\ln 2} = \frac{8}{\ln 2} - \frac{4}{\ln 2} \).

Step 6: Simplify the Result

Simplifying \( \frac{8}{\ln 2} - \frac{4}{\ln 2} \) gives us \( \frac{4}{\ln 2} \). So, the value of the integral is \( \frac{4}{\ln 2} \).

Key Concepts

Definite Integrals

A definite integral is a way to calculate the accumulation of quantities. It is represented by integrating a function over a specific interval, in this case from 1 to 4. You can think of it as finding the area under the curve of the function within these boundaries. It differs from an indefinite integral in that it produces a specific numerical value rather than a general form plus a constant of integration.

When you see the notation \( \int_{1}^{4} \frac{2^{\sqrt{x}}}{\sqrt{x}} \, dx \), it's a definite integral because of the specified limits of integration (1 and 4). This tells us to find the total value summed between these boundaries of the function provided.

• The lower limit of integration (1) is where you start accumulation, and
• The upper limit (4) is where you stop.

Through definite integrals, you can solve real-world problems involving quantities like distance, area, and more.

Substitution Method

The substitution method is a powerful technique in calculus for simplifying integrals and making them easier to evaluate. When dealing with complicated integrals, replacing a part of the integral with a simpler variable (often \(u\)) can streamline the calculation process.

In our problem, we used \(u = \sqrt{x}\). This choice simplifies the integral significantly. By substituting \(x = u^2\), we also express \(dx\) in terms of \(du\) as \(dx = 2u \, du\). Don't forget to adjust the limits accordingly:

• For \(x = 1\), we found \(u = 1\),
• And for \(x = 4\), \(u = 2\).

After substitution, our integral reforms into a simpler expression: \( 2 \int_{1}^{2} 2^u \, du \). This reduction makes integration and further calculations more straightforward.

Exponential Functions

Exponential functions, such as \( 2^u \), are significant in calculus due to their distinctive properties and roles in various real-life applications. An exponential function is characterized by a constant base raised to a variable exponent.

The integration of basic exponential functions often results in terms involving natural logarithms. For example, the integral of an exponential function \(a^x\) is \(\frac{a^x}{\ln a} + C\), where \(C\) is the constant of integration. In our exercise, integrating \(2^u\) gives us \(\frac{2^u}{\ln 2}\).

Exponential functions grow rapidly, and their derivatives and integrals produce results featuring the original function. This unique nature makes them essential in diverse fields such as finance, physics, and population dynamics, where processes occur continuously and compounding relationships are present.

Through understanding these functions and their properties, evaluating such integrals becomes a more intuitive task.