Problem 38 - Thomas Calculus

Question

Evaluate the integrals. \begin{equation}\int 2 e^{(2 x-1)} d x\end{equation}

Step-by-Step Solution

Answer

The integral evaluates to \( e^{(2x-1)} + C \).

Step 1: Identify the Integral Formula

We need to evaluate the integral \( \int 2 e^{(2x - 1)} \, dx \). This is an exponential integral, and it can be solved using the formula \( \int e^{ax+b} \, dx = \frac{1}{a}e^{ax+b} + C \).

Step 2: Determine the Coefficient

In our integral, the term inside the exponential function is \( 2x - 1 \), so \( a = 2 \) and \( b = -1 \). We identify that \( \int e^{(2x-1)} \, dx = \frac{1}{2} e^{(2x-1)} + C \).

Step 3: Apply the Coefficient of 2

Since the integral is \( \int 2 e^{(2x - 1)} \), multiply the result by 2. Thus, we have \( 2 \times \left( \frac{1}{2} e^{(2x-1)} + C \right) \).

Step 4: Simplify the Expression

After multiplying, we simplify: \( 2 \times \frac{1}{2} e^{(2x-1)} = e^{(2x-1)} \). So, the integral results in \( e^{(2x-1)} + C \).

Key Concepts

Integral Formula

When faced with an integral, identifying the correct formula to use is crucial for solving it. In this case, we are dealing with an exponential integral, specifically a type that involves the exponential function to a linear power. The integral formula that applies here is for integrals of the form: \( \int e^{ax+b} \, dx \). This formula can be transformed into a simpler expression: \[ \int e^{ax+b} \, dx = \frac{1}{a} e^{ax+b} + C \]Here, "\(a\)" and "\(b\)" are coefficients of the variables in the exponent, and "\(C\)" is the constant of integration. Identifying these coefficients accurately will pave the way for solving any integral involving exponential functions. In our exercise, we have \( a = 2 \) and \( b = -1 \), making it important to plug these values into the formula properly to reach the solution.

Exponential Function

Exponential functions are a foundational concept in calculus and often appear in integrals. These functions have the form \( e^{ax+b} \), where \(e\) is the base of natural logarithms, and \(ax+b\) is the exponent. The beauty of exponential functions lies in their sensitive growth, where even small changes in "\(a\)" or "\(b\)" result in significant changes to the function's value. In integration, exponentials often simplify the process, especially when accompanied by constants or linear exponents. For the integral \( \int 2 e^{(2x - 1)} \, dx \), the focus is on the linear expression within the exponent, \(2x-1\). Recognizing that the derivative of this exponent is a simple constant (\(2\)) is crucial for simplifying the integration process.Learning to bridge the gap between different sets of function and integral transformations can significantly enhance your problem-solving toolkit.

Simplification of Expressions

After applying the integral formula, simplification of expressions is often necessary to obtain a neat and concise result. Simplification involves reducing an expression to its simplest form, often by combining like terms or factoring.In our specific problem, once the integral is evaluated using the exponential formula, we multiply the outcome by the external coefficient "\(2\)."The expression becomes:\[ 2 \times \left( \frac{1}{2} e^{(2x-1)} + C \right) \]By distributing and simplifying, we simplify it to:\[ e^{(2x-1)} + C \]Simplification helps in recognizing patterns and makes the solution more readable and easy to understand. Done correctly, it also verifies that the solution aligns with the expected behavior of mathematical properties. Understanding these steps in simplification can greatly aid in getting to the final solution with confidence.