For What Value Of B Does The Integral

For What Value of b Does the Integral ∫₀¹ (bx² + 2x - 1) dx = 0? Solving a Definite Integral Problem

This article explores the mathematical problem of finding the value of b for which the definite integral ∫₀¹ (bx² + 2x - 1) dx equals zero. We will delve into the step-by-step solution, providing a detailed explanation of the integration process and algebraic manipulations involved. Understanding this problem enhances your comprehension of definite integrals, their applications, and how to solve equations involving integrals. This is a fundamental concept in calculus with applications in various fields, including physics, engineering, and economics.

Understanding Definite Integrals

Before we embark on solving the problem, let's briefly review the concept of definite integrals. A definite integral is a mathematical object that represents the signed area of the region bounded by the graph of a function and the x-axis, within specified limits of integration. In our case, the limits of integration are 0 and 1. The integral ∫₀¹ (bx² + 2x - 1) dx calculates the area under the curve of the function f(x) = bx² + 2x - 1 between x = 0 and x = 1. The value of this area is dependent on the value of b.

Solving the Integral

Our task is to find the value of b such that the definite integral ∫₀¹ (bx² + 2x - 1) dx = 0. To achieve this, we must first evaluate the integral. We can do this by applying the power rule of integration:

∫xⁿ dx = (xⁿ⁺¹)/(n+1) + C, where n ≠ -1 and C is the constant of integration.

Applying this rule to each term in our integrand (bx² + 2x - 1), we get:

∫₀¹ (bx² + 2x - 1) dx = [ (b/3)x³ + x² - x ]₀¹

The notation "[ ]₀¹" indicates that we need to evaluate the expression within the brackets at the upper limit (x = 1) and subtract the value of the expression at the lower limit (x = 0).

Evaluating the expression at the upper limit (x = 1):

(b/3)(1)³ + (1)² - (1) = b/3

Evaluating the expression at the lower limit (x = 0):

(b/3)(0)³ + (0)² - (0) = 0

Therefore, the definite integral becomes:

∫₀¹ (bx² + 2x - 1) dx = b/3 - 0 = b/3

Finding the Value of b

Now, we set the result of the integration equal to zero, as specified in the problem:

b/3 = 0

To solve for b, we simply multiply both sides of the equation by 3:

b = 0

Therefore, the value of b for which the integral ∫₀¹ (bx² + 2x - 1) dx = 0 is b = 0.

Graphical Interpretation

Let's consider the graphical interpretation of this result. The function f(x) = bx² + 2x - 1 represents a parabola. When b = 0, the function simplifies to f(x) = 2x - 1. The integral represents the area under this line between x = 0 and x = 1. If we sketch this line, we'll find that the area above the x-axis is equal to the area below the x-axis, resulting in a net area of zero. This confirms our solution.

Extending the Understanding: Different Values of b

While we've solved for the specific case where the integral equals zero, let's explore what happens with different values of b:

• b > 0: The parabola opens upwards. The area under the curve may be positive or negative depending on the specific value of b. A larger positive b generally leads to a larger positive area.

• b < 0: The parabola opens downwards. Similar to the positive case, the area can be positive or negative depending on the value of b. A more negative b generally leads to a larger negative area (or a smaller positive area).

This exploration highlights the dependency of the definite integral's value on the parameter b.

Practical Applications

The concept of finding the value of a constant that makes a definite integral equal to zero has numerous practical applications. For instance:

• Physics: In physics, integrals are frequently used to calculate work, displacement, or other physical quantities. Setting the integral to zero can represent a condition of equilibrium or a specific point in a physical process.

• Engineering: Similar to physics, engineers use integrals to analyze structures, calculate forces, and model various systems. Setting the integral to zero might represent a condition of stability or a point of balance.

• Economics: Economic models often employ integrals to represent total revenue, cost, or consumer surplus. Setting an integral to zero could signify a break-even point or a market equilibrium condition.

Frequently Asked Questions (FAQ)

Q1: What if the limits of integration were different?

A1: If the limits of integration were different, the result would also be different. The value of b that makes the integral zero would change accordingly. You would need to re-evaluate the definite integral with the new limits to find the corresponding value of b.

Q2: Can this problem be solved using numerical methods?

A2: While we solved this analytically, numerical methods like the trapezoidal rule or Simpson's rule could also be used to approximate the definite integral for different values of b. This would allow you to iteratively find a value of b that makes the integral close to zero. However, the analytical approach provides an exact solution.

Q3: Are there other techniques to solve this type of problem?

A3: Yes, other techniques could be used, depending on the complexity of the integrand. For more complex functions, techniques like integration by parts or substitution might be necessary before solving for the value of the constant.

Q4: What if the integrand contained other variables?

A4: If the integrand contained other variables besides b, the solution would involve solving a system of equations, where the number of equations would depend on the number of variables. The approach would be similar, but require more complex algebraic manipulation.

Conclusion

In conclusion, we have successfully determined that the value of b for which the definite integral ∫₀¹ (bx² + 2x - 1) dx equals zero is b = 0. This problem demonstrates a fundamental application of definite integrals and highlights the importance of understanding the power rule of integration and the process of evaluating definite integrals. The solution provides a solid foundation for tackling more complex problems involving definite integrals and parameters within the integrand. Remember that the graphical interpretation helps build intuition and understanding, and the various applications demonstrate the importance of this concept across different disciplines.