Problem 75 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. $$\int_{2}^{2} \frac{e^{x}}{\sqrt{1+x}} d x=0$$

Step-by-Step Solution

Answer

The given statement is true. The integral \(\int_{2}^{2} \frac{e^{x}}{\sqrt{1+x}} dx = 0\) because the limits of integration are the same, 2 in this case. According to the fundamental theorem of calculus, the definite integral with the same limits always evaluates to 0.

Step 1: Identify the given definite integral

We are given the definite integral: \[\int_{2}^{2} \frac{e^{x}}{\sqrt{1+x}} dx\]

Step 2: Evaluate the definite integral using the fundamental theorem of calculus

According to the fundamental theorem of calculus, a definite integral with both limits the same, i.e., \(\int_a^a f(x)\,dx\) is always equal to 0, regardless of the function f(x). So we can write: \[\int_{2}^{2} \frac{e^{x}}{\sqrt{1+x}} dx = 0\]

Step 3: Conclusion

The given statement is true. The integral \(\int_{2}^{2} \frac{e^{x}}{\sqrt{1+x}} dx = 0\) because the limits of integration are the same, 2 in this case. According to the fundamental theorem of calculus, the definite integral with the same limits always evaluates to 0.

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concepts of differentiation and integration, two main pillars of calculus. It states that differentiation and integration are inverse processes. This theorem is composed of two parts.

• The first part tells us how to use antiderivatives to evaluate definite integrals. If a function is continuous over an interval \(a, b\), and \ F(x) \ is an antiderivative of \ f(x) \ on that interval, the definite integral from \ a \ to \ b \ of \ f(x) \ is given by \ F(b) - F(a) \.
• The second part states that if \ F(x) \ is defined as the integral of \ f(t) \ from \ a \ to \ x, then \ F'(x) = f(x) \.

In our original exercise, a specific property is used: when the limits of integration are the same, the integral evaluates to zero because \( F(a) - F(a) = 0 \). This property simplifies problems tremendously when both limits of integration are identical.

Integral Calculus

Integral Calculus focuses on the concept of accumulation of quantities, and areas under or between curves. It comes in two main forms: definite and indefinite integrals.

• **Definite integrals** have specific limits and calculate the net area under a curve within these limits.
• **Indefinite integrals** find antiderivatives of functions and do not possess specific limits.

To calculate the area or accumulated value using definite integrals, you apply the process described in the Fundamental Theorem of Calculus. While our exercise used a basic scenario where the limits were the same, integral calculus often requires more steps to evaluate a given integral. Understanding these basics helps in comprehending the accumulation processes and area calculations pertinent in physics, engineering, and other fields.

Limits of Integration

Limits of integration are the values that define the interval over which we integrate a function. They are denoted as the lower and upper bounds in a definite integral, commonly written as \int_{a}^{b} f(x) \, dx\. Here's why they matter:

• The lower limit \(a\) represents the starting point, and the upper limit \(b\) marks the endpoint of integration.
• Changing these limits changes the area being calculated, affecting the result.

In the example given, since \ a = b \, the integral simplifies into zero as no interval exists to calculate area or accumulation. Understanding how these limits work helps answer problems quickly, without performing unnecessary calculations. In more complex problems, careful attention to these limits ensures accurate problem-solving.