Problem 34 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{-1}^{1}\left(x^{2}-1\right)^{2} d x$$

Step-by-Step Solution

Answer

The definite integral \(\int_{-1}^{1} (x^2 - 1)^2 dx\), after finding the antiderivative and applying the Fundamental Theorem of Calculus, is given by \(\boxed{\frac{8}{15}}\).

Step 1: Find the antiderivative of the function

First, let's find the antiderivative (indefinite integral) of the function \(\left(x^2-1\right)^2\). We need to expand the square and then integrate term by term. The expanded form of \(\left(x^2 - 1\right)^2\) is: \((x^4 - 2x^2 + 1)\). Now let's find the antiderivative of each term: \[\int (x^4 - 2x^2 + 1) dx = \int x^4 dx - 2 \int x^2 dx + \int 1 dx\] Using power rule for integration: \[\frac{1}{5}x^5 - \frac{2}{3}x^3 + x + C\] Here, C represents the constant of integration.

Step 2: Evaluate the definite integral using the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, to find the definite integral of a function f(x) in the interval [a, b], we first find the antiderivative F(x), and then compute the difference F(b) - F(a). So for our given problem, the definite integral comprises of: \[\int_{-1}^{1} \left(x^2 - 1\right)^2 dx = \left[\frac{1}{5}x^5 - \frac{2}{3}x^3 + x\right]_{-1}^1\] Now we will compute F(1) and F(-1): F(1) = \(\frac{1}{5}(1)^5 - \frac{2}{3}(1)^3 + (1)\) \\ F(-1) = \(\frac{1}{5}(-1)^5 - \frac{2}{3}(-1)^3 + (-1)\) Now, find F(1) - F(-1): \(\int_{-1}^{1} \left(x^2 - 1\right)^2 dx = \left[\frac{1}{5} - \frac{2}{3} + 1\right] - \left[-\frac{1}{5} + \frac{2}{3} - 1\right]\) After simplifying and adding the terms, the definite integral value is: \[\boxed{\frac{8}{15}}\]

Key Concepts

Antiderivative

Understanding the concept of an antiderivative is crucial when solving problems involving integration. An antiderivative of a function f(x) is another function F(x) such that the derivative of F(x) is f(x). In other words, if you differentiate F(x), you get back the original function f(x). The process of finding an antiderivative is called indefinite integration.

When finding the antiderivative in our example, we're looking for a function whose derivative gives us back the integrand (x^2 - 1)^2. One essential characteristic of antiderivatives is that they include a constant of integration, represented by C, because the derivative of a constant is zero, and thus the original function may have had any constant value added to it. When we are working with a definite integral, this constant cancels out in the final evaluation.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus bridges the concept of differentiation with that of integration and is key to solving definite integrals. This theorem allows us to evaluate definite integrals using antiderivatives. The fundamental theorem comes in two parts, but for the calculation of definite integrals, the second part is used. It states that if F(x) is an antiderivative of a continuous function f(x) over an interval [a, b], then the definite integral of f(x) from a to b is given by F(b) - F(a).

In practice, this means that once we've found the antiderivative of the function we're integrating, we simply evaluate that antiderivative at the upper and lower bounds of the integral and subtract the latter from the former, as shown in the solution for the given exercise. This theorem is what makes the computation of areas under curves and other practical integrations possible.

Power Rule for Integration

The power rule for integration is one of the most fundamental rules in calculus and is used to find the antiderivative of power functions. The rule states that for any real number n not equal to -1, the integral of x^n is given by (1/(n+1))x^{n+1} plus a constant C. Symbolically, it’s expressed as \[\int x^n dx = \frac{1}{n+1}x^{n+1} + C.\]

Applying this rule simplifies the process of integrating polynomials term by term, as seen in the given exercise. It's essential for students to note that this rule can only be used directly on monomials. For polynomials with multiple terms, as in our example, you may have to employ the power rule on each term separately and then combine the results, which is done after expanding the initial expression, if needed, as with (x^2 - 1)^2.