Problem 31 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the definite integral. $$\int_{0}^{4} x\left(x^{2}-1\right) d x$$

Step-by-Step Solution

Answer

The short answer based on the provided step-by-step solution is: Using expansion, antiderivative calculation, and the Fundamental Theorem of Calculus, the value of the definite integral \(\int_{0}^{4} x(x^{2}-1) dx\) is equal to \(56\).

Step 1: Expand the integrand function

First, let's expand the integrand function: $$x(x^{2}-1) = x^{3}-x$$

Step 2: Find the antiderivative

Now, let's find the antiderivative of the expanded function: $$\int (x^{3}-x) dx = \frac{1}{4}x^{4} - \frac{1}{2}x^{2} + C$$ However, since this is a definite integral, we will only need the antiderivative without the constant (C).

Step 3: Apply the fundamental theorem of calculus

Now, let's apply the fundamental theorem of calculus to evaluate the definite integral: $$\int_{0}^{4} (x^{3}-x) dx = \left[\frac{1}{4}x^{4} - \frac{1}{2}x^{2}\right]_{0}^{4}$$

Step 4: Evaluate the antiderivative at the limits of integration

Next, we will evaluate the antiderivative at the upper limit and lower limit of the integration, and find the difference between the two: $$= \left(\frac{1}{4}(4)^{4} - \frac{1}{2}(4)^{2}\right) - \left(\frac{1}{4}(0)^{4} - \frac{1}{2}(0)^{2}\right)$$

Step 5: Simplify and find the definite integral value

Finally, let's simplify and find the value of the definite integral: $$= \left(64 - 8\right) - \left(0 - 0 \right) = 56$$ Thus, the value of the definite integral is: $$\int_{0}^{4} x(x^{2}-1) dx = 56$$

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) serves as a bridge between differentiation and integration, two core concepts in calculus. It ensures that we can analyze the area under a curve—or the integral of a function—by primarily using antiderivatives.

FTC is divided into two parts: the first part provides a way to evaluate a definite integral using an antiderivative, while the second part asserts that the derivative of an integral function is the original function. For example, in the exercise we evaluated the definite integral \(\int_{0}^{4} x(x^{2}-1) dx\), the FTC helps in simplifying the process by substituting the upper and lower limits of integration into the antiderivative.

By applying this theorem, we calculate the difference between the values of the antiderivative at the upper limit (4) and lower limit (0), which gives us the exact area under the curve, or in this case, the value 56.

Antiderivative

An antiderivative, also known as an indefinite integral, is essentially the reverse process of differentiation. If we're given a function f(x), an antiderivative F(x) is a function whose derivative is f(x): \( F'(x) = f(x) \).

In the context of the given exercise, we found the antiderivative of \(x^3-x\) as \(\frac{1}{4}x^4 - \frac{1}{2}x^2\), which is used to evaluate the definite integral over the interval [0, 4]. The constant of integration (C) usually associated with the indefinite integral is omitted when calculating a definite integral, since it cancels out when taking the difference of the antiderivative's values at the upper and lower limits of integration.

Integration Techniques

While facing a problem that involves finding integrals, various integration techniques can be employed based on the form of the function to be integrated. These range from simple antiderivative rules to more complex strategies like substitution, integration by parts, partial fraction decomposition, and trigonometric integration, to name a few.

For the provided exercise, we used an elementary technique of expanding and simplifying the integrand before finding the antiderivative. We saw \(x(x^2-1)\) and expanded it to \(x^3-x\), which is a form more conducive to direct integration.

Often, the chosen technique depends on the integrand's complexity; starting with a straightforward method, such as expanding an algebraic expression, can sometimes make the problem much easier to deal with, as it did in this case, leading to a simpler antiderivative and an effortlessly computed definite integral.