Problem 24 - College Algebra and Calculus: An Applied Approach

Question

Evaluate the definite integral. $$ \int_{1}^{7} 3 v d v $$

Step-by-Step Solution

Answer

The value of the definite integral from 1 to 7 of the function \(3v\) is \(72\).

Step 1: Find the Antiderivative of the Function

The integral of a function is its antiderivative. In this case, the antiderivative of \(3v\) is \(1.5v^2\). The integral from one limit a to another limit b is given by \(F(b) - F(a)\), where \(F\) is the antiderivative of the function.

Step 2: Evaluate the Antiderivative at the Upper Limit

Substitute the upper limit \(7\) into the antiderivative function: \(1.5(7^2) = 73.5\).

Step 3: Evaluate the Antiderivative at the Lower Limit

Substitute the lower limit \(1\) into the antiderivative function: \(1.5(1^2) = 1.5\).

Step 4: Subtract the Lower Limit Result from the Upper Limit Result

To obtain the definite integral, subtract the result calculated at the lower limit from the result calculated at the upper limit: \(73.5 - 1.5 = 72\)

Key Concepts

Antiderivative

One of the most fundamental concepts in calculus is that of the antiderivative. Simply put, finding the antiderivative, also known as the indefinite integral, involves determining the original function that was differentiated to get the given function. For example, if we take the function represented by the derivative 3v, its antiderivative is the function from which it was derived through differentiation.

In the exercise, the antiderivative was correctly identified as 1.5v². A helpful tip when determining antiderivatives is to use the reverse of differentiation rules. For instance, the power rule for differentiation states that if you have vⁿ, its derivative is nv^n-1; the antiderivative then reverses this, adding 1 to the exponent and dividing by the new exponent.

When working with antiderivatives, it's essential to remember to include the constant of integration, which represents the indefinite nature of antiderivatives. However, for definite integrals, as in this exercise, the constant cancels out when we evaluate the limits, and thus it is not included in the final steps.

Integral Evaluation

Integral evaluation is a process whereby we calculate the area under the curve of a function between two points on the x-axis (referred to as 'limits' when dealing with definite integrals). This process is central to finding the definite integral of a function.

In the exercise, we're evaluating the integral from the lower limit of 1 to the upper limit of 7 for the function 3v. The main steps in this process involve first finding the antiderivative of the function, as mentioned previously, and then applying the evaluation theorem, which tells us to evaluate the antiderivative at each limit.

The simplification process might involve standard arithmetic and algebraic manipulation. Once we have these evaluations (as seen in the solution, 73.5 from the upper limit and 1.5 from the lower limit), we find the difference between the two. This difference gives us the value of the definite integral, which in the context of a graph, represents the net area between the function and the x-axis over the interval [1, 7].

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) bridges the concepts of differentiation and integration, showing that they are essentially inverse processes. It is divided into two parts: the first part guarantees that the integral of a function can be reversed by differentiation, while the second part allows us to evaluate definite integrals using the antiderivative.

The exercise applies the second part of FTC, where the calculation of the definite integral is performed by taking the difference between the values of the antiderivative at the upper and lower limits of integration. Following the FTC, we substitute 7 into the antiderivative to get 73.5, and 1 into the antiderivative to get 1.5, then subtract the latter from the former to obtain the result 72.

Understanding the FTC is essential since it not only simplifies the computations involved in finding areas under curves but also lays the groundwork for solving a wide range of problems in mathematics, physics, engineering, and beyond. Remembering that differentiation and integration are connected operations helps to better grasp the entirety of calculus.