Problem 34 - Essential Calculus

Question

In Exercises 31-36, evaluate the integral using the following values. $$\int_{2}^{4} x^{3} d x=60, \quad \int_{2}^{4} x d x=6, \quad \int_{2}^{4} d x=2$$ $$ \int_{2}^{4} 15 d x $$

Step-by-Step Solution

Answer

The value of \( \int_{2}^{4} 15 d x \) is 30.

Step 1: Identifying the Constant and the Interval

The aim of this integral is to find the area under the function \( f(x) = 15 \) from \( x = 2 \) to \( x = 4 \). Now 15 is a constant, and the interval of integration is from 2 to 4.

Step 2: Applying the Rule for Integrating a Constant

The integral of a constant function \( f(x) = k \) over an interval [a, b] is given by \( \int_{a}^{b} k d x = k(b - a) \).

Step 3: Calculating the Integral

Applying the formula from Step 2, the integral becomes: \( \int_{2}^{4} 15 d x = 15(4 - 2) = 30 \).

Key Concepts

Integration by Substitution

Integration by substitution, often referred to as u-substitution, is a method used to simplify the process of evaluating integrals. This technique involves substituting a part of the integral with a new variable, typically denoted as 'u', which simplifies the integrand into a form that's easier to integrate.

For example, if we have an integral with a function of the form \( g(f(x))f'(x) dx \), we might let \( u = f(x) \), hence \( du = f'(x) dx \). Substituting these into the integral, we can often find the antiderivative more straightforwardly. After integrating with respect to 'u', we substitute back to the original variable x to get the final answer.

To apply this method effectively, it's crucial to identify an inner function whose derivative is present elsewhere in the integrand. The selection of 'u' greatly depends on this observation. For the given textbook exercise, the integral \( \int_{2}^{4} 15 dx \) is already straightforward to evaluate without the need for substitution since it involves a constant function.

Definite Integral

A definite integral is a fundamental concept in calculus that represents the net area between the graph of a function and the x-axis, within the limits of integration. The definite integral has both a lower and an upper limit, and it is represented as \( \int_{a}^{b} f(x) dx \), where 'a' is the lower limit, 'b' is the upper limit, and \( f(x) \) is the function being integrated.

When evaluating a definite integral, the result is a number which can be interpreted as the signed area; positive area above the x-axis and negative area below the x-axis between the curve of \( f(x) \) and the x-axis, from x = a to x = b. It is essential to understand this concept, as it doesn't just calculate the area, but gives a sense of accumulation or the net result of a continuous quantity over an interval.

Returning to our exercise, the integral \( \int_{2}^{4} 15 dx \) with definite limits from 2 to 4 calculates the area of a rectangle with height 15 and width equal to the interval length, which is \( 4 - 2 \).

Area Under a Curve

The 'area under a curve' is an expression typically used to describe one of the applications of definite integrals in calculus. This concept is associated with finding the region occupied under the graph of a given function \( f(x) \) between two vertical lines \( x = a \) and \( x = b \) and the x-axis.

Graphically, to evaluate the area, one usually looks for the integral of the absolute value of the function if the function crosses the x-axis within the considered interval. Otherwise, with definite integrals, if the function remains above the x-axis within the interval, the area and the integral are numerically the same. Conversely, if the function is below the x-axis, the integral is negative, representing the area with a sign indicative of its position relative to the x-axis.

In our textbook solution's context, the function is a constant \( f(x) = 15 \) from \( x = 2 \) to \( x = 4 \) which does not cross the x-axis. The area under this curve is simply the area of the rectangle formed by the height (the value of the constant function) and the length of the interval on the x-axis, reflecting how straightforward the concept can be for certain functions.