Question

A function \(f\) is defined below. Use geometric formulas to find \(\int_{0}^{8} f(x) d x\) $$f(x)=\left\\{\begin{array}{ll}4, & x<4 \\ x, & x \geq 4\end{array}\right.$$

Short answer

The value of the integral of the given function over the interval [0, 8] is 40.

Step-by-step solution

Define the intervals

Separate the interval [0, 8] into [0, 4] and [4, 8] according to the definition of the function \(f(x)\). Calculate the integral over these two intervals separately.

Calculate the integral over the first interval

The function on the interval [0, 4] is described by the constant 4. Calculate the integral over this interval using the formula for the integral of a constant: \(\int \alpha dx = \alpha x\), where \(\alpha\) is a constant. The integral from 0 to 4 of \(4dx\) is \(4x\) evaluated from 0 to 4, which is \(4*4 - 4*0 = 16\).

Calculate the integral over the second interval

The function on the interval [4, 8] is described by \(x\). Calculate the integral over this interval using the formula for the integral of \(x\): \(\int x dx = 1/2 * x^2\). The integral from 4 to 8 of \(xdx\) is \(1/2 * x^2\) evaluated from 4 to 8, which is \(1/2*8^2 - 1/2*4^2 = 24\).

Sum the integrals

The integral over the whole interval [0, 8] is the sum of the integrals over the intervals [0, 4] and [4, 8]. Therefore, \(\int_{0}^{8} f(x) dx = 16 + 24 = 40\).

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on the processes of integration and its applications. It is primarily concerned with finding the total size or value, such as area under a curve, displacement given velocity, or volume of a solid. The basic idea of an integral calculus is the concept of accumulation - adding pieces to determine the whole. It's encapsulated in the Definite Integral, which is used to calculate exact areas or accumulated quantities and is represented by the notation \( \int_{a}^{b} f(x) dx \), where \(f(x)\) is the function to integrate over the interval from \(a\) to \(b\).

In our particular exercise, we utilized integral calculus to find the area under the piecewise function \(f(x)\) from \(x = 0\) to \(x = 8\). By breaking it down into simpler shapes corresponding to the intervals defined by the piecewise function, we applied geometric formulas to sum up the individual areas and find the total.

Piecewise Functions

Piecewise functions are defined by different expressions for different intervals of the domain. They often represent scenarios where a given rule changes over different intervals. When dealing with piecewise functions in integral calculus, it is crucial to address each piece individually - a technique that allows for the integration of complex, discontinuous functions.

In the given exercise, \( f(x) \) was a piecewise function with two expressions: \(4\) when \(x<4\), and \(x\) when \(x \geq 4\). Correspondingly, the definite integral was resolved by splitting the integration interval into two parts, [0, 4] and [4, 8], and applying appropriate integration techniques to each interval.

Geometric Formulas

Geometric formulas are mathematical equations that relate to the measurements of geometric figures and shapes, such as areas of squares, rectangles, and triangles, as well as the volumes of solids like cylinders, spheres, and cones. These formulas are incredibly useful in integral calculus because many functions can be graphically represented, and their integrals interpreted as areas or volumes.

For instance, the first part of our exercise involved the integral of a constant function, which geometrically represents the area of a rectangle with width \(4\) and height \(4\). Thus, the geometric formula for the area of a rectangle (length \times width) could be directly applied to find the integral.

Integration Techniques

There are several techniques for performing integrals in calculus, such as substitution, integration by parts, partial fractions, and others. The technique used often depends on the type of function one is dealing with. For simple functions, a basic approach often suffices. For more complex functions, specialized techniques are necessary.

In the context of our exercise, two basic integration techniques were employed. The first one was the integration of a constant function over an interval, and the second was the integration of \(x\) with respect to \(x\). The first integral was straightforward, as the integral of a constant is simply the constant multiplied by the interval width. The second integral required using the power rule, where the integral of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\), applied to the function \(x\) over the interval from 4 to 8.

Understanding when and how to apply these integration techniques is pivotal in solving many calculus problems effectively.