Question

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ y=\frac{1}{x^{2}}, \quad y=0, \quad x=1, \quad x=5 $$

Short answer

The area of the bounded region is \(\frac{4}{5}\) unit squares.

Step-by-step solution

Understanding the Problem

Start by understanding what each of the given equations represent. \( y = \frac{1}{{x^2}} \) is a graph with a hyperbolic shape which never touches the x-y axes. \( y = 0 \) is the x-axis. \( x = 1 \) and \( x = 5 \) are vertical lines that cross the x-axis at points 1 and 5 respectively. Here, it is important to visualize the bounded region. Sketching the graphs will give a preliminary understanding of where the equation needs to be evaluated to find the area.

Sketching the Graphs

First, sketch the graph for \( y = \frac{1}{{x^2}} \). This function curves downwards as x moves away from 0, only existing in the first and third quadrants. When \( y = 0 \), it represents the x-axis. On the other hand, \( x = 1 \) and \( x = 5 \) are vertical lines that intersect x-axis at 1 and 5 respectively. The region to find is thus a vertical stripe running from x=1 to x=5 under the curve \( y = \frac{1}{{x^2}} \) and above the x-axis.

Calculating the Area

To calculate the area of a region bounded by the x-axis and a function plot, the formula is \( \int_{a}^{b} f(x) \, dx \) where \( a \) and \( b \) are the x-values bounding the region and \( f(x) \) is the function within these bounds. Here, the region is contained within \( x = 1 \) and \( x = 5 \), and under the function \( y = \frac{1}{{x^2}} \), so the integral representing the area of the region is \( \int_{1}^{5} \frac{1}{{x^2}} \, dx \). Solving this will give the area.

Solving the Integral

Solve the integral \( \int_{1}^{5} \frac{1}{{x^2}} \, dx \). The antiderivative of \( \frac{1}{{x^2}} \) is \( -\frac{1}{x} \). Applying the limits 1 and 5 gets \( \left[-\frac{1}{{x}}\right]_{1}^{5} \) which simplifies to \( -\frac{1}{5} - (-\frac{1}{1}) = -\frac{1}{5} + 1 = \frac{4}{5} \) unit squares.

Key concepts

Definite Integral

When tasked with finding the area beneath a curve on a graph, the definite integral is the mathematical tool we usually employ. It sums up an infinite number of infinitely small bits of area under the curve between two specified limits. These limits are the x-values that define the interval over which we calculate the integral. For instance, in our problem, we're dealing with the area under the curve of the function \( y = \frac{1}{{x^2}} \) between the vertical lines \( x = 1 \) and \( x = 5 \). The definite integral that represents this is \( \[\int_{1}^{5} \frac{1}{{x^2}} \, dx\] \), and its value will give us the exact area of the region bounded by the curve and the x-axis within these limits.

The process is essentially 'summing up' the area of infinitely many rectangles, each having a height equal to the function's value at a specific point and an infinitesimally small width, as we move along the x-axis from the lower limit to the upper limit.

Sketching Graphs of Functions

Before attempting to find the area between curves, it's often helpful to sketch the functions to understand how they interact. In our example, sketching the graph of \( y = \frac{1}{{x^2}} \) shows that as \( x \) increases, \( y \) rapidly decreases, producing a steep downward curve in the first quadrant (where both x and y are positive). Meanwhile, \( y = 0 \) is simply the x-axis, and \( x = 1 \) and \( x = 5 \) are vertical lines, each cutting through the x-axis at their respective x-value. Drawing these graphs allows us to visualize the 'strip' we're calculating the area of, making it easier to set up the integral properly. This visual aid is a powerful step in solving many algebraic problems.

While the calculations are paramount, often the insight gained from a simple sketch can be invaluable, preventing mistakes and clarifying the problem's parameters. This process turns abstract equations into concrete geometric figures we can easily comprehend.

Hyperbolic Function

A hyperbolic function, which appears in our exercise as \( y = \frac{1}{{x^2}} \), is named due to its relationship to hyperbolas, similar to how trigonometric functions are related to circles. This function, in particular, is known as the reciprocal squared function and it has a distinct curve that approaches both the x-axis and y-axis but never touches them, residing solely in the first quadrant under these conditions. The relationship that hyperbolic functions have to their corresponding hyperbolas is not just visual but also algebraic, as these functions can represent sections of a hyperbola in coordinate geometry.

Understanding the shape and behavior of hyperbolic functions is critical when sketching the graph of such a function and when finding the area under them, as it dictates the region that will be considered for integration.

Integral Calculus

Integral calculus is one of the two principal branches of calculus, with the other being differential calculus. The main idea behind integral calculus is accumulation, which, as mentioned earlier, can be used to calculate areas, volumes, and other concepts where summing infinitely small quantities is required. It encompasses a range of techniques and theorems to handle different kinds of problems, like finding the antiderivative or evaluating definite integrals as we have in our example.

The process of integration is the reverse of differentiation. Given a function's rate of change, or derivative, we can find the original function – although it is not always straightforward due to constants of integration. In solving our given problem, we used an antiderivative to evaluate the definite integral, which is a fundamental method in integral calculus for finding bounded areas under curves on a graph.