Question

Find the area of the region between the curves \(y=1 / x\) and \(y=1 /\left(x^{3}+x\right)\) for \(0

Short answer

The area is \(\frac{1}{2}\ln 2\).

Step-by-step solution

Find Points of Intersection

First, identify where the curves intersect by setting the equations equal to each other.Solve for these values:\[\frac{1}{x} = \frac{1}{x^3 + x}\]Cross multiply:\[x^3 + x = x\]This simplifies to:\[x^3 = 0\]This implies:\[x = 0\]But since we are looking in the range \(0 < x \leq 1\), the curves only intersect when \(x = 0\), which lies outside our interval. Hence, check behavior at the end of the interval.

Determine Which Function is On Top

Evaluate the functions at a value within the interval, such as at \(x = 0.5\).Calculate:\[f(0.5) = \frac{1}{0.5} = 2\]\[g(0.5) = \frac{1}{0.5^3 + 0.5} = \frac{1}{0.125 + 0.5} = \frac{1}{0.625} \approx 1.6\]Since \(f(0.5) > g(0.5)\), the function \(y = \frac{1}{x}\) is above \(y = \frac{1}{x^3 + x}\) on \(0 < x \leq 1\).

Setup the Integral to Find the Area

The area between the curves is found using the integral:\[\text{Area} = \int_{0}^{1} \left( \frac{1}{x} - \frac{1}{x^3 + x} \right) dx\]

Simplify the Integral Expression

Write the integrand as:\[\frac{1}{x} - \frac{1}{x^3 + x} = \frac{1}{x} - \frac{1}{x(x^2 + 1)}\]Simplify:\[= \frac{x^2 + 1 - 1}{x(x^2 + 1)}= \frac{x^2}{x(x^2 + 1)} = \frac{x}{x^2 + 1}\]

Integrate the Simplified Expression

Compute the integral:\[\int_{0}^{1} \frac{x}{x^2 + 1} \, dx\]Use substitution. Let \( u = x^2 + 1\), then \( du = 2x \, dx\), or \(\frac{1}{2} du = x \, dx\).The limits change as follows:\[x = 0 \rightarrow u = 1\]\[x = 1 \rightarrow u = 2\]Thus, the integral becomes:\[\frac{1}{2} \int_{1}^{2} \frac{1}{u} \, du\]Integrate:\[= \frac{1}{2} [\ln|u|]_{1}^{2}\]\[= \frac{1}{2} (\ln 2 - \ln 1)\]\[= \frac{1}{2} \ln 2\]

Conclude With the Area Calculation

Thus, the area between the curves over the given interval is:\[\frac{1}{2} \ln 2\]

Key concepts

Definite Integrals

Definite integrals are a crucial tool for calculating the area under a curve or between curves, like in our exercise. When we use a definite integral, we integrate a function over a specified range, providing us with a numerical result. This result represents the accumulated quantity, which in geometric terms, can be the area under a graph.

To compute a definite integral:

• Identify the bounds of integration, also known as the limits. These specify the interval over which we calculate the area.
• Determine the integrand, which is the function you integrate. In our exercise, the integrand is the difference between the two given functions that define the top and bottom of the area.
• Calculate the integral by finding the antiderivative of the function, then evaluate it at the upper and lower bounds.

The outcome tells us the total area. In our example, the definite integral provides us with the area between two curves from 0 to 1, leading to \(\frac{1}{2} \ln 2\). This is the simplest way to find such specific areas mathematically.

Area Between Curves

Finding the area between two curves is an essential application of integration. It involves determining which curve is above the other within the given interval, and then calculating the area sandwiched between them.

When attempting to find the area between the curves, you must:

• Identify points of intersection, as this helps anchor the limits of your integral setup.
• Determine which curve is above within the interval. In our problem, solving for some test values such as \(x = 0.5\) reveals that \(y = \frac{1}{x}\) remains the upper curve between 0 and 1.
• Subtract the equation of the lower curve from the equation of the upper curve to form the integrand.

This process allows you to set up a proper integral for finding the area between the curves. It ensures that you are effectively capturing the enclosed space on a graph which can sometimes appear irregular or non-intuitive at first glance.

Antiderivatives

Antiderivatives, also known as indefinite integrals, are functions that reverse the process of differentiation. Finding an antiderivative of a function means determining a function whose derivative is the given function.

In integration, finding the antiderivative is a fundamental step. In our problem, to find the area between two curves, we must calculate:\[\int \left( \frac{1}{x} - \frac{1}{x^3 + x} \right) \, dx\]By simplifying this expression to \(\frac{x}{x^2 + 1}\), we can better identify the technique needed: substitution.

Using substitution involves:

• Selecting a part of the expression to rewrite in terms of a new variable, typically chosen to make integration straightforward. Here, let \( u = x^2 + 1 \), simplifying the task.
• Rewriting the integral in terms of \(u\) , adjusting the limits this variable applies to for definite integrals.
• Finding the antiderivative in terms of the new variable and substituting back to the original variable if necessary.

Understanding how to find antiderivatives effectively unlocks the potential of calculus to solve vastly differing applied mathematics and engineering problems.