Problem 147 - Thomas Calculus

Question

Find the area of the region between the curve \(y=2 x /\left(1+x^{2}\right)\) and the interval \(-2 \leq x \leq 2\) of the \(x\) -axis.

Step-by-Step Solution

Answer

The area between the curve and the x-axis is 0.

Step 1: Set Up the Integral for Area

The area between the curve and the x-axis from \(-2\) to \(2\) is given by the integral \( \int_{-2}^{2} \frac{2x}{1+x^2} \, dx \). This integral will help us find the required area under the curve.

Step 2: Simplify the Integral

Notice that the integrand, \( \frac{2x}{1+x^2} \), can be related to a known derivative. The derivative of \( \ln|1+x^2| \) is \( \frac{2x}{1+x^2} \), which makes the integration straightforward.

Step 3: Integrate the Function

Apply the antiderivative to the integrand: \[ \int \frac{2x}{1+x^2} \, dx = \ln|1+x^2| + C \] Evaluate the definite integral: \[ \left[ \ln|1+x^2| \right]_{-2}^{2} \].

Step 4: Evaluate the Definite Integral

Compute the definite integral from the antiderivative: \[ \ln|1+2^2| - \ln|1+(-2)^2| = \ln|5| - \ln|5| = 0 \].

Step 5: Interpret the Result

After evaluating the definite integral, we find that the area is \(0\). This indicates symmetrical areas above and below the x-axis canceling each other out.

Key Concepts

Definite integral

The concept of a definite integral is foundational in calculus, especially when dealing with problems involving area calculation under curves. Essentially, the definite integral of a function over a specific interval gives the net area between the graph of the function and the x-axis, taking into account whether the area is above or below the axis.
To compute a definite integral, you need to evaluate the antiderivative at two points, the bounds of your interval, and then subtract the two results. In our problem concerning the function \( y=\frac{2x}{1+x^2} \) over the interval \(-2 \leq x \leq 2\), we set up the integral \( \int_{-2}^{2} \frac{2x}{1+x^2} \, dx \). This represents the total signed area from \(-2\) to \(2\).

• The limits \(-2\) and \(2\) are the lower and upper bounds, respectively.
• True to the definite nature, the evaluation uses only these bounds, unlike indefinite integrals which include an arbitrary constant.

Understanding definite integrals is pivotal not only in solving area problems but also in many applications in physics and engineering.

Antiderivative

An antiderivative, in calculus, is a function whose derivative is the given function. When we discuss finding the antiderivative or "integral" of a function, we are essentially reversing the process of differentiation. This is critical when solving integrals, as the fundamental theorem of calculus links antiderivatives and definite integrals.
In the original exercise, we identified the antiderivative of \( \frac{2x}{1+x^2} \) by recognizing it as the derivative of \( \ln|1+x^2| \). Therefore, the indefinite integral:
\[ \int \frac{2x}{1+x^2} \, dx = \ln|1+x^2| + C \]
Where \(C\) is the constant of integration. However, when calculating a definite integral, this constant cancels out during the evaluation process.

• An essential skill in integration is recognizing forms that match standard derivatives, simplifying the process significantly.
• Using substitution or recognizing patterns helps to find these antiderivatives quickly.

Integral calculus

Integral calculus is a major part of calculus that focuses on the process of integration. As opposed to differential calculus, which breaks down functions into their rates of change, integral calculus combines values to determine quantities like areas, volumes, and other totals.
In the given exercise, integral calculus allowed us to determine the area between a curve and the x-axis. This requires understanding both definite integrals for calculating exact areas and antiderivatives when dealing with indefinite integrals.

• Integral calculus provides tools for a wide range of applications, from calculating the mass of an object given its density function to solving complex differential equations in engineering.
• It involves techniques like substitution and integration by parts to handle more complicated integrals.

Understanding the basic principles of integral calculus and its methods makes it a powerful tool in mathematics and sciences.