Problem 58 - Calculus: Early Transcendentals

Question

(a) Find the number a such that the line \( x = a \) bisects the area under the curve \( y = \frac{1}{x^2} \), \( 1 \le x \le 4 \).

Step-by-Step Solution

Answer

The line that bisects the area is at \( x = \frac{8}{5} \).

Step 1: Understand the Problem

We need to find a vertical line \( x = a \) that divides the area under the curve \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 4 \) into two equal parts. To do this, we must calculate the total area under the curve and then find \( a \) such that the area from \( x = 1 \) to \( x = a \) is half of the total area.

Step 2: Calculate the Total Area

The total area under the curve \( y = \frac{1}{x^2} \) between \( x = 1 \) and \( x = 4 \) is given by the definite integral: \[ \int_{1}^{4} \frac{1}{x^2} \, dx. \]We can solve this integral by finding the antiderivative of \( \frac{1}{x^2} \), which is \( -\frac{1}{x} \). After integrating, we evaluate from 1 to 4:\[ \left[-\frac{1}{x}\right]_1^4 = -\frac{1}{4} - (-1) = 1 - \frac{1}{4} = \frac{3}{4}. \]

Step 3: Set Up the Equation for Bisected Area

To find \( a \), we need one-half of the total area under the curve, which is \( \frac{3}{8} \). We set up the following integral to find \( a \):\[ \int_{1}^{a} \frac{1}{x^2} \, dx = \frac{3}{8}. \]Using the antiderivative \( -\frac{1}{x} \), we can write:\[ \left[-\frac{1}{x}\right]_1^a = \frac{3}{8}. \]This simplifies to:\[ -\frac{1}{a} - (-1) = \frac{3}{8}, \]or\[ 1 - \frac{1}{a} = \frac{3}{8}. \]

Step 4: Solve for a

Rearrange the equation from Step 3:\[ \frac{1}{a} = 1 - \frac{3}{8} = \frac{5}{8}. \]Take the reciprocal of both sides to solve for \( a \):\[ a = \frac{8}{5}. \]

Step 5: Validate the Solution

Verify that this value of \( a = \frac{8}{5} \) provides the correct area partition.Calculate the integral from 1 to \( \frac{8}{5} \): \[ \int_{1}^{\frac{8}{5}} \frac{1}{x^2} \, dx = \left[-\frac{1}{x}\right]_1^{\frac{8}{5}} = -\frac{5}{8} - (-1) = 1 - \frac{5}{8} = \frac{3}{8}. \]This confirms that the area from 1 to \( \frac{8}{5} \) is indeed \( \frac{3}{8} \), matching one-half of the total area.

Key Concepts

Definite Integral

A definite integral refers to the calculation of the area under a curve within a specified range on the x-axis. In this exercise, the definite integral allows us to determine the total area under the curve of the function \( y = \frac{1}{x^2} \) from \( x = 1 \) to \( x = 4 \).
This is accomplished by evaluating the integral \[\int_{1}^{4} \frac{1}{x^2} \, dx,\]revealing an area of \( \frac{3}{4} \).

Understanding definite integrals is crucial since it provides a way to calculate the accumulation of quantities, such as areas under curves, which is essential in various applied mathematics fields.

Antiderivative

The antiderivative is a fundamental concept in calculus directly linked to the concept of integration. It represents a function whose derivative results in the original function provided.
In the exercise, the antiderivative of \( \frac{1}{x^2} \)is required to compute the definite integral. We find it to be\(-\frac{1}{x},\)meaning that differentiating \(-\frac{1}{x}\)yields \(\frac{1}{x^2}.\)

Antiderivatives are vital because they provide the reverse operation of differentiation, which helps determine functions for integration purposes. Knowing how to find antiderivatives is essential for solving integrals and is a key step in determining areas under curves.

Area Under a Curve

Calculating the area under a curve is a common application of definite integrals and involves finding the region enclosed by the graph of a function from a start to an endpoint along the x-axis. Given the function \( y = \frac{1}{x^2}, \)the problem requires obtaining the total area between \( x = 1 \)and \( x = 4 \).
Using integration, we confidently determined this area to be\( \frac{3}{4}. \) The goal was to bisect this area, bifurcating it equally at \( x = a.\)

This exercise illustrates how definite integrals allow us to calculate areas, which is useful across many fields like physics and engineering, where such calculations model real-world phenomena.

Integration Techniques

Different techniques exist for calculating integrals, each suited for various types of functions and situations. In this specific exercise, simple integration was employed due to the straightforward nature of the problem's function, \( \frac{1}{x^2}.\)
To solve the problem, finding the antiderivative was sufficient. This basic approach exemplifies one of the integration techniques called direct integration or elementary integration.

However, there are other methods like

• Substitution technique,
• Integration by parts,
• Partial fraction decomposition.

These are crucial in tackling more complex integrations.
Mastery of these techniques broadens one's ability to solve a wide range of calculus problems smoothly.