Question

Find the area (in square units) bounded by each curve, the given lines, and the \(x\) axis. Sketch the curve for some of these, and try to make a quick estimate of the area. Also check some graphically or by calculator. \(y=1 / \sqrt{x}\) from \(x=\frac{1}{2}\) to 8

Short answer

The area under the curve from \(x=\frac{1}{2}\) to \(x=8\) is approximately 4.24 square units.

Step-by-step solution

Understand the Area Under the Curve

To find the area under the curve of the function y = \(\frac{1}{\sqrt{x}}\) between \(x=\frac{1}{2}\) and \(x=8\), we will use the definite integral. The area under the curve and above the x-axis can be calculated as \(\int_{\frac{1}{2}}^{8} \frac{1}{\sqrt{x}} dx\).

Integrate the Function

Integrate the function with respect to \(x\) from \(\frac{1}{2}\) to 8. \(\int_{\frac{1}{2}}^{8} \frac{1}{\sqrt{x}} dx = \int_{\frac{1}{2}}^{8} x^{-\frac{1}{2}} dx = \left[ 2x^{\frac{1}{2}} \right]_{\frac{1}{2}}^{8}\).

Evaluate the Definite Integral

Plug in the limits of the integral to find the area: \(2x^{\frac{1}{2}}\Big|_{\frac{1}{2}}^{8} = 2(8^{\frac{1}{2}}) - 2\left(\left(\frac{1}{2}\right)^{\frac{1}{2}}\right) = 2(\sqrt{8}) - 2\left(\frac{1}{\sqrt{2}}\right) = 4\sqrt{2} - \sqrt{2} = 3\sqrt{2}\).

Calculate the Numerical Value

Evaluate the exact value of the area: \(3\sqrt{2}\) is approximately 4.24 square units. Now, check this result graphically or with a calculator if needed.

Key concepts

Area Under the Curve

Understanding the area under a curve in integral calculus is essential for solving a wide range of real-world problems. The region sandwiched between the graph of a function and the x-axis, limited by two vertical lines is referred to as the 'area under the curve'. This concept not only applies to physical areas but also other quantities like distance, probability, or volume.

To visualize this, imagine plotting the function on a graph. The curve represents the function, while the area below it, above the x-axis, and between two points along the x-axis captures what we're interested in. In practice, this is a common way to determine distances traveled when given a speed-time graph, or to calculate the total growth over time from a rate of growth graph.

Visual Estimation and Graphical Validation

Before diving into calculations, a quick sketch can be insightful. It provides a rough estimate of the area and helps in visualizing the problem. Furthermore, using a graphing calculator or software can serve as a check, ensuring that the calculated area reasonably fits the visually approximated or graphically computed area.

Integral Calculus

Integral calculus is a part of mathematics that deals with integrals and the properties of integrable functions. It allows you to calculate the accumulation of quantities, such as areas under curves. There are two primary types of integrals in calculus: indefinite and definite integrals.

An indefinite integral, also known as an antiderivative, represents a family of functions and includes a constant of integration. It shows the general form without specific limits. On the other hand, a definite integral computes the net value of a function within a certain interval. This is the tool that enables us to find the exact area under the curve between two points.

Definite Integral and Area Calculation

The definite integral has upper and lower limits, which directly correspond to the boundaries of the area we're calculating. Once we integrate the function to get an antiderivative, we then evaluate this antiderivative at the upper and lower limits and take the difference to find the definite integral's value. The result is the precise area under the curve between these two points.

Function Integration

Function integration involves finding the integral or antiderivative of a function. When integrating a function, certain rules and methods, such as substitution or integration by parts, can be applied to simplify complex expressions. For functions with a simple antiderivative, integration is straightforward.

For the function given in the exercise, y = \frac{1}{\(\sqrt{x}\)}, or written with exponents, y = x^{-\frac{1}{2}}, the antiderivative is found by adding 1 to the exponent and dividing by the new exponent. This results in the integrated function, 2x^\frac{1}{2}, which must then be evaluated between the given limits to calculate the area under the curve.

Applying the Power Rule for Integration

In this particular example, the Power Rule of Integration is utilized to find the antiderivative. For any real number n different from -1, the Power Rule states that the integral of x^n is \(\frac{1}{n+1}\)x^{n+1}. This proves to be an effective and commonly used technique in evaluating integrals, especially for polynomial functions or when dealing with functions that are powers of x.