Problem 11 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$y=4 \sqrt{x}, x=0, y=1, y=3$$

Step-by-Step Solution

Answer

The area bounded by the curves is \(\frac{13}{24}\) square units.

Step 1: Interpret the Problem

We need to find the area bounded by the curves. Specifically, this means finding the region enclosed between the curve \(y = 4\sqrt{x}\), the vertical line \(x = 0\), and the horizontal lines \(y = 1\) and \(y = 3\).

Step 2: Express x in Terms of y

Since the curve is given as \(y = 4\sqrt{x}\), we first solve for \(x\) in terms of \(y\):\[ y = 4\sqrt{x} \implies \sqrt{x} = \frac{y}{4} \implies x = \left(\frac{y}{4}\right)^2 = \frac{y^2}{16}. \]

Step 3: Establish Limits of Integration

From the problem, we need to find the area between \(y = 1\) and \(y = 3\). Thus, the limits of integration for \(y\) will be from 1 to 3.

Step 4: Setup the Integral for Area

The area of interest will be the integral of \(x\) in terms of \(y\) from \(y = 1\) to \(y = 3\). Therefore, you set up the integral as follows:\[ A = \int_{1}^{3} \frac{y^2}{16} \, dy. \]

Step 5: Calculate the Integral

Evaluate the integral \(\int \frac{y^2}{16} \, dy\):\[ \int \frac{y^2}{16} \, dy = \frac{1}{16} \int y^2 \, dy = \frac{1}{16} \cdot \frac{y^3}{3} = \frac{y^3}{48}. \]

Step 6: Evaluate the Definite Integral

Apply the limits 1 to 3 to the antiderivative to find the area:\[ A = \left[ \frac{y^3}{48} \right]_{1}^{3} = \frac{3^3}{48} - \frac{1^3}{48} = \frac{27}{48} - \frac{1}{48} = \frac{26}{48} = \frac{13}{24}. \]

Step 7: Finalize the Solution

The area bounded by the curve and the specified lines is \(\frac{13}{24}\) square units.

Key Concepts

Definite Integral

In calculus, a definite integral is a fundamental concept used to compute the area under a curve within a certain interval. The definite integral of a function is represented by an expression like \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are known as the limits of integration. These limits define the interval over which you're calculating the area. By integrating a function over this interval, you essentially sum up infinitely small rectangles under the curve.

To solve a definite integral, follow these steps:

• Find the antiderivative: Determine the antiderivative of the function \( f(x) \). This is the function \( F(x) \) whose derivative is \( f(x) \).
• Apply the limits: Evaluate the antiderivative at the upper bound \( b \) and at the lower bound \( a \) and then subtract: \( F(b) - F(a) \). This gives you the net area.

Using definite integrals, you can calculate physical properties such as the work done by a force and the accumulated distance an object has traveled. It is an essential tool in both physics and engineering to solve problems related to continuous change.

Area between Curves

Calculating the area between curves is an extension of definite integrals, to find the region that lies between two curves on a graph. When two functions \( f(x) \) and \( g(x) \) are plotted, the area between them from \( x = a \) to \( x = b \) is calculated by integrating the difference of the functions' values throughout that interval.

The process for finding the area between curves involves:

• Identify points of intersection: Determine the x-values where the curves intersect to establish bounds for integration.
• Setup Integrals: Subtract the lower function from the upper one: \( \int_{a}^{b} [f(x) - g(x)] \), where \( f(x) \) is the upper function and \( g(x) \) is the lower function.
• Integrate and Evaluate: Compute the integral and apply the limits of integration to find the area between the curves.

This method can be adapted when working with functions expressed in terms of \( y \) by switching the order of variables, as demonstrated in the original exercise, showing how versatile the technique can be for geometric applications in mathematics.

Integration

Integration, often referred to as the inverse of differentiation, is a core operation in calculus for combining small quantities to measure a whole. Fundamentally, integration involves summing infinitely small parts, and it can be understood visually as finding the area beneath or between curves on a graph.

There are two main types of integration:

• Indefinite Integration: Finds the general form of the antiderivative of a function, represented as \( \int f(x) \, dx = F(x) + C \), where \( C \) is the constant of integration.
• Definite Integration: Finds the specific value of the area under the curve from one point to another, removing the need for the constant of integration.

Integration is a powerful tool for solving a variety of problems involving area, volume, and even more complex applications such as electricity and fluid dynamics. By understanding the fundamentals of integration, students can apply this concept in multiple areas of science and engineering.