Problem 14 - Basic Technical Mathematics with Calculus

Question

Find the areas bounded by the indicated curves. $$x=y^{2}-4 y, x=0$$

Step-by-Step Solution

Answer

The area bounded by the curves is \(\frac{32}{3}\) square units.

Step 1: Understand the Problem

We are given two curves: \(x = y^2 - 4y\) and \(x = 0\). We are tasked to find the area between these curves. The curve \(x = 0\) represents the y-axis.

Step 2: Determine Intersection Points

Set the equations equal to solve for \(y\):\(y^2 - 4y = 0\). Factor this equation to \(y(y-4) = 0\), giving solutions \(y = 0\) and \(y = 4\). These are the y-values where the curves intersect.

Step 3: Set Up the Integral for Area

Since the area is bounded between the curve \(x = y^2 - 4y\) and the line \(x = 0\), we integrate with respect to \(y\) from \(y = 0\) to \(y = 4\). The integral is:\[\int_{0}^{4} (y^2 - 4y) \, dy\].

Step 4: Calculate the Integral

First, find the antiderivative:\[\int (y^2 - 4y) \, dy = \frac{1}{3}y^3 - 2y^2 + C\].Then evaluate from \(y = 0\) to \(y = 4\):\[\left[\frac{1}{3}(4)^3 - 2(4)^2\right] - \left[\frac{1}{3}(0)^3 - 2(0)^2\right]\].

Step 5: Simplify the Calculation

Calculate \(\frac{1}{3}(4)^3 = \frac{1}{3} \times 64 = \frac{64}{3}\) and \(2(4)^2 = 32\).Thus, the area is:\[\frac{64}{3} - 32\].

Step 6: Finalize the Area

Convert \(32\) to thirds: \(32 = \frac{96}{3}\).Thus, \(\frac{64}{3} - \frac{96}{3} = -\frac{32}{3}\).Since area must be positive, take the absolute value: \(\frac{32}{3}\).

Key Concepts

Integration

Integration is a fundamental concept in calculus, and it is crucial for calculating areas under curves, among other applications. It is essentially the reverse process of differentiation. By integrating a function, we can find the accumulated total or the net change, such as the total area under a curve.

The process of integration involves finding the antiderivative of a function. The antiderivative is another function, which represents the area accumulated from a starting value to any point along the curve. In this exercise, the antiderivative of the function \( y^2 - 4y \) is given by \( \frac{1}{3}y^3 - 2y^2 \), which is found using basic integration rules.

Once the antiderivative is determined, we evaluate it over a particular interval to find definite values, such as the area between curves or under a single curve, as seen in this example.

Definite Integral

A definite integral is a type of integral that involves limits of integration, which define the interval over which the function is being evaluated. The definite integral of a function \( f(x) \) from \( a \) to \( b \) is denoted as \( \int_{a}^{b} f(x) \, dx \).

In the case of finding the area between curves, the definite integral captures the net area between the specified curves over the interval defined by their intersection points. The integration is done with respect to one variable, and these limits of integration are crucial as they tell us the range of values over which the function's antiderivative should be evaluated.

In our exercise, the definite integral \( \int_{0}^{4} (y^2 - 4y) \, dy \) represents the exact area between the curve \( x = y^2 - 4y \) and the y-axis from \( y = 0 \) to \( y = 4 \), capturing the area segment of interest.

Area between curves

Calculating the area between curves is an important application of definite integrals in calculus. To find this area, you need to integrate the difference between the two functions that define the curves over the specified interval.

The basic idea is to determine the intersection points of the curves, as these define the interval of integration. In this example, the curves are \( x = y^2 - 4y \) and \( x = 0 \), which intersect at \( y = 0 \) and \( y = 4 \).

With these boundaries set, the area between the curves is given by integrating the function that lies above minus the function below, which simplifies to integrating \( y^2 - 4y \) from \( y = 0 \) to \( y = 4 \). This yields the net area, which is computed by evaluating the antiderivative at these points and taking the difference. The final step ensures the area is expressed as a positive value, as area cannot be negative.