Problem 36 - Calculus for Biology and Medicine

Question

Find the areas of the regions bounded by the lines and curves by expressing \(x\) as a function of \(y\) and integrating with respect to \(y .\) \(y=x, y=x^{2}\)

Step-by-Step Solution

Answer

The area between the curves is \(\frac{1}{6}\).

Step 1: Identify the Intersection Points

To find the limits of integration, identify the points where the functions intersect by setting them equal: \(y = x\) and \(y = x^2\). This gives us the equation \(x = x^2\). Solve for \(x\) by factoring: \(x^2 - x = 0\) which leads to \(x(x - 1) = 0\), giving intersection points at \(x = 0\) and \(x = 1\). Therefore, these values correspond to \(y = 0\) and \(y = 1\).

Step 2: Express x as a Function of y

For \(y = x\), express \(x\) as \(x = y\). For \(y = x^2\), express \(x\) as \(x = \sqrt{y}\), using only the positive root because the curves intersect in the first quadrant where both \(x\) and \(y\) are non-negative.

Step 3: Set Up the Integral

The area between the curves as functions of \(y\) is given by the integral \( \int_{0}^{1} [(\text{right curve}) - (\text{left curve})] \, dy \). Here, the right curve is \(x = y\) and the left curve is \(x = \sqrt{y}\). Therefore, set up the integral as \( \int_{0}^{1} [y - \sqrt{y}] \, dy \).

Step 4: Integrate with Respect to y

Evaluate the integral \( \int_{0}^{1} (y - \sqrt{y}) \, dy \). This splits into two integrals: \( \int_{0}^{1} y \, dy - \int_{0}^{1} \sqrt{y} \, dy \). Compute each: \( \int_{0}^{1} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1}{2} \), and \( \int_{0}^{1} \sqrt{y} \, dy = \left[ \frac{2}{3} y^{3/2} \right]_{0}^{1} = \frac{2}{3} \).

Step 5: Calculate the Area

Subtract the results of the two integrals computed: \( \frac{1}{2} - \frac{2}{3} = \frac{3}{6} - \frac{4}{6} = -\frac{1}{6} \). The area is actually the absolute value, so \( \frac{1}{6} \).

Key Concepts

Area Under a Curve

The area under a curve represents the region enclosed between the graph of a function and the axis over a specified interval. It essentially measures the space beneath the curve. In the context of calculus, this is typically calculated using integration. Consider any two curves, when you need to find the area between them, the task involves evaluating the space that's contained in between the two graphs.Understand the problem:

• The first step is visualizing the two curves: here, these curves are described by equations such as \(y = x\) and \(y = x^2\).
• The region whose area we're interested in lies between these two curves on a graph.

If we were to simply look at these functions on a standard x-y plane, the intersection of these functions defines the boundaries of the area we need to calculate. The entire task can then be translated into the application of integration within these limits to find the precise area. Remember, known as the "signed area," it is always positive when considering regions between curves.

Integration

Integration is a fundamental concept in calculus that involves adding up infinitesimal slices of an area under a curve to find the whole. Think of it as a really detailed way of measuring area. Integration can tackle complex shapes by breaking them down into smaller, more manageable pieces and taking the sum of these parts.In this scenario:

• We express the functions in terms of \(y\), such as \(x = y\) and \(x = \sqrt{y}\).
• This allows us to set up an integral between two limits that refer to the intersection points on the y-axis.

After setting up the integral, which evaluates the difference between the two functions (right curve minus left curve), we integrate over the designated limits. This means computing the definite integral from the lower point to the upper point, considering the function expression derived from our equations.Integration essentially automates and simplifies the computational process, providing the complete area between the curves efficiently.

Intersection Points

Intersection points are crucial in determining the parameters and limits for integration when calculating the area between curves. These are the points where the curves meet and cross one another, serving as the boundaries for the region of interest.Finding these points involves:

• Equating the two functions, such as \(y = x\) and \(y = x^2\).
• Solving the resulting equation gives the specific \(x\) values, which can then be translated into \(y\) values.

For example, setting \(x^2 = x\) and solving gives us \(x = 0\) and \(x = 1\). This means our intersection points, where the curves meet, occur at these x-values. Translating these x-values into y-values (since the functions are y-based), they both correspond to the points \(y = 0\) and \(y = 1\) respectively. Hence these points guide the range of integration, ensuring the computed area remains within the bounds defined by the intersections.