Question

Write an iterated integral of a continuous function \(f\) over the following regions. $$R=\\{(x, y): 0 \leq x \leq y(1-y)\\}$$

Short answer

Question: Determine the iterated integral of a continuous function \(f(x, y)\) over the region \(R\) described by the inequality \(0 \leq x \leq y(1-y)\) and \(0 \leq y \leq 1\). Answer: $$\int_{0}^{1} \int_{0}^{y(1-y)} f(x, y) dx dy$$

Step-by-step solution

Determine the boundary points for \(y\)

To determine the boundaries for \(y\), we can rewrite the inequality as \(y(1-y) \geq 0\). This quadractic inequality has two roots: \(y=0\) and \(y=1\). Therefore, the range for \(y\) is \(0 \leq y \leq 1\).

Set up the iterated integral

Now that we have the bounds for both \(x\) and \(y\), we can set up the iterated integral. Since \(x\) depends on \(y\), we should integrate with respect to \(x\) first: $$\int_{0}^1 \int_{0}^{y(1-y)} f(x, y) dx dy$$

Evaluate the iterated integral

To find the volume of the region under the continuous function \(f(x, y)\), we need to evaluate the iterated integral. However, since the exercise did not provide a specific function for \(f(x, y)\), we cannot perform the integration. The iterated integral, as shown in Step 2, is the complete answer to the exercise: $$\int_{0}^{1} \int_{0}^{y(1-y)} f(x, y) dx dy$$

Key concepts

Continuous Functions

In calculus, continuous functions are those that are smooth and unbroken over a certain range. They do not have abrupt changes, jumps, or holes. For a function to be continuous at a point, three conditions must be satisfied:

• The function is defined at that point.
• The limit of the function as it approaches the point from either side exists.
• The value of the function at that point is equal to the limit.

For example, the function provided in the exercise, denoted simply as \( f(x, y) \), is assumed to be continuous across the given region. This assumption is crucial because it ensures that the iterated integral can be computed over this region without encountering undefined behavior. Continuous functions are essential in many mathematical applications because they enable predictable and stable calculations when integrating or differentiating over a region. By integrating these functions over a particular domain, we can accurately assess areas, volumes, or other accumulated quantities.

Quadratic Inequality

A quadratic inequality, like the one seen in the exercise, involves a quadratic expression, which is any polynomial expression of the form \( ax^2 + bx + c \). The exercise uses the inequality \( y(1-y) \geq 0 \). Solving this inequality can help identify bounds for integration.To address this type of inequality:

• First, identify the roots of the corresponding equation, obtained by setting the inequality to zero, \( y(1-y) = 0 \).
• The roots, or solutions, are \( y = 0 \) and \( y = 1 \), which tells us where the quadratic expression changes sign.
• Next, determine where the inequality holds. In this context, \( y(1-y) \geq 0 \) means either both factors are non-negative or both are non-positive, which happens between the roots, inclusive.

This approach helps establish the valid range \( 0 \leq y \leq 1 \) for integration. Finding these bounds forms the basis of setting up iterated integrals, guiding which region of the function is evaluated.

Integration Bounds

Integration bounds define the limits over which we calculate an integral. When setting these up for iterated integrals, understanding the dependency between variables is key.For the given region, the integration bounds are:

• For \( y \), the bounds are determined by the solutions of the quadratic inequality: \( 0 \leq y \leq 1 \).
• Once the bounds for \( y \) are known, the bounds for \( x \) must be established considering its dependency on \( y \): \( 0 \leq x \leq y(1-y) \).

To set up the iterated integral correctly:
- Integrate with respect to \( x \) first since \( x \) depends on \( y \). Thus, integrate from \( 0 \) to \( y(1-y) \) within each slice of \( y \).
- Then, integrate with respect to \( y \) from \( 0 \) to \( 1 \).
These bounds completely specify the region over which the function \( f(x, y) \) is integrated, ensuring that all parts of the described area between the curves are accounted for in the integral. Setting integration bounds properly is fundamental to calculating areas and volumes accurately in multi-variable calculus.