Problem 52 - Essential Calculus

Question

In Exercises \(49-52,\) use the limit process to find the area of the region between the graph of the function and the \(y\) -axis over the given \(y\) -interval. Sketch the region. $$ h(y)=y^{3}+1,1 \leq y \leq 2 $$

Step-by-Step Solution

Answer

The area under the curve \(h(y)=y^{3}+1\) for \(1 \leq y \leq 2\) is \((3/2)(2^{2/3}-1)\)

Step 1: Express the function in terms of \(y\)

This is already accomplished because the function is framed as \(h(y)=y^{3}+1\).

Step 2: Find the inverse of the function

To find the inverse of \(h(y)=y^{3}+1\), set \(y=x^{3}+1\). Then, solve for \(x\) to obtain the inverse function \(g(y) = \sqrt[3]{y-1}\).

Step 3: Identify x-coordinates parallel to the given y-interval

Next, establish the x-coordinates by substituting the y-interval values for \(y\) in function \(g(y)\). That is, for \(y=1\) and \(y=2\), the resulting \(x\) values are \(g(1)=\sqrt[3]{1-1}=0\) and \(g(2)=\sqrt[3]{2-1}=1\), respectively.

Step 4: Compute the definite integral

Given the x-coordinates \(0 \leq x \leq 1\) and the inverse function \(g(y)\), the definite integral to find the area between the y-axis and the function graph is: \[ \int_{0}^{1} g(y) \, dx \] This integral can be computed as: \[ \int_0^1 \sqrt[3]{x-1} \, dx \]

Step 5: Evaluate the definite integral

A standard integration formula is applied to integrate \(\sqrt[3]{x-1}\). The solution will be computed as \((3/2)(2^{2/3}-1)\).

Key Concepts

Definite Integral

When discussing the concept of a definite integral, imagine it as a tool that lets us calculate the area under a curve. If you have a function represented graphically, and you need to know how much space is enclosed between this curve and the x-axis over a specific interval, definite integrals come to the rescue. They help in finding precise areas without approximations.

For example, consider integrating the function discussed in the exercise \(h(y) = y^3 + 1\). By establishing intervals like those between \(x = 0\) and \(x = 1\), the definite integral provides that enclosed area. This is particularly useful not only for verifying the shapes and sizes of sections beneath curves but also in various applications where space or quantity descriptions are needed.

To compute the integral, you apply integration rules, which are systematic methods to find how much area exists between a curve and the axis over the specified limits of integration.

Inverse Function

One fascinating aspect of functions is their inverses. The inverse of a function essentially reverses the role of inputs and outputs. If you start with a function that takes a certain input to produce an output, the inverse lets you start with the output to find what input would give it.

From the exercise, the function \(h(y) = y^3 + 1\) is used. We find its inverse by reversing roles: Normally \(y = x^3 + 1\), but to find the inverse, we look for the value of \(x\) when given \(y\). After shifting terms, solve by \(x = \sqrt[3]{y-1}\), which is the inverse function. Understanding inverses is key in solving equations and transforming data back to its original form.

Whenever you encounter a situation needing a switch between dependent and independent variables, the concept of inverse functions proves to be very powerful.

Area Under a Curve

Calculating the area under a curve is a common task in mathematics, particularly when interpreting real-world phenomena through functions. This area is vital as it often translates to quantities necessary in physics, engineering, and other sciences.

In the exercise, finding the area under the curve for \(h(y) = y^3 + 1\) between limits \(1 \leq y \leq 2\) implies determining how much space exists between this graph and the y-axis. The curve represents a function's behavior, and the area gives insights into such behavior's magnitude over the interval. Here, reversing the coordinates and focusing on \(x\) helps meander through changes made clearer by the expression for \(g(y) = \sqrt[3]{y - 1}\).

Graphically representing these calculations serves as a solid way to visualize and comprehend the results. It translates abstract numbers into concrete imagery, reinforcing the role of area under a curve in practical scenarios."