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. $$ h(y)=y^{3}+1,1 \leq y \leq 2 $$

Short 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-by-step solution

Express the function in terms of \(y\)

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

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 \(g(y) = \sqrt[3]{y-1}\).

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.

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: \[ \int_0^1 \sqrt[3]{x-1} \, dx \]

Evaluate the definite integral

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