Problem 17 - College Algebra and Calculus: An Applied Approach

Question

Sketch the region bounded by the graphs of the functions and find the area of the region. $$ f(x)=\sqrt[3]{x}, g(x)=x $$

Step-by-Step Solution

Answer

The area of the region bounded by the functions \(f(x) = \sqrt[3]{x}\) and \(g(x) = x\) is 0.25 square units.

Step 1: Sketch the graphs of the functions

Start with sketching the graphs of the two given functions, \(f(x) = \sqrt[3]{x}\) and \(g(x) = x\). Both these function are easily graphed as they are basic functions. For function \(f(x) = \sqrt[3]{x}\), it's a cubic root function that increases slowly. For function \(g(x) = x\), it's a linear function with a positive slope. You will notice that these two functions intersect at two points: (0,0) and (1,1).

Step 2: Identify the bounded region

Now, identify the region bounded by the two curves. It is the area in the 1st quadrant enclosed between the two functions. In this case, it is the region between the x-axis, the cubic root curve above, and the line to its right.

Step 3: Set up integral for area

The area of the region bounded by two functions \(f(x)\) and \(g(x)\) is given by computing the definite integral of the absolute value of the difference between the two functions, in the interval where they intersect. In this case, the functions intersect at x=0 and x=1. The area A of the region bounded by \(f(x)\), \(g(x)\), and the x-axis, from x=0 to x=1, is the integral of \(|f(x) - g(x)|dx\) from 0 to 1. Here \(g(x) > f(x)\) for 0 <= x <= 1. So, set up the integral \( A = \int_0^1 (g(x) - f(x)) dx \).

Step 4: Compute the integral

Now, compute the integral. Substitute the functions into the integral, \( A = \int_0^1 (x - \sqrt[3]{x}) dx \). You calculate this integral by computing the antiderivatives of each function, then evaluating at the endpoints and subtracting. This gives: \( A = [0.5 * x^2 - 0.75 * x^{4/3} ]_0^1 \), which simplifies to \( A = 0.5 * 1^2 - 0.75 * 1^{4/3} - (0.5 * 0^2 - 0.75 * 0^{4/3}) = 0.5 - 0.75 = -0.25 \). However, area cannot be negative, so the absolute value is taken, hence, the area is 0.25 square units.