Problem 37 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Sketch the graph and find the area of the region completely enclosed by the graphs of the given functions \(f\) and \(g\). $$f(x)=x^{2}\( and \)g(x)=x^{3}$$

Step-by-Step Solution

Answer

The enclosed area between the graphs of the functions \(f(x) = x^{2}\) and \(g(x) = x^{3}\) is \(\frac{1}{12}\).

Step 1: Identify the Points of Intersection

To find the points of intersection between the functions f(x) and g(x), set them equal to each other and solve for x: $$f(x) = g(x)$$ $$x^{2} = x^{3}$$ Since the exponents in the x terms differ, we can solve for x by factoring x from the equation: $$x(x^{2} - x^{3}) = 0$$ Now, we can find the three points of intersection: $$x = 0, x^{2} = x^{3}$$ We can find the other two points of intersection: $$x^{2} - x^{3} = 0$$ $$x^{2}(1-x)=0$$ $$x = 0, 1$$ So, we have found the three points of intersection: \((0,0)\), \((1,1)\)

Step 2: Sketch the Graphs

Now that we have identified the points of intersection, we can sketch the graphs of the functions f(x) and g(x), paying special attention to their points of intersection: - f(x) = \(x^{2}\) is a parabola with its vertex at the origin. - g(x) = \(x^{3}\) is a cubic function that passes through the origin and is symmetric with the line y = x. With these characteristics, we can sketch both functions and identify the region enclosed by the two graphs.

Step 3: Determine the Area of the Enclosed Region

To calculate the area of the enclosed region, we'll subtract the integral of g(x) from the integral of f(x) on the interval where the enclosed region exists. Since we have found the points of intersection to be at (0,0) and (1,1), we define the interval as [0,1]: $$A = \int_{0}^{1}(f(x)-g(x))dx$$ Where A represents the area of the enclosed region. Now, we can plug in the functions for f(x) and g(x) and evaluate the integral: $$A = \int_{0}^{1} (x^{2} - x^{3}) dx$$ To solve the integral, we need to find the antiderivative of the function inside the integral: $$F(x) = \frac{x^{3}}{3} - \frac{x^{4}}{4} + C$$ Now, we can calculate the area A by evaluating F(x) at the limits of the interval (0 and 1): $$A = F(1) - F(0)$$ $$A = \left(\frac{1^{3}}{3} - \frac{1^{4}}{4}\right) - \left(\frac{0^{3}}{3} - \frac{0^{4}}{4}\right)$$ $$A = \frac{1}{3} - \frac{1}{4}$$ $$A = \frac{1}{12}$$ The area of the region completely enclosed by the graphs of the functions f and g is \(\frac{1}{12}\).

Key Concepts

Integration

Integration is a fundamental concept in calculus that refers to the process of finding the antiderivative or the area under a curve. In this exercise, we use integration to find the area of the region enclosed between two curves, specifically the functions \( f(x) = x^2 \) and \( g(x) = x^3 \).
To perform integration, we first need to identify the interval over which the area is to be computed. In our case, this is determined by the points of intersection of the two graphs, given as \( x=0 \) and \( x=1 \).
Once the interval is identified, the integral is set up as the difference between the two functions over that interval:

• \( A = \int_{0}^{1}(f(x)-g(x))dx \)
• This involves computing the antiderivative of the difference \( x^2 - x^3 \).

Ultimately, the integral gives us an exact numeric value for the area under the curve in that region.

Graph Sketching

Graph sketching involves drawing a visual representation of the function's behavior over a certain interval. In this exercise, it's essential to understand the shape and intersection points of \( f(x) = x^2 \) and \( g(x) = x^3 \).
The parabola \( f(x) = x^2 \) has its vertex at the origin and opens upwards, forming a symmetric U-shape.
Cubic function \( g(x) = x^3 \) also passes through the origin but increases and decreases more sharply since it has an inflection point at zero.

• Both graphs start at \( (0, 0) \), coincide at that point, and extend beyond.
• They intersect again at \( (1, 1) \).

To sketch these functions accurately, plot key points like intersections and vertex, and draw the curves through these points. Doing so helps identify regions enclosed by the graphs that you might need to analyze further.

Function Intersection

Finding where two functions intersect means identifying the values of \( x \) where the functions meet, i.e., have the same y-value. In mathematical terms, this is done by setting \( f(x) = g(x) \) and solving for \( x \).
In our case, finding the intersection of \( f(x) = x^2 \) and \( g(x) = x^3 \), we set them equal and solve for \( x \):

• \( x^2 = x^3 \) leads to \( x(x^2 - x) = 0 \), giving points \( x=0 \) and \( x=1 \).
• Each intersection corresponds to a point on the graph: \( (0, 0) \) and \( (1, 1) \).

These points demarcate the boundaries of the region we are interested in when calculating the area under the curves.

Area under the Curve

The area under the curve is a measure of the total 'space' between the graph of a function and the x-axis over a given interval. When considering the area between two curves, it involves integrating the difference between the higher and lower functions across the interval.

• This is represented as \( A = \int_{a}^{b} [f(x) - g(x)] \, dx \), where \( a \) and \( b \) are the points of intersection.
• In our specific problem, the area between \( x=0 \) and \( x=1 \) of \( x^2 \, \text{and} \, x^3 \) is calculated in this way.

Ultimately, the process of finding this area highlights the application of integration in determining areas between functions, which is a critical skill in calculus.