Problem 13 - Essential Calculus

Question

In Exercises \(13-26,\) sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ y=\frac{1}{2} x^{3}+2, y=x+1, x=0, x=2 $$

Step-by-Step Solution

Answer

The area of the region bounded by the graphs of the functions \( y=\frac{1}{2} x^{3}+2, y=x+1, x=0, x=2 \) is 0

Step 1: Plot the Graphs

First, plot the graphs of the functions \( y=\frac{1}{2} x^{3}+2, y=x+1 \) and the lines \( x=0, x=2 \). After plotting these, one can easily observe that the region bounded by these graphs is between the x-axis (x=0) and the vertical line (x=2)

Step 2: Identify the Edges of the Bounded Region

Identify the functions that make up the lower and upper edges of the region: the cubic function \( y=\frac{1}{2} x^{3}+2 \) and the linear function \( y=x+1 \) respectively. And the region is also bounded by vertical lines \( x=0, x=2 \)

Step 3: Set Up the Integral

The area of the region bounded by these functions is given by the definite integral \( \int_{a}^{b} (f(x)-g(x))dx \), where \( f(x) \) is the upper function, \( g(x) \) is the lower one, and [a, b] is the interval on the x-axis over which the region extends. Here, \( f(x) = x+1, g(x) = \frac{1}{2} x^{3}+2 \) and the interval is [0, 2]. So, set up the integral: \( \int_{0}^{2} ((x+1)-(\frac{1}{2} x^{3}+2))dx \)

Step 4: Evaluate the Integral

Simplify the integrand and then use the fundamental theorem of calculus to evaluate the integral to find the area of the region. \( \int_{0}^{2} (x+1-\frac{1}{2} x^{3}-2))dx = \int_{0}^{2} (x-\frac{1}{2} x^{3})dx \). Evaluate this integral by applying the power rule in reverse (adding 1 to the exponent and dividing by the new exponent), to each term and taking the antiderivative, then substitute the limits and subtract the lower limit result from the upper limit result: \( [\frac{1}{2} x^{2} - \frac{1}{8} x^{4}]_{0}^{2} = (2 - 2) - (0 - 0) = 0 \)

Key Concepts

Definite Integral

A definite integral is a powerful tool in calculus that allows us to calculate the area under a curve between two specified points on the x-axis. It is represented as \( \int_a^b f(x) \, dx \), where \( f(x) \) is the function being integrated, and \( a \) and \( b \) are the lower and upper limits of integration.
The process of finding a definite integral involves calculating the antiderivative of the function, substituting the upper and lower limits, and finding the difference.

The definite integral gives us the net area, accounting for areas above the x-axis as positive and those below as negative.

• Calculate the antiderivative.
• Substitute the limits \( a \) and \( b \).
• Subtract the value at \( a \) from the value at \( b \).

This tool is crucial for problems where you need exact areas, such as finding areas between curves as seen in the given problem.

Area under a Curve

The area under a curve represents the total space between the graph of a function and the x-axis over a certain interval. To find this area, the definite integral is used, which sums infinitely small rectangles from one boundary to another.
For the exercise at hand, where two functions intersect, we compute the area between them by integrating the difference of the two functions over the given interval.

The basic steps are:

• Determine the interval over x (where the area calculation starts and ends).
• Identify the upper and lower functions over this interval.
• Set up the integral of the difference \( f(x) - g(x) \) where \( f(x) \) is the top function.
• Evaluate the integral to obtain the area.

This procedure allows us to understand the spatial distribution of the functions and their interaction on a graph.

Graph Sketching

Graph sketching is an essential skill when dealing with calculus problems, particularly those involving areas under curves. A sketch can provide a visual understanding of the functions and how they interact.
To sketch the graph of functions like \( y = \frac{1}{2}x^3 + 2 \) and \( y = x + 1 \), start by plotting key points and observing their behavior at critical values.

A few tips for graph sketching:

• Find the intercepts and plot them.
• Identify and mark turning points or inflection points.
• Determine where curves start and end for a given interval.
• Understand the behavior as x approaches positive and negative infinity.

With a good sketch, you can adequately identify the boundaries and intersections, making it easier to set up integrals for area calculations.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects the procedures of differentiation and integration, two central operations in calculus. It establishes that the process of integrating a function can be reversed by differentiation.
This theorem is essential for calculating definite integrals. It states that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then:
\[\int_a^b f(x)\, dx = F(b) - F(a)\]The theorem tells us how to evaluate a definite integral by finding an antiderivative of the integrand.

• Find the antiderivative \( F(x) \) of \( f(x) \).
• Subtract \( F(a) \) from \( F(b) \).

This not only saves time but also simplifies solving complex integrals, just like in our exercise where it was used to find the area between two curves.