Problem 41 - Applied Calculus

Question

Sketch each parabola and line on the same graph and find the area between them from \(x=0\) to \(x=3\). \(y=x^{2}+4\) and \(y=2 x+1\)

Step-by-Step Solution

Answer

The area between the given curves from \(x=0\) to \(x=3\) is 9 square units.

Step 1: Set Up the Problem

We have two functions to consider: a parabola given by \(y = x^2 + 4\) and a line given by \(y = 2x + 1\). We need to find the area between these two curves from \(x = 0\) to \(x = 3\). The first step is to determine the points of intersection, as they will define the bounds of integration. However, since we are given bounds \(x = 0\) and \(x = 3\), we will focus on integration over this range.

Step 2: Determine the Intersection Points

To determine if there are intersection points within the range, set \(x^2 + 4 = 2x + 1\). Rearrange this to \(x^2 - 2x + 3 = 0\). Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), here \(a = 1\), \(b = -2\), \(c = 3\), giving a discriminant \(\Delta = (-2)^2 - 4 \cdot 1 \cdot 3 = -8\). Since \(\Delta < 0\), no real solutions exist, confirming no intersection within \(x = 0\) to \(x = 3\).

Step 3: Determine Which Function is Above the Other

Evaluate both functions at a point, say \(x = 1\). For the parabola: \(y = 1^2 + 4 = 5\). For the line: \(y = 2(1) + 1 = 3\). Thus, for \(0 \leq x \leq 3\), the parabola is above the line.

Step 4: Set Up the Integral to Find the Area

The area between the curves is given by the integral of the difference between the top function and the bottom function over the interval \([0, 3]\). Thus, the area \(A\) is: \[ A = \int_0^3 ((x^2 + 4) - (2x + 1)) \, dx \]

Step 5: Simplify the Integrand

Simplify the expression inside the integral: \((x^2 + 4) - (2x + 1) = x^2 - 2x + 3\). The integral becomes: \[ A = \int_0^3 (x^2 - 2x + 3) \, dx \]

Step 6: Integrate the Function

Find the indefinite integral: \(\int (x^2 - 2x + 3) \, dx = \frac{x^3}{3} - x^2 + 3x + C\). Now evaluate this from 0 to 3: \[\left[ \frac{x^3}{3} - x^2 + 3x \right]_0^3\]

Step 7: Evaluate the Definite Integral

Compute the definite integral: \[\left( \frac{3^3}{3} - 3^2 + 3(3) \right) - \left( \frac{0^3}{3} - 0^2 + 3(0) \right)\] Simplifying, we get: \[\left( 9 - 9 + 9 \right) = 9\]. Therefore, the area between the curves from \(x=0\) to \(x=3\) is 9 square units.

Key Concepts

Integration

Integration is a method used in calculus to find the area under a curve. In this exercise, we're finding the area between two curves. This involves calculating the integral of the difference between the two functions over a given interval.
This integral accumulates the difference in area at each point along an interval, providing the total area between the two curves.
Understand that integration reverses the process of differentiation, allowing us to sum up infinitely small values over an interval to yield a total value. When dealing with complex problems like these, it's crucial to set up the integral correctly, using proper boundaries and ensuring that the functions are in their simplest forms.

Parabolas

A parabola is a symmetrical, U-shaped curve represented by a quadratic equation of the form \(y = ax^2 + bx + c\). In this exercise, the parabola is given by the equation \(y = x^2 + 4\).
Parabolas have unique properties, such as a vertex, axis of symmetry, and the fact that they can open upwards or downwards.
The vertex is the highest or lowest point depending on the parabola's orientation. They have a unique connection to quadratic equations and play a key role in geometry and algebra. This particular parabola opens upwards because the coefficient of \(x^2\) is positive.

Lines

Lines are straight and can be expressed in slope-intercept form, \(y = mx + b\), where \(m\) is the slope and \(b\) is the y-intercept.
In this scenario, the line is described by \(y = 2x + 1\). This representation helps us visualize how steep the line is and where it crosses the y-axis.
The slope of a line indicates its steepness and direction, which impacts how it intersects other curves. For analyzing areas between curves, knowing the line's position relative to other functions is crucial, as it affects which function lies above or below within the desired interval.

Definite Integral

A definite integral calculates the area under a curve within specified limits, giving a numeric value as a result.
For this problem, the definite integral is set from \(x=0\) to \(x=3\). It accumulates the difference between the two functions over this range.
The process involves integrating the simplified function and evaluating the result within these bounds. The value you obtain represents the total area between the curves over the chosen interval. Definite integrals provide solutions to geometric problems involving areas, ensuring precise calculations within specific parameters.

Difference of Functions

The concept of the difference of functions is central to determining the area between curves. This difference comes from subtracting one function from another, specifically the function that lies beneath from the function above over the interval of integration.
In mathematical terms, if \(f(x)\) is the upper function and \(g(x)\) is the lower, then the area sought is defined by \( \int_a^b (f(x) - g(x)) \, dx\). By simplifying this expression, the problem becomes more manageable, as we can then integrate a single function instead of two separate ones.
This process requires determining which function is above within the chosen interval, which affects how we set up the integral accurately.