Problem 43 - Applied Calculus

Question

Sketch each parabola and line on the same graph and find the area between them from \(x=0\) to \(x=3\). \(y=3 x^{2}-3\) and \(y=2 x+5\)

Step-by-Step Solution

Answer

The area between the parabola and the line from \(x=0\) to \(x=3\) is 14 square units.

Step 1: Identify the Functions

We are given two functions: the parabola \( y = 3x^2 - 3 \) and the line \( y = 2x + 5 \). We'll need to sketch these to understand their intersection and behavior between \( x = 0 \) and \( x = 3 \).

Step 2: Find Points of Intersection

To find where the parabola and the line intersect, set their equations equal: \( 3x^2 - 3 = 2x + 5 \). Rearrange to form: \( 3x^2 - 2x - 8 = 0 \). Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 3, b = -2, c = -8 \). Compute \( x = \frac{2 \pm \sqrt{(-2)^2 - 4 \times 3 \times (-8)}}{2 \times 3} \). This simplifies to \( x = \frac{2 \pm \sqrt{4 + 96}}{6} = \frac{2 \pm \sqrt{100}}{6} = \frac{2 \pm 10}{6} \). This gives intersection points at \( x = 2 \) and \( x = -\frac{4}{3} \).

Step 3: Determine the Integration Bounds

Given \( x = 0 \) and \( x = 3 \), the bounds for integration are clear. The intersection at \( x = 2 \) lies within this interval, affecting the part of the area to consider between curves. Use the interval \([0, 2] \) and \([2, 3] \) to calculate separate areas.

Step 4: Calculate the Area from x = 0 to x = 2

From \( x = 0 \) to \( x = 2 \), the line lies above the parabola. Calculate the area difference: \( \int_{0}^{2} [(2x + 5) - (3x^2 - 3)] \, dx = \int_{0}^{2} (-3x^2 + 2x + 8) \, dx \). Compute the integral: \[ \int_{0}^{2} (-3x^2 + 2x + 8) \, dx = \left[ -x^3 + x^2 + 8x \right]_{0}^{2} = 0 - (-(8) + 4 + 16) = 12. \]

Step 5: Calculate the Area from x = 2 to x = 3

From \( x = 2 \) to \( x = 3 \), the line is still above the parabola. Calculate: \( \int_{2}^{3} [(2x + 5) - (3x^2 - 3)] \, dx = \int_{2}^{3} (-3x^2 + 2x + 8) \, dx \). Compute this integral: \[ \int_{2}^{3} (-3x^2 + 2x + 8) \, dx = \left[ -x^3 + x^2 + 8x \right]_{2}^{3} = (-(27) + 9 + 24) - (-(8) + 4 + 16) = -6 + 4 = -2. \] Note this area is treated as positive, so 2 square units do contribute.

Step 6: Add the Areas Together

Calculate the total area between the parabola and the line from \( x = 0 \) to \( x = 3 \) by adding areas from both sections. Result: \( 12 + 2 = 14 \).

Key Concepts

Parabola

The parabola is a U-shaped curve that is symmetrically centered around its vertex. In the context of our problem, we deal with the quadratic equation \( y = 3x^2 - 3 \). This depicts a parabola that opens upwards, as indicated by the positive coefficient of the \( x^2 \) term. Quadratic equations generally have the form \( ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants. For our specific parabola:

• \( a = 3 \) implies the parabola is relatively steep.
• \( c = -3 \) suggests the vertex is located below the x-axis, specifically at the point \((0, -3)\).

To sketch this parabola accurately, locate the vertex and plot additional points by choosing different \( x \) values, then calculate their corresponding \( y \) values using the equation. The shape is symmetrical around the vertical line passing through the vertex.

Linear Equation

A linear equation represents a straight line and can be expressed as \( y = mx + c \), where \( m \) is the slope and \( c \) is the y-intercept. For the linear equation \( y = 2x + 5 \) given in our problem:

• \( m = 2 \) indicates the line has a steep positive incline.
• \( c = 5 \) shows the line crosses the y-axis at \( y = 5 \).

To sketch this line, start by plotting the y-intercept \((0, 5)\). Then use the slope to determine another point: from \( (0, 5) \), move up two units and one unit to the right to reach \( (1, 7) \). Draw a straight line through these points. The nature of a linear equation ensures the line continues indefinitely in both directions.

Integration

Integration is a fundamental concept in calculus used to find areas, among other applications. It involves summing infinitesimal quantities to determine the whole and is denoted by the integral symbol \( \int \). In our exercise, we use integration to find the area between the parabola and the line from \( x = 0 \) to \( x = 3 \). We calculate the definite integrals over specified bounds:

• For \( x = 0 \) to \( x = 2 \), we find \( \int_{0}^{2} ((2x + 5) - (3x^2 - 3)) \, dx \).
• For \( x = 2 \) to \( x = 3 \), it's \( \int_{2}^{3} ((2x + 5) - (3x^2 - 3)) \, dx \).

This allows us to quantify the area where the line is above the parabola, represented mathematically by the difference in functions \((2x + 5) - (3x^2 - 3)\). The obtained values from these definite integrals are added to find the total area, which is 14 square units in this case.

Intersection Points

Intersection points are crucial for determining where two graphs meet. They help us understand how the functions relate in the given domain. To find these points for the parabola \( y = 3x^2 - 3 \) and line \( y = 2x + 5 \), set the equations equal:\[ 3x^2 - 3 = 2x + 5 \]Rearrange to form a quadratic equation: \( 3x^2 - 2x - 8 = 0 \). Solving this using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) gives:

• Two solutions: \( x = 2 \) and \( x = -\frac{4}{3} \).

In this exercise, since we only consider \( x \) values from 0 to 3, we focus on \( x = 2 \) as the point of intersection within this range. This point divides our interval and affects the bounds used in integration to calculate the definite areas between the curves.