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Chapter 5: Problem 44

Sketch each parabola and line on the same graph and find the area between them from \(x=0\) to \(x=3\). \(y=3 x^{2}-12\) and \(y=2 x-11\)

Short Answer

Expert verified
The area between the curves from \(x=0\) to \(x=3\) is 20 square units.

Step by step solution

01

Understand the Equations

The given functions are a parabola: \(y = 3x^2 - 12\) and a line: \(y = 2x - 11\). We need to find their intersection points between \(x=0\) and \(x=3\), then determine the area enclosed by these two curves.
02

Find Points of Intersection

To find the intersection points, set the two equations equal: \(3x^2 - 12 = 2x - 11\). Rearrange the equation to form a quadratic: \(3x^2 - 2x - 1 = 0\). Solve using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a=3\), \(b=-2\), \(c=-1\). Calculate the discriminant: \(b^2 - 4ac = (-2)^2 - 4(3)(-1) = 4 + 12 = 16\). So \(x = \frac{2 \pm \sqrt{16}}{6} = \frac{2 \pm 4}{6}\). Thus, \(x = 1\) and \(x = -\frac{1}{3}\). Only \(x=1\) is in the given range \([0,3]\).
03

Calculate the Points at Limits

Calculate the y-values of both functions at \(x=0\) and \(x=3\). For \(x=0\), \(y_{parabola} = 3(0)^2 - 12 = -12\), and \(y_{line} = 2(0) - 11 = -11\). At \(x=3\), \(y_{parabola} = 3(3)^2 - 12 = 15\), and \(y_{line} = 2(3) - 11 = -5\).
04

Determine Area Between the Curves

The region of interest is between \(x=0\) and \(x=3\). Calculate the area using definite integrals from \(x=0\) to \(x=1\) for when the line is above, and from \(x=1\) to \(x=3\) for when the parabola is above. Express each as an integral: 1. From \(0\) to \(1\), integrate \((2x - 11) - (3x^2 - 12)\).2. From \(1\) to \(3\), integrate \((3x^2 - 12) - (2x - 11)\). 3. Add the absolute values of these two integrals to find the total area.
05

Integrate to Find Area

1. Integrate from 0 to 1: \(\int_{0}^{1} ((2x - 11) - (3x^2 - 12)) \; dx = \int_{0}^{1} (-3x^2 + 2x + 1)\; dx\).Evaluating, we get: \( \left[ -x^3 + x^2 + x \right]_{0}^{1} = -(1)^3 + (1)^2 + (1) = 1\).2. Integrate from 1 to 3: \(\int_{1}^{3} (3x^2 - 2x - 1) \; dx\).Evaluating, we get: \( \left[ x^3 - x^2 - x \right]_{1}^{3} = ((3)^3 - (3)^2 - (3)) - ((1)^3 - (1)^2 - (1)) = 19\).Add the two areas: \(1 + 19 = 20\).
06

Sketch the Curves and Mark the Area

Plot the curves and identify the points of intersection and the region between them. Shade the area enclosed between \(x=0\) and \(x=3\) on the graph to visually represent the area calculated.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Parabola equation
A parabola is a symmetrical curve represented by a quadratic equation. It has the form \( y = ax^2 + bx + c \). In this exercise, our equation is \( y = 3x^2 - 12 \), which means:
  • The coefficient \(a = 3\) determines how "open" or "closed" the parabola is.
  • The parabola opens upwards because \(a > 0\).
  • The vertex is the lowest point of the curve because it opens upward.
To sketch the parabola, note that at \(x = 0\), the y-intercept is \( -12 \). Additionally, as \(x\) moves from negative to positive, \(y\) increases significantly because the \(3x^2\) term dominates the expression. Understanding these characteristics helps in visualizing and solving problems involving parabolas.
Line equation
Lines are straight and are represented by the first-degree equation \( y = mx + b \). For our line, \( y = 2x - 11 \):
  • The slope \(m = 2\) indicates the line rises steadily as \(x\) increases.
  • The y-intercept is \(-11\), where the line crosses the y-axis.
  • The line is linear, meaning it doesn鈥檛 curve or change direction.
Visualizing this line helps to determine the section where it intersects and the regions of interest with the parabola. The slope shows the consistency of the increase or decrease in \(y\) values over \(x\), simplifying decisions about the upper or lower bound in area calculations.
Definite integral
The definite integral is a fundamental tool in calculus used to find the area under a curve. When comparing two functions, it's used to determine the area between them by calculating:
  • \( \int (\text{upper curve} - \text{lower curve}) \, dx \).
For between \(x = 0\) and \(x = 1\), the line \(2x - 11\) is above the parabola \(3x^2 - 12\). In this range, you integrate the difference to get the area:
  • \( \int_{0}^{1} ((2x - 11) - (3x^2 - 12)) \, dx \)
From \(x = 1\) to \(x = 3\), the parabola is above, so:
  • \( \int_{1}^{3} ((3x^2 - 12) - (2x - 11)) \, dx \)
Adding these integrals鈥� absolute values gives the total area. The definite integral not only solves such area problems but also extends to real-world applications wherever accumulative totals are required.
Intersection points
Finding intersection points is crucial when calculating the area between two curves. For any two equations, setting them equal determines where they cross. For this parabola \(y = 3x^2 - 12\) and line \(y = 2x - 11\), equating gives:
  • \( 3x^2 - 12 = 2x - 11 \)
  • Rearrange to form the quadratic \( 3x^2 - 2x - 1 = 0 \).
  • Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
This results in \( x = 1 \) and \( x = -\frac{1}{3} \), but only \( x = 1 \) fits our given range \([0,3]\). The intersections help divide the curve into segments for easier area calculations. More broadly, they are key in determining when one curve sits above or below another, which is essential for integrals.

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