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Magazine Chapter 5: Problem 32
Find the area between the graph of \(y=x^{2}-2\) and the \(x\) -axis, between \(x=0\) and \(x=3.\)
Short Answer
Expert verified
The area is \( 3 + \frac{8\sqrt{2}}{3} \).
Step by step solution
01
Understand the Problem
You need to find the area between the curve given by the function \( y = x^2 - 2 \) and the \( x \)-axis. The boundaries for this area are where \( x = 0 \) and \( x = 3 \).
02
Determine the Intersection Points
Find where the curve intersects the \( x \)-axis by setting \( y = 0 \) in the equation \( y = x^2 - 2 \): \( x^2 - 2 = 0 \). Solving this gives \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). However, only \( x = \sqrt{2} \) is in our region of interest \([0, 3]\).
03
Set Up the Integral
Since the function \( y = x^2 - 2 \) lies below the \( x \)-axis for some parts of \([0, 3]\), we need to account for negative values. Thus, the area is the integral of the absolute value: \( \int_{0}^{3} |x^2 - 2| \, dx \).
04
Split the Interval Based on Sign Change
The function changes sign at \( x = \sqrt{2} \). For \( 0 \leq x < \sqrt{2} \), \( x^2 - 2 < 0 \). For \( \sqrt{2} < x \leq 3 \), \( x^2 - 2 \geq 0 \). Split the integral: \( \int_{0}^{\sqrt{2}} (2 - x^2) \, dx + \int_{\sqrt{2}}^{3} (x^2 - 2) \, dx \).
05
Calculate the First Integral
Calculate \( \int_{0}^{\sqrt{2}} (2 - x^2) \, dx \):\[ \int 2 \, dx = 2x \]\[ \int x^2 \, dx = \frac{x^3}{3} \]Evaluate from 0 to \( \sqrt{2} \): \[ \left[ 2x - \frac{x^3}{3} \right]_{0}^{\sqrt{2}} = \left( 2\sqrt{2} - \frac{(\sqrt{2})^3}{3} \right) - (0) \]\[ = 2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3} \].
06
Calculate the Second Integral
Calculate \( \int_{\sqrt{2}}^{3} (x^2 - 2) \, dx \):\[ \int x^2 \, dx = \frac{x^3}{3} \]\[ \int 2 \, dx = 2x \]Evaluate from \( \sqrt{2} \) to 3: \[ \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{3} = \left( \frac{27}{3} - 6 \right) - \left( \frac{(\sqrt{2})^3}{3} - 2\sqrt{2} \right) \]\[ = 3 - \left( \frac{2\sqrt{2}}{3} - 2\sqrt{2} \right) \]\[ = 3 - \left( -\frac{4\sqrt{2}}{3} \right) \]\[ = 3 + \frac{4\sqrt{2}}{3} \].
07
Add Both Integrals
Add the results of both integrals to find the total area:\[ \frac{4\sqrt{2}}{3} + \left( 3 + \frac{4\sqrt{2}}{3} \right) = 3 + \frac{8\sqrt{2}}{3} \]. Thus, the total area is \( 3 + \frac{8\sqrt{2}}{3} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
A definite integral can be thought of as the area under a curve within a specific range of values. In essence, it sums up tiny, infinite slices between the curve and the x-axis for a specified interval. The definite integral of a function from point \( a \) to \( b \) is denoted by \( \int_{a}^{b} f(x) \, dx \).
- The limits of integration \( a \) and \( b \) indicate the start and end points on the x-axis, framing the segment where you analyze the area.
- When the area lies above the x-axis, the definite integral is positive, whereas it becomes negative if the area is below the x-axis.
- Calculating a definite integral involves the Fundamental Theorem of Calculus, which connects differentiation with integration.
Function Intersections
When seeking the area between a curve and the x-axis, identifying where the function intersects the axis is vital. These intersection points often serve as boundaries for integration.
- To find these intersections, set the function \( y = f(x) \) equal to zero, solving for \( x \).
- The solutions to this equation, which occur when \( y = 0 \), are the x-coordinates where the curve meets the x-axis.
- In the given problem, the equation \( x^2 - 2 = 0 \) provides the intersections at \( x = \sqrt{2} \) and \( x = -\sqrt{2} \). However, focus only on the interval \([0, 3]\), so \( x = \sqrt{2} \) is of interest.
Integral of Absolute Value
The integral of an absolute value function deals with capturing areas without sign-dependent issues. It simplifies the calculation when a function crosses the x-axis within an interval.
- To integrate \( |f(x)| \), break the integral into parts across intervals where the function maintains consistent positivity or negativity.
- For each segment where the function is negative, integrate \(-f(x)\) to account for the absolute value.
- In the exercise, the curve \( y = x^2 - 2 \) becomes negative between \( x = 0 \) and \( x = \sqrt{2} \), and positive from \( \sqrt{2} \) to \( 3 \).
- Thus, the integral splits into \( \int_{0}^{\sqrt{2}} (2 - x^2) \, dx \) and \( \int_{\sqrt{2}}^{3} (x^2 - 2) \, dx \).
