Problem 32 - Applied Calculus

Question

Find the area between the graph of \(y=x^{2}-2\) and the \(x\) -axis, between \(x=0\) and \(x=3.\)

Step-by-Step Solution

Answer

The area is \( 3 + \frac{8\sqrt{2}}{3} \).

Step 1: 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 \( x = 3 \).

Step 2: 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 \([0, 3]\).

Step 3: 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: \( \int_{0}^{3} |x^2 - 2| \, dx \).

Step 4: 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: \( \int_{0}^{\sqrt{2}} (2 - x^2) \, dx + \int_{\sqrt{2}}^{3} (x^2 - 2) \, dx \).

Step 5: 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 \( \sqrt{2} \): \[ \left[ 2x - \frac{x^3}{3} \right]_{0}^{\sqrt{2}} = \left( 2\sqrt{2} - \frac{(\sqrt{2})^3}{3} \right) - (0) \]\[ = 2\sqrt{2} - \frac{2\sqrt{2}}{3} = \frac{4\sqrt{2}}{3} \].

Step 6: 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: \[ \left[ \frac{x^3}{3} - 2x \right]_{\sqrt{2}}^{3} = \left( \frac{27}{3} - 6 \right) - \left( \frac{(\sqrt{2})^3}{3} - 2\sqrt{2} \right) \]\[ = 3 - \left( \frac{2\sqrt{2}}{3} - 2\sqrt{2} \right) \]\[ = 3 - \left( -\frac{4\sqrt{2}}{3} \right) \]\[ = 3 + \frac{4\sqrt{2}}{3} \].

Step 7: 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 \( 3 + \frac{8\sqrt{2}}{3} \).

Key Concepts

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.

Ultimately, the definite integral gives a precise measurement of the net area between the curve and the x-axis. This is crucial in solving problems related to areas between curves, which can involve intersecting functions or multiple curve segments.

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.

These intersection points matter because they determine where the function changes its position relative to the x-axis, essential for correctly setting up integrals for accurate area calculation between the curve and the axis.

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 \( \int_{\sqrt{2}}^{3} (x^2 - 2) \, dx \).

By working with absolute values in this way, we ensure correct area measurements, fully accounting for the regions below the x-axis where the function is negative. This approach avoids negative areas that could lead to calculation errors.