Problem 32 - Applied Calculus

Question

Use an integral to find the specified area. Above the curve \(y=x^{4}-8\) and below the \(x\) -axis.

Step-by-Step Solution

Answer

The area is \(\frac{96\sqrt{2}}{5}\).

Step 1: Determine the Points of Intersection

To find the area above the curve \(y = x^4 - 8\) and below the \(x\)-axis, we need to find where the curve intersects the \(x\)-axis, i.e., where \(y = 0\). Set the equation equal to zero: \(x^4 - 8 = 0\). Solve for \(x\): \(x^4 = 8\), so \(x = \pm \sqrt[4]{8}\), which simplifies to \(x = \pm 2\sqrt{2}\). Thus, the points of intersection are \((-2\sqrt{2}, 0)\) and \((2\sqrt{2}, 0)\).

Step 2: Set Up the Integral for the Area

The area between the curve \(y = x^4 - 8\) and the \(x\)-axis from \(-2\sqrt{2}\) to \(2\sqrt{2}\) can be found by taking the integral of \(-f(x)\) which is \(-(x^4 - 8)\) or \(8 - x^4\) over the interval, because the curve is below the \(x\)-axis. The integral expression is: \[ \int_{-2\sqrt{2}}^{2\sqrt{2}} (8 - x^4) \, dx \].

Step 3: Calculate the Definite Integral

Calculate the integral: \[ \int (8 - x^4) \, dx = \int 8 \, dx - \int x^4 \, dx = 8x - \frac{x^5}{5} + C \]. Evaluate this from \(-2\sqrt{2}\) to \(2\sqrt{2}\). First, at \(x = 2\sqrt{2}\): \(8(2\sqrt{2}) - \frac{(2\sqrt{2})^5}{5}\). Second, at \(x = -2\sqrt{2}\): \(8(-2\sqrt{2}) - \frac{(-2\sqrt{2})^5}{5}\).

Step 4: Simplify and Combine Results

Simplifying the expressions: \(8(2\sqrt{2}) = 16\sqrt{2}\) and \(8(-2\sqrt{2}) = -16\sqrt{2}\). Calculate \((2\sqrt{2})^5 = 32\sqrt{2}\), hence \(-\frac{32\sqrt{2}}{5}\). Combine values: \[ (16\sqrt{2} - \frac{32\sqrt{2}}{5}) - (-16\sqrt{2} - \frac{32\sqrt{2}}{5}) = 32\sqrt{2} - \frac{64\sqrt{2}}{5} \] which simplifies to \(\frac{96\sqrt{2}}{5}\) as the final value, giving the area.

Key Concepts

Definite Integrals

In calculus, a definite integral is a crucial concept that aids in calculating the "net" area under a curve. It represents the accumulation of quantities, where the starting and ending points are specified on the graph. In our problem, the definite integral is used to find the area between the curve represented by the function \( y = x^4 - 8 \) and the x-axis.

To solve this, we integrate the function over a specified interval. This interval is determined by the points where the curve intersects the x-axis, providing natural boundaries for our calculation. The solution involves calculating:

• The integral \( \int (8 - x^4) \, dx \), where \( 8 - x^4 \) is the adjusted form of the function due to its position below the x-axis.
• Evaluating this integral gives us the actual numerical area, which in this case is calculated from \(-2\sqrt{2}\) to \(2\sqrt{2}\).
• The definite integral then amounts to \( \frac{96\sqrt{2}}{5} \).

Mastering definite integrals allows students to tackle real-world problems involving accumulated quantities like area and volume.

Area under a curve

The area under a curve in the context of calculus can be interpreted as the total space between a curve and a given axis. It can describe quantities in physics, engineering, and probability.

For this particular exercise, the curve defined by \( y = x^4 - 8 \) dips below the x-axis, indicating that we are finding the area above the curve and below the x-axis.

• Typically, if a function is entirely above the axis, the integral itself provides the area.
• If the curve dips below the x-axis, like in our case, adjustments are made by taking the negative of the function, leading to \( 8 - x^4 \).

This adjustment ensures that our result reflects the true magnitude of the area, avoiding negative area values in our calculations. Calculating the area under curves is essential in numerous scientific calculations that involve summing over continuously varying quantities.

Points of Intersection

Points of intersection between a function and the axes, specifically here with the x-axis, are vital as they define the bounds for integrating to find an area under a curve.

To identify these intersection points for \( y = x^4 - 8 \), we solved the equation \( x^4 - 8 = 0 \). Solving gives us:

• \( x^4 = 8 \)
• Thus, \( x = \pm \sqrt[4]{8} = \pm 2\sqrt{2} \)
• These results in points \((-2\sqrt{2}, 0)\) and \((2\sqrt{2}, 0)\).

These points help in setting up a definite integral. They're crucial because each intersection point marks the boundary of an area where the curve changes its behavior in relation to the axis. Finding these intersections accurately allows for the precise application of integration techniques, ensuring that no parts of the curve are overlooked in calculations.