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Magazine Chapter 5: Problem 27
Use an integral to find the specified area. Above the curve \(y=x^{4}-8\) and below the \(x\)-axis.
Short Answer
Expert verified
The area is \( 16 \cdot 2^{3/4} - \frac{2 \cdot (2^{3/2} \cdot 2^{1/4})}{5} \).
Step by step solution
01
Identify the Relevant Region
We need to find the area above the curve \( y = x^4 - 8 \) and below the \( x \)-axis. This means we are looking for the interval where the curve is below the \( x \)-axis.
02
Solve for the Boundary Points
Set \( y = x^4 - 8 = 0 \) to find the \( x \)-coordinates where the curve intersects the \( x \)-axis.\[x^4 = 8 \x = \pm \sqrt[4]{8} \]The points where the curve intersects the \( x \)-axis are \( x = \pm \sqrt[4]{8} \). Since the curve opens upwards, it is below the \( x \)-axis between these points.
03
Set Up the Integral
We will integrate \( |y| = |x^4 - 8| \) over the interval from \( x = -\sqrt[4]{8} \) to \( x = \sqrt[4]{8} \). Since \( y \leq 0 \) within this interval, the expression \( |x^4 - 8| \) simplifies to \( 8 - x^4 \). The integral becomes:\[\int_{-\sqrt[4]{8}}^{\sqrt[4]{8}} (8 - x^4) \, dx\]
04
Calculate the Integral
First, find the antiderivative of \( 8 - x^4 \):\[\int (8 - x^4)\, dx = 8x - \frac{x^5}{5} + C\]Evaluate this from \( x = -\sqrt[4]{8} \) to \( x = \sqrt[4]{8} \):\[\left[ 8x - \frac{x^5}{5} \right]_{-\sqrt[4]{8}}^{\sqrt[4]{8}}\]
05
Evaluate the Definite Integral
Substitute \( x = \sqrt[4]{8} \) and \( x = -\sqrt[4]{8} \) into the antiderivative:\[\left(8(\sqrt[4]{8}) - \frac{(\sqrt[4]{8})^5}{5}\right) - \left(8(-\sqrt[4]{8}) - \frac{(-\sqrt[4]{8})^5}{5}\right)\]Calculate:\[= \left(8 \cdot 8^{1/4} - \frac{8^{5/4}}{5}\right) - \left(-8 \cdot 8^{1/4} + \frac{-8^{5/4}}{5}\right)\]This simplifies to:\[16 \cdot 8^{1/4} - \frac{2 \cdot 8^{5/4}}{5}\]
06
Simplify the Expression
Calculate \( 8^{1/4} = \sqrt[4]{8} = \sqrt{2} \cdot \sqrt[4]{2} = 2^{3/4} \), and substitute back:\[16 \cdot 2^{3/4} - \frac{2 \cdot (2^{3/2} \cdot 2^{1/4})}{5}\]Calculate each term separately and obtain the final value.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
area between curves
The concept of finding the area between curves is a fascinating aspect of integral calculus. It helps determine the space enclosed between a curve and a reference line, often the x-axis. In our scenario, we are interested in the area above the curve given by the equation \( y = x^4 - 8 \) and below the x-axis. This means the curve dips below the x-axis, and we're looking at that "negative" area.
- First, identify the region of interest between the curves. This requires knowing where the curve intersects the x-axis. In our case, this intersection occurs where \( x^4 = 8 \), leading to \( x = \pm \sqrt[4]{8} \).
- The region between these points where the curve is below the x-axis is what we're analyzing for area calculation.
- Set up the integral of the absolute value of the curve function on this interval. Here, it's \( \int_{-\sqrt[4]{8}}^{\sqrt[4]{8}} (8 - x^4) \, dx \), noting that the curve's expression is negated because it is below the x-axis.
definite integral
A definite integral is a fundamental tool used in calculus to calculate the accumulation of a quantity, such as area, between certain bounds. The main feature of a definite integral is that it provides a numerical result, unlike the indefinite integral, which includes a constant.
- In our exercise, the definite integral \( \int_{-\sqrt[4]{8}}^{\sqrt[4]{8}} (8 - x^4) \, dx \) is used to find the area between the curve \( y = x^4 - 8 \) and the x-axis over an interval.
- This integral gives us the accumulated area while taking into account that the curve dips below the x-axis, hence the use of absolute value functions.
- The bounds \( -\sqrt[4]{8} \) to \( \sqrt[4]{8} \) specify the segment where the area under consideration lies.
antiderivative
The antiderivative, also known as the primitive function, reverses the process of differentiation. It's a crucial step in solving definite integrals, as it helps in finding the accumulation over an interval.
- We calculate the antiderivative of the function \( 8 - x^4 \) in our exercise. This function's antiderivative can be expressed as \( 8x - \frac{x^5}{5} + C \), where \( C \) is the constant of integration, which isn't necessary for definite integrals.
- The next step is to substitute the upper and lower bounds of our interval into this antiderivative expression.
- By evaluating \( \left[ 8x - \frac{x^5}{5} \right]_{-\sqrt[4]{8}}^{\sqrt[4]{8}} \), we can find the accumulated value of the definite integral.
calculus
Calculus serves as the backbone of mathematics, especially in its application to real-world problems. It encompasses tools and techniques such as derivatives and integrals which help us solve complex problems.
- Integral calculus, specifically, involves concepts like finding areas or volumes, which are otherwise difficult to compute using basic geometric formulas.
- In our exercise, calculus allows us to transition from understanding the shape and behavior of the function \( y = x^4 - 8 \) to calculating the precise area it encompasses.
- This is achieved through the formulation of integrals, solving through antiderivatives, and evaluating them over definite boundaries.
