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91Ó°ÊÓ

Use any method (including geometry) to find the area of the following regions. In each case, sketch the bounding curves and the region in question. The region in the first quadrant bounded by \(y=x^{2 / 3}\) and \(y=4\)

Short Answer

Expert verified
The area of the region is \(\frac{64}{5}\) square units.

Step by step solution

01

Sketch the Bounding Curves

Start by sketching the two curves in the first quadrant. The graph of \(y = x^{2/3}\) is the curve where the power is a fractional exponent, and \(y=4\) is a horizontal line.
02

Determine the Points of Intersection

To find the points of intersection, set the two equations equal to each other: \(x^{2/3} = 4\) Raise both sides to the power of \(3/2\) to solve for x: \(x = (4)^{3/2} = 8\) Now that we have the x-coordinate of the intersection, we can find the y-coordinate by plugging x back into either equation. Since \(y = 4\) is simpler, we'll use that: \(y = 4\) So, the point of intersection is \((8, 4)\).
03

Set Up the Integral Expression

We will use integration to find the area between the curves. Since the area of the region is in the first quadrant and both functions are continuous and non-negative over the interval, we can use the following integral expression: Area = \(\int_{a}^{b} [(y_{2} - y_{1}) dx]\) where \(a\) and \(b\) are the x-coordinates of the points of intersection, and \(y_{1}\) and \(y_{2}\) are the functions of the bounding curves. In this case, \(a = 0\), \(b = 8\), \(y_{1} = x^{2/3}\), and \(y_{2} = 4\). Thus, we have: Area = \(\int_{0}^{8} [(4 - x^{2/3}) dx]\)
04

Evaluate the Integral

Now, we will evaluate the integral to find the area: Area = \(\int_{0}^{8} [(4 - x^{2/3}) dx] = 4\int_{0}^{8} [1 - \frac{1}{4}x^{2/3}] dx\) Using the power rule and the linearity property, we can break up the integral into individual parts: Area = \(4[x - \frac{1}{4} \cdot \frac{3}{5}x^{5/3} ] \Big|_{0}^{8} = 4[(8 - \frac{1}{4} \cdot \frac{3}{5} (8)^{5/3} ) - (0 - \frac{1}{4} \cdot \frac{3}{5} (0)^{5/3})]\) After evaluating the expression, we get: Area = \(4(8 - \frac{3}{5} (8) ) = 4(8 - \frac{24}{5}) = 4(\frac{16}{5}) = \frac{64}{5}\) So, the area of the region bounded by the curves \(y = x^{2/3}\) and \(y=4\) in the first quadrant is \(\frac{64}{5}\) square units.

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus, often introduced as the reverse process of differentiation. Think of it as a way to calculate the total accumulation of quantities, such as area under a curve. The integration process is akin to piecing together infinitely small pieces (infinitesimals) to find a whole - a continuous sum, if you will. It's crucial for calculating areas, as well as other concepts such as volume, displacement, and work.

For instance, the area under a curve might represent the distance traveled by a car over time, if the curve described its speed. To visualize this, imagine drawing rectangles under the curve; integration lets you sum the area of these rectangles, even as they become infinitesimally thin, thus giving the exact area under the curve.
Definite Integrals
Definite integrals are a specific type of integration where the start and end points are defined, allowing for the calculation of an exact value. They are notated with the integral symbol, a particular function, and limits of integration. The limits, often referred to as the bounds of integration, are the x-values that 'bookend' the area we're interested in.

In our exercise, these bounds are from the x-value at which the area starts, to the x-value where it ends. When you evaluate a definite integral, you're finding the net area between the curve and the x-axis, within these bounds. This 'net' concept is important when the curve dips below the x-axis, which can result in negative values. However, in the given problem, both functions are non-negative in the interval considered, so the integral gives the actual physical area.
Finding Points of Intersection
Finding points of intersection between curves is like solving a puzzle where two patterns overlap. It involves equating the functions representing the curves and solving for the variable, typically x or y. This process identifies the exact points where the curves cross each other.

The challenge here often lies in solving equations that can be quite complex, involving algebraic manipulation, exponent rules, or even special factorizations. Once these solutions are found, they determine the bounds for definite integrals (like in our exercise example), mark the endpoints of the region whose area we want to calculate, or simply let us understand the relationship between two geometric shapes.
Sketching Curves
Sketching curves is an essential tool for visualizing problems in calculus. A good sketch helps us understand the shape of the curve, its behavior, and its relationship to other curves or axes. In other words, it gives us a map of where we're going to work.

Start with basic functions you know (lines, parabolas, etc.) and seek out key features such as intercepts, asymptotes, maxima, and minima. In our exercise, for instance, sketching allows us to see at a glance where the function meets the horizontal line y=4, and how the area whose size we're calculating is shaped. Good sketching, combined with a clear understanding of the underlying functions, makes it easier to set up and solve calculus problems, such as finding areas between curves.

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Most popular questions from this chapter

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