Chapter 26: Problem 25
Find the areas bounded by the indicated curves. $$y=\frac{1}{2} x^{5}, x=-1, x=2, y=0$$
Short Answer
Expert verified
The area bounded by the curves is \( \frac{21}{4} \).
Step by step solution
01
Understand the Problem
We are tasked with finding the area between the curve defined by the function \( y = \frac{1}{2}x^5 \), the vertical lines \( x = -1 \) and \( x = 2 \), and the horizontal line \( y = 0 \) (the x-axis).
02
Set Up the Integral
Since we want the area between the curve \( y = \frac{1}{2}x^5 \) and the x-axis from \( x = -1 \) to \( x = 2 \), we can set up the integral as follows:\[\int_{-1}^{2} \frac{1}{2}x^5 \, dx\]
03
Integrate the Function
To find the integral of \( \frac{1}{2}x^5 \), use the power rule for integration:Any integral of the form \( \int x^n \, dx \) is \( \frac{x^{n+1}}{n+1} \). In our case,\[\int \frac{1}{2}x^5 \, dx = \frac{1}{2} \cdot \frac{x^6}{6} = \frac{x^6}{12}\]
04
Evaluate the Integral
Substitute the limits of integration \( x = -1 \) and \( x = 2 \) into the antiderivative, \( \frac{x^6}{12} \):\[\left[ \frac{x^6}{12} \right]_{-1}^{2} = \frac{2^6}{12} - \frac{(-1)^6}{12}\]Calculate the values:\( \frac{2^6}{12} = \frac{64}{12} \) and \( \frac{1}{12} \).The area is:\[\frac{64}{12} - \frac{1}{12} = \frac{63}{12} = \frac{21}{4}\]
05
Interpret the Result
Since the integral calculated the "net" area, and the curve remains above the x-axis from \( x = -1 \) to \( x = 2 \), the result \( \frac{21}{4} \) is the total area between the curve and the x-axis over the given interval.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Definite Integral
The concept of a definite integral is central to finding the area under a curve between two specified points on the x-axis. It effectively gives us the total area enclosed by the curve and the x-axis within the bounds we choose. The definite integral is expressed as
The bounds, \( a \) and \( b \), define where our calculation begins and ends. In the exercise, these bounds are \( x = -1 \) and \( x = 2 \). The goal is to evaluate the entire expression from these points, which delivers a numerical result that represents the total enclosed area. As long as the curve of our function does not dip below the x-axis within the bounds (causing negative values), the integral in this context fully reflects the true physical area.
- \( \int_{a}^{b} f(x) \, dx \)
- where \( a \) and \( b \) are the limits of integration, and \( f(x) \) is the function that defines the curve.
The bounds, \( a \) and \( b \), define where our calculation begins and ends. In the exercise, these bounds are \( x = -1 \) and \( x = 2 \). The goal is to evaluate the entire expression from these points, which delivers a numerical result that represents the total enclosed area. As long as the curve of our function does not dip below the x-axis within the bounds (causing negative values), the integral in this context fully reflects the true physical area.
Area under Curve
When we discuss the "area under a curve," we're talking about the total space between the curve and the x-axis. This can be visualized as the shaded area in a graph where the bound lines slice across the curve. Integrals are the mathematical tools we use for calculating such areas.
Let's break it down:
The calculated result of the integral, \( \frac{21}{4} \), reveals the exact area beneath our curve of \( y = \frac{1}{2}x^5 \) from \( x = -1 \) to \( x = 2 \), in this context perfectly capturing the enclosed space on the coordinate plane.
Let's break it down:
- Suppose you have a curve given by the function \( y = \frac{1}{2}x^5 \). To find the area between this curve and the x-axis, we calculate its integral.
- The exercise shows that the integral \[ \int_{-1}^{2} \frac{1}{2}x^5 \, dx \]captures the total area of interest.
The calculated result of the integral, \( \frac{21}{4} \), reveals the exact area beneath our curve of \( y = \frac{1}{2}x^5 \) from \( x = -1 \) to \( x = 2 \), in this context perfectly capturing the enclosed space on the coordinate plane.
Power Rule for Integration
The power rule for integration is a powerful tool that simplifies the process of integrating functions that are polynomials. It is a specific rule within calculus that transforms a function for easy integration. Here's how it works:
The process simplifies polynomials and allows for immediate identification of antiderivatives, thereby streamlining the task of finding areas or solving differential equations.
- When you have a function of the form \( \int x^n \, dx \), the integral is calculated as:\[ \frac{x^{n+1}}{n+1} + C \]where \( n \) is not equal to \(-1\).
- The exercise applies this rule to integrate \( \frac{1}{2}x^5 \). We treat \( \frac{1}{2} \) as a constant factor and focus on the polynomial \( x^5 \).
- Using the power rule, \[ \int \frac{1}{2}x^5 \, dx = \frac{1}{2} \cdot \frac{x^{6}}{6} = \frac{x^6}{12}. \]
The process simplifies polynomials and allows for immediate identification of antiderivatives, thereby streamlining the task of finding areas or solving differential equations.