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Magazine Chapter 5: Problem 32
Find the area under the graph of \(f\) corresponding to the interval \(\mid\) a) [1,3 and (b) \([a, b]\). \(f(x)=x^{3}+2\)
Short Answer
Expert verified
For [1,3], area = 21.75. For [a,b], area = \(\frac{b^4}{4} + 2b - \frac{a^4}{4} - 2a\).
Step by step solution
01
Identify the function and interval
The function given is \( f(x) = x^3 + 2 \). We are asked to find the area under the curve over two intervals: (a) for \([1, 3]\) and (b) for a general interval \([a, b]\).
02
Find the antiderivative of \(f(x)\)
To find the area under the curve of \(f(x)\), we need to find its antiderivative. The antiderivative of \(f(x) = x^3 + 2\) is \( F(x) = \int (x^3 + 2) \, dx = \frac{x^4}{4} + 2x + C \), where \(C\) is the constant of integration.
03
Apply the Fundamental Theorem of Calculus for interval [1,3]
To find the area from \([1,3]\), plug these values into the antiderivative:\[A = F(3) - F(1) = \left(\frac{3^4}{4} + 2\times3\right) - \left(\frac{1^4}{4} + 2\times1\right)\]Calculate each term to get:\[A = \left(\frac{81}{4} + 6\right) - \left(\frac{1}{4} + 2\right) = \frac{81}{4} + \frac{24}{4} - \frac{1}{4} - \frac{8}{4}\]\[A = \frac{96}{4} - \frac{9}{4} = \frac{87}{4} = 21.75\]
04
Generalize the solution for interval [a,b]
For a general interval \([a, b]\), use the antiderivative:\[A = F(b) - F(a) = \left(\frac{b^4}{4} + 2b\right) - \left(\frac{a^4}{4} + 2a\right)\]Simplify the expression:\[A = \frac{b^4}{4} + 2b - \frac{a^4}{4} - 2a\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Integration
Integration is like assembling the pieces of a puzzle to find the total area under a curve. It is used in calculus to find quantities such as areas, volumes, and much more.
In this context, we're interested in determining the area under the curve of a given function over a specific interval. Integration serves as the mathematical tool to achieve just that.
This process involves summing up infinitely small slices of area, each represented as a tiny rectangular strip, to obtain the total area. This is what enables us to find exact solutions to problems involving continuous change rather than just approximations.
In this context, we're interested in determining the area under the curve of a given function over a specific interval. Integration serves as the mathematical tool to achieve just that.
This process involves summing up infinitely small slices of area, each represented as a tiny rectangular strip, to obtain the total area. This is what enables us to find exact solutions to problems involving continuous change rather than just approximations.
- Key idea: Summing up smaller parts to find a total measure
- Utilizes the antiderivative to calculate the exact area
- Geometrically represents applications in physics and engineering
Antiderivative
An antiderivative is a function that reverses the process of differentiation, helping us undo a derivative. For the function in our exercise, the antiderivative is crucial in computing the area underneath the curve.
We are looking at the function: \(f(x) = x^3 + 2\). Its antiderivative translates the problem of finding an area under the curve into a straightforward calculation using its primitive form, \(F(x) = \frac{x^4}{4} + 2x + C\).
Here, especially note the appearance of \(C\), a constant of integration. While it forms part of the general antiderive solution, in the context of definite integrals, it usually cancels out.
We are looking at the function: \(f(x) = x^3 + 2\). Its antiderivative translates the problem of finding an area under the curve into a straightforward calculation using its primitive form, \(F(x) = \frac{x^4}{4} + 2x + C\).
Here, especially note the appearance of \(C\), a constant of integration. While it forms part of the general antiderive solution, in the context of definite integrals, it usually cancels out.
- Key Tool: Provides the function for calculating areas precisely
- Reverses differentiation
- Includes an arbitrary constant \(C\) representing infinite potential shifts vertically
Fundamental Theorem of Calculus
The Fundamental Theorem of Calculus is a bridge connecting differentiation and integration. It allows us to evaluate a definite integral by using an antiderivative.
In our step-by-step solution, it is used to find the area specifically for the interval [1,3]. This theorem states, in simple terms, that if you have a continuous function \(f(x)\), then the integral of \(f\) from \(a\) to \(b\) can be computed using an antiderivative \(F(x)\): \[A = F(b) - F(a)\]
This powerful theorem simplifies the problem significantly by reducing the task to simple subtraction.
In our step-by-step solution, it is used to find the area specifically for the interval [1,3]. This theorem states, in simple terms, that if you have a continuous function \(f(x)\), then the integral of \(f\) from \(a\) to \(b\) can be computed using an antiderivative \(F(x)\): \[A = F(b) - F(a)\]
This powerful theorem simplifies the problem significantly by reducing the task to simple subtraction.
- Central Idea: Connects derivatives and integrals
- Allows quick evaluation of complex definite integrals
- Reduces finding areas to evaluating endpoints on the antiderivative
Definite Integral
A definite integral helps us calculate the real, tangible area under a curve between two specified points. This differs from indefinite integrals, which represent a family of functions.
To compute the area under \(f(x) = x^3 + 2\) from 1 to 3, we used a definite integral. Through the steps, we took our antiderivative and evaluated it at the endpoints a and b:\[A = F(b) - F(a)\]. This gives us the exact numerical area trapped between the curve, x-axis, and the lines x = 1 and x = 3.
Definite integrals have various applications beyond area calculation, including in probability and estimates of physical quantities.
To compute the area under \(f(x) = x^3 + 2\) from 1 to 3, we used a definite integral. Through the steps, we took our antiderivative and evaluated it at the endpoints a and b:\[A = F(b) - F(a)\]. This gives us the exact numerical area trapped between the curve, x-axis, and the lines x = 1 and x = 3.
Definite integrals have various applications beyond area calculation, including in probability and estimates of physical quantities.
- Practical Use: Calculates areas, masses, volumes, and more
- Specific to given limits: Represents actual quantities
- Reflection of real-world applications in science and engineering
