Chapter 5: Problem 27
The area \(A(x)\) under the graph of \(f\) and over the interval \([a, x]\) is given. Find the function \(f\) and the value of \(a\). $$A(x)=x^{2}-4$$
Short Answer
Expert verified
Function \( f(x) = 2x \); \( a = \pm 2 \).
Step by step solution
01
Understand the Problem
We need to find the function \( f \) such that the area under the curve of \( f \) from \( a \) to \( x \) is \( A(x) = x^2 - 4 \). Additionally, we have to determine the value of \( a \).
02
Differentiate the Given Area Function
Differentiate \( A(x) = x^2 - 4 \) with respect to \( x \) to find \( f(x) \). The derivative \( A'(x) = \frac{d}{dx}(x^2 - 4) = 2x \). Therefore, \( f(x) = 2x \).
03
Interpret the Fundamental Theorem of Calculus
According to the Fundamental Theorem of Calculus, if \( A(x) \) represents the area from \( a \) to \( x \) under the function \( f(x) \), then \( A(x) = \int_{a}^{x} f(t) \, dt \) and \( f(x) = A'(x) \). Since we found \( f(x) = 2x \), this supports our differentiation.
04
Determine the Constant of Integration and \( a \)
The given area function is \( A(x) = \int_{a}^{x} 2t \, dt = x^2 - a^2 \). We compare \( x^2 - a^2 \) to the given \( x^2 - 4 \). Therefore, \( a^2 = 4 \), leading to \( a = \pm 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Area under a curve
The concept of the area under a curve is crucial in calculus and often represents accumulated quantities, such as distance or total cost, depending on the context. For a given function \( f(x) \), the area under the curve from point \( a \) to point \( x \) on the x-axis, can be expressed as the integral \( \int_{a}^{x} f(t) \, dt \). This integral calculates the total area bounded by the function \( f(x) \), the x-axis, and vertical lines at \( x = a \) and \( x = x \).
Understanding the area under a curve enables us to interpret it as the net sum of infinitesimally small areas \((f(t) \, dt)\) from \( a \) to \( x \). It transforms a seemingly complex summation into a mathematically elegant solution using integration.
Understanding the area under a curve enables us to interpret it as the net sum of infinitesimally small areas \((f(t) \, dt)\) from \( a \) to \( x \). It transforms a seemingly complex summation into a mathematically elegant solution using integration.
- Visualizing this area can help in comprehending physical situations, like finding the total distance traveled when speed varies over time.
- The area is positive if the curve lies above the x-axis and negative if it lies below, providing the concept of net area.
Definite integral
The definite integral is a specific type of integral with defined upper and lower limits, often represented as \( \int_{a}^{b} f(x) \, dx \). It calculates the "exact" area under the curve of \( f(x) \) between the two limits \( a \) and \( b \).
The key use of a definite integral is to compute the accumulated value of a changing quantity, which can be seen in the solution of the given problem. Here, the definite integral from \( a \) to \( x \) of \( f(t)\) is linked to the area function \( A(x) \), specifically indicating the function's behavior and total change over the interval.
The key use of a definite integral is to compute the accumulated value of a changing quantity, which can be seen in the solution of the given problem. Here, the definite integral from \( a \) to \( x \) of \( f(t)\) is linked to the area function \( A(x) \), specifically indicating the function's behavior and total change over the interval.
- The process of evaluating \( \int_{a}^{b} f(x) \, dx \) involves finding an antiderivative of \( f(x) \) and using the difference \( F(b) - F(a) \), where \( F \) is an antiderivative.
- This numerical result helps understand the initial problem, translating it to a physical meaning like total accumulation or displacement.
Differentiation
Differentiation is a foundational concept in calculus involving finding the derivative of a function, which represents its rate of change. In the context of the Fundamental Theorem of Calculus, differentiation is used to "invert" integration.
In the exercise, differentiating the area function \( A(x) = x^2 - 4 \) with respect to \( x \) helps us find \( f(x) \). We get \( A'(x) = 2x \), indicating that \( f(x) \) is the derivative of the area function, thus the original function representing the curve.
By understanding differentiation, we grasp how a function changes at any x-value, providing insights into slopes and instantaneous rates of change.
In the exercise, differentiating the area function \( A(x) = x^2 - 4 \) with respect to \( x \) helps us find \( f(x) \). We get \( A'(x) = 2x \), indicating that \( f(x) \) is the derivative of the area function, thus the original function representing the curve.
By understanding differentiation, we grasp how a function changes at any x-value, providing insights into slopes and instantaneous rates of change.
- Differentiation transforms a problem of summation across intervals into finding particular values, giving precise information about dynamics.
- In calculus, it pairs with integration to form a powerful toolkit, establishing relationships between curves and their tangents.