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Magazine Chapter 4: Problem 3
Find the area under the given curve over the indicated interval. $$ y=2 x ; \quad[1,3] $$
Short Answer
Expert verified
The area is 8.
Step by step solution
01
Identify the curve and interval
The curve given is a linear function: \( y = 2x \). We need to find the area under this curve from \( x = 1 \) to \( x = 3 \).
02
Set up the integral
The area under the curve can be found using a definite integral. Set up the integral of the function \( y = 2x \) over the interval from \( x = 1 \) to \( x = 3 \): \[ A = \int_{1}^{3} 2x \, dx \]
03
Compute the antiderivative
Find the antiderivative of the function \( 2x \). The antiderivative of \( 2x \) is \( x^2 + C \), where \( C \) is the constant of integration. For definite integrals, the constant \( C \) can be ignored.
04
Evaluate the definite integral
Evaluate the integral by finding the difference of the antiderivative evaluated at the upper and lower bounds:\[A = \left[ x^2 \right]_{1}^{3} = (3^2) - (1^2)\]
05
Calculate the numeric solution
Compute \( 3^2 = 9 \) and \( 1^2 = 1 \). Therefore, the area is:\[A = 9 - 1 = 8\]
06
State the final answer
The area under the curve \( y = 2x \) from \( x = 1 \) to \( x = 3 \) is \( 8 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Antiderivative
In calculus, the concept of an antiderivative is crucial. An antiderivative of a function is another function whose derivative is the original function.
For example, if you have a function given by \( f(x) = 2x \), its antiderivative, denoted as \( F(x) \), would be a function such that when you differentiate \( F(x) \), you get \( f(x) \) back.
For the function \( 2x \), its antiderivative is \( x^2 + C \), where \( C \) is the constant of integration. The presence of \( C \) indicates that there are infinitely many antiderivatives for any given function, differing by a constant. This is because adding a constant does not affect the derivative, as constant terms vanish during differentiation.
When calculating definite integrals, this constant of integration \( C \) is not of concern, as it cancels out when considering the difference at the bounds of the integration.
For example, if you have a function given by \( f(x) = 2x \), its antiderivative, denoted as \( F(x) \), would be a function such that when you differentiate \( F(x) \), you get \( f(x) \) back.
For the function \( 2x \), its antiderivative is \( x^2 + C \), where \( C \) is the constant of integration. The presence of \( C \) indicates that there are infinitely many antiderivatives for any given function, differing by a constant. This is because adding a constant does not affect the derivative, as constant terms vanish during differentiation.
When calculating definite integrals, this constant of integration \( C \) is not of concern, as it cancels out when considering the difference at the bounds of the integration.
Area under a curve
The area under a curve in a plane for a function, especially in the context of a graph, offers insights into total accumulation or other practical applications.
The most common way to find the area under a curve involves using the definite integral.
In the mathematical exercise at hand, the focus is on finding the area under the equation \( y = 2x \) between \( x = 1 \) and \( x = 3 \). Setting up a definite integral, \( \int_{1}^{3} 2x \, dx \), allows us to calculate the total area.
By evaluating the antiderivative \( x^2 \) at these boundary points, we realize this area represents the physical space between the curve and the x-axis over the interval.
Finding this area is especially useful in physics and engineering for calculating things like work done by a force.
Being able to visualize this area aids tremendously in understanding the significance of integrals in calculus.
The most common way to find the area under a curve involves using the definite integral.
In the mathematical exercise at hand, the focus is on finding the area under the equation \( y = 2x \) between \( x = 1 \) and \( x = 3 \). Setting up a definite integral, \( \int_{1}^{3} 2x \, dx \), allows us to calculate the total area.
By evaluating the antiderivative \( x^2 \) at these boundary points, we realize this area represents the physical space between the curve and the x-axis over the interval.
Finding this area is especially useful in physics and engineering for calculating things like work done by a force.
Being able to visualize this area aids tremendously in understanding the significance of integrals in calculus.
Linear function integration
Understanding linear function integration is fundamental for mastering calculus concepts. A linear function, like \( y = 2x \), creates a straight line on a graph, characterized by its constant rate of change.
To find the area under a linear curve, we use integration. For the function \( y = 2x \), the integration transforms the rate of change into an overall accumulation, which is the area.
The process starts by identifying the function and interval: \( \int_{1}^{3} 2x \, dx \). The antiderivative is then found, which is \( x^2 \) in this instance.
Evaluating the antiderivative at \( x = 3 \) and \( x = 1 \), and finding their difference \( 9 - 1 \), provides the total area covered.
Linear integration is quite straightforward compared to more complex functions, as linear functions inherently have simpler antiderivatives. This examination underscores how integration helps translate a graph's linear slope into a calculable quantity, aiding understanding and application of calculus concepts.
To find the area under a linear curve, we use integration. For the function \( y = 2x \), the integration transforms the rate of change into an overall accumulation, which is the area.
The process starts by identifying the function and interval: \( \int_{1}^{3} 2x \, dx \). The antiderivative is then found, which is \( x^2 \) in this instance.
Evaluating the antiderivative at \( x = 3 \) and \( x = 1 \), and finding their difference \( 9 - 1 \), provides the total area covered.
Linear integration is quite straightforward compared to more complex functions, as linear functions inherently have simpler antiderivatives. This examination underscores how integration helps translate a graph's linear slope into a calculable quantity, aiding understanding and application of calculus concepts.
