Problem 15
Question
Bounded by the line \(y=2 x-1\), the \(x\) axis, and the lines \(x=1\) and \(x=5\).
Step-by-Step Solution
Verified Answer
The area of the region is 20 square units.
1Step 1: Understand the Region
Identify the region bounded by the given lines. The region is under the line given by the equation \(y = 2x - 1\), above the \(x\)-axis, and between the vertical lines \(x = 1\) and \(x = 5\).
2Step 2: Set Up the Integral
To find the area of the region, set up the integral. The area can be calculated using the definite integral of the function \(y = 2x - 1\) from \(x = 1\) to \(x = 5\). The integral to solve is: \int_{1}^{5} (2x - 1) \, dx.
3Step 3: Integrate the Function
Find the antiderivative of the function \(2x - 1\). The antiderivative is: \( \int (2x - 1) dx = x^2 - x + C \).
4Step 4: Evaluate the Definite Integral
Evaluate the definite integral from \(x = 1\) to \(x = 5\). Substitute the limits into the antiderivative: \[ \left[ x^2 - x \right]_{1}^{5} = (5^2 - 5) - (1^2 - 1) = (25 - 5) - (1 - 1) = 20 - 0 = 20 \].
5Step 5: Conclude the Solution
The calculated value from the definite integral represents the area of the region. The area of the region is 20 square units.
Key Concepts
Area under a curveAntiderivativeBounded region
Area under a curve
The area under a curve is a fundamental concept in calculus. It refers to the region enclosed between the graph of a function and the x-axis. In this exercise, we want to find the area under the line given by the equation \(y = 2x - 1\).
The graph of this function is a straight line. To find the area under this curve between two points, we use definite integrals.
Definite integrals allow us to accurately calculate this enclosed area using calculus.
When we set up our integral, we use the x-axis limits (in this case, from \(x=1\) to \(x=5\)). By integrating the function \(2x - 1\), we sum all the tiny slices of area under the curve within this interval. This process provides us with a precise total area.
The graph of this function is a straight line. To find the area under this curve between two points, we use definite integrals.
Definite integrals allow us to accurately calculate this enclosed area using calculus.
When we set up our integral, we use the x-axis limits (in this case, from \(x=1\) to \(x=5\)). By integrating the function \(2x - 1\), we sum all the tiny slices of area under the curve within this interval. This process provides us with a precise total area.
Antiderivative
An antiderivative is the reverse process of differentiation. It is the function whose derivative is the given function.
For example, if we have a function \(f(x) = 2x - 1\), its antiderivative is the function whose derivative gives us \(2x - 1\) again.
The antiderivative of a function \(f(x)\) is usually represented as \(F(x) + C\), where \(C\) is the constant of integration.
In this exercise, the antiderivative of \(2x - 1\) is calculated as:
\[ \int (2x - 1) \, dx = x^2 - x + C \] When we evaluate a definite integral, the constant \(C\) cancels out, leaving us with a concrete area value after evaluating the boundaries.
For example, if we have a function \(f(x) = 2x - 1\), its antiderivative is the function whose derivative gives us \(2x - 1\) again.
The antiderivative of a function \(f(x)\) is usually represented as \(F(x) + C\), where \(C\) is the constant of integration.
In this exercise, the antiderivative of \(2x - 1\) is calculated as:
\[ \int (2x - 1) \, dx = x^2 - x + C \] When we evaluate a definite integral, the constant \(C\) cancels out, leaving us with a concrete area value after evaluating the boundaries.
Bounded region
The bounded region refers to the specific area enclosed by the function and additional boundary lines. In this example, we have the line \(y = 2x - 1\), the x-axis, and the vertical lines at \(x = 1\) and \(x = 5\).
To visualize it, imagine a line that cuts through the plane and creates a trapezoid-like shape between these points. Determining the bounded region is crucial as it clearly defines where we will calculate the area.
Understanding the limits and the shape of the region helps in correctly setting up the definite integral. This ensures that our area calculation stays accurate and consistent with the geometric boundaries set by the problem.
To visualize it, imagine a line that cuts through the plane and creates a trapezoid-like shape between these points. Determining the bounded region is crucial as it clearly defines where we will calculate the area.
Understanding the limits and the shape of the region helps in correctly setting up the definite integral. This ensures that our area calculation stays accurate and consistent with the geometric boundaries set by the problem.
Other exercises in this chapter
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