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7.5. Solved Examples
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Today, we'll explore the concept of calculating the area under a curve. Who can remind us of the definition of the area under the curve?
It's the integral of the function from one limit to another, right?
Exactly! So, if we have a function y = f(x), the area from x = a to x = b is given by the integral from a to b of f(x) dx. Let's try a quick example: How would we find the area under the curve y = x² from x = 0 to x = 2?
We integrate x² from 0 to 2!
Correct! The integral gives us [x³/3] evaluated from 0 to 2. Can anyone calculate that?
That would be (2³/3) - (0³/3) = 8/3.
Well done! So, the area under the curve is 8/3. Remember, this is a positive area since the curve is above the x-axis.
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Now let's move on to finding the area between two curves. If we have the curves y = x and y = x², can someone tell me how we would set up the integral to find the area between them from x = 0 to x = 1?
We need to integrate the difference, right? So, it would be ∫ from 0 to 1 of (x - x²) dx.
Excellent! And can anyone calculate that for us?
The integral evaluates to [x²/2 - x³/3] from 0 to 1, which gives (1/2 - 1/3) = 1/6.
That's right! The area between these two curves is 1/6. This method of integration allows us to find areas even when curves intersect each other.
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Create a free accountTo wrap up, let’s summarize what we learned. Can anyone explain why we sometimes take the absolute value when calculating areas?
Because if the function is below the x-axis, the integral gives a negative value.
Exactly! We consider the absolute value to get the actual area. Now, for some practice, could someone explain how we might approach finding the area bounded by y = √x, the x-axis, and the vertical lines x = 1 and x = 4?
We would integrate √x from 1 to 4, right?
Yes! That will give us the bounded area. Good job, everyone! Remember to always sketch your curves and identify the upper and lower functions.
Overview
Short Summary
This section provides worked solutions for problems related to areas under curves and between curves, reinforcing integral applications in geometry.
Medium Summary
The 'Solved Examples' section includes detailed solutions to problems demonstrating how to find areas under and between curves using integrals. By working through these examples, students can understand the practical applications of calculus in geometric contexts.
Detailed Summary
Solved Examples
In this section, we focus on two important aspects of integration: finding the area under a curve and finding the area between two curves.
Example 1: Area Under a Curve
We calculate the area under the curve defined by the function y = x² from x = 0 to x = 2. By applying the definite integral formula, we find:
Area = ∫ from 0 to 2 (x²) dx = [x³/3] from 0 to 2 = (8/3) - (0) = 8/3.
This calculation shows us the geometric interpretation of integration as the accumulation of area.
Example 2: Area Between Two Curves
Next, we find the area between the curves y = x and y = x² over the interval [0, 1]. Here, we first determine that y = x is the upper curve and y = x² the lower curve.
Using the integral of the difference between these two functions, we compute: Area = ∫ from 0 to 1 (x - x²) dx = [x²/2 - x³/3] from 0 to 1 = (1/2 - 1/3) = 1/6.
These worked examples exemplify the application of integrals to solve geometric problems efficiently and accurately.
Audio Book
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Create a free accountExample 1: Area under a curve Find the area under the curve 𝑦 = 𝑥² from 𝑥 = 0 to 𝑥 = 2. Solution: Area = ∫ 𝑥² 𝑑𝑥 from 0 to 2 = [𝑥³/3] from 0 to 2 = (2³/3) - (0³/3) = 8/3.
Detailed Explanation
In this example, we are tasked with finding the area underneath the curve defined by the equation y = x². The process involves calculating the definite integral of the function from x = 0 to x = 2. Here's how it works:
- Set up the integral: We write the integral symbol, followed by the function and 'dx' to indicate that we are integrating with respect to x.
- Calculate the integral: The indefinite integral of x² is (x³)/3.
- Evaluate the definite integral: Substitute the upper limit (2) and lower limit (0) into the integral. This gives us (2³/3) - (0³/3) = 8/3. Thus, the area under the curve from x = 0 to x = 2 is 8/3 square units.
Examples & Analogies
Imagine you're filling a swimming pool where the bottom is shaped like the graph of y = x². To understand how much water you need for the section of the pool from 0 to 2 meters, you can think of calculating how much space that section occupies beneath the curve, which is precisely what we did with the integral!
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Create a free accountExample 2: Area between two curves Find the area between the curves 𝑦 = 𝑥 and 𝑦 = 𝑥² from 𝑥 = 0 to 𝑥 = 1. Solution: Here, 𝑦 = 𝑥 is above 𝑦 = 𝑥² on [0, 1]. Area = ∫ (𝑥 − 𝑥²) 𝑑𝑥 from 0 to 1 = [𝑥²/2 − 𝑥³/3] from 0 to 1 = (1/2 - 1/3) = 1/6.
Detailed Explanation
In this example, we find the area between two functions, y = x and y = x² over the interval from 0 to 1. The first step is to identify which curve is on top between 0 and 1, which is y = x in this case. We then calculate the integral of the difference of the upper function and the lower function:
- Set up the integral: The area can be computed using the integral of (upper function - lower function), which is (x - x²).
- Calculate the integral: The antiderivatives of x and x² are (x²/2) and (x³/3), respectively.
- Evaluate the definite integral: We substitute the limits into the antiderivative function: (1/2 - 1/3) = 1/6. Therefore, the area between the two curves over the interval [0, 1] is 1/6 square units.
Examples & Analogies
Think of two rivers where one is wider (y = x) and the other is narrower (y = x²) between two points (0 and 1 km). The area between them represents the land that separates these rivers. Finding the area using integration gives you the exact measurement of that land.
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Key Concepts
Core takeaways and short definitions to help you quickly recall the key ideas from this section.
Definite Integral: The integration of functions over specific limits to calculate the area.
Area Under a Curve: Represents the accumulation of area beneath the curve and above the x-axis.
Area Between Two Curves: The integral of the difference between two functions over specific limits.
Examples
Step-by-step examples to apply the section's ideas and test your understanding.
To find the area under y = x² from x = 0 to x = 2, we compute the integral ∫ from 0 to 2 (x²) dx = 8/3.
To find the area between y = x and y = x² from x = 0 to x = 1, we calculate the integral ∫ from 0 to 1 (x - x²) dx = 1/6.
Memory Aids
Interactive tools to help you remember key concepts
Stories
Flash Cards
Glossary
Definite Integral
An integral that computes the accumulation of quantities, such as area, over a specific interval.
Area Under a Curve
The region enclosed between a function and the x-axis over a specified interval.
Area Between Two Curves
The area enclosed by two functions where one is above the other over a specified interval.