Definite Integration and Its Applications
Hey everyone! Welcome to the exciting world of Definite Integration. If you've ever wondered how we can find the exact area of weird, curvy shapes, you've come to the right place. Don't worry if "integration" sounds intimidating. We're going to break it down step-by-step.
In this chapter, we'll learn a powerful technique to calculate exact areas, see how it's magically connected to differentiation, and use it to solve some interesting real-world problems. Think of it as upgrading your maths toolkit with a super-precise measuring tool!
1. What is Definite Integration? The Big Idea
The Area Problem
Finding the area of a square (length × width) or a triangle (½ × base × height) is easy. But what about the area under a curve, like the one shown below? There's no simple formula for that!
Imagine you want to find the area of land under a winding river path on a map. How would you do it?
Slicing It Up: The Rectangle Idea
The core idea is simple: let's approximate the area using something we know – rectangles! We can slice the area under the curve into lots of thin vertical rectangles and add up their areas.
Think about slicing a loaf of bread. One slice is a simple shape. All the slices together make up the whole loaf.
The more rectangles we use, and the thinner we make them, the closer our approximation gets to the true area. Definite integration is the concept of using an infinite number of infinitely thin rectangles to get the exact area.
The Notation for Definite Integration
This whole process is represented by a special notation. Don't let it scare you; let's break it down:
$$ \int_{a}^{b} f(x) \,dx $$- The Integral Sign ($$ \int $$): This is just a stretched-out 'S', which stands for 'Sum'. It tells us we're adding things up (the areas of all those tiny rectangles).
- The Limits of Integration (a and b): These numbers tell us where to start (a, the lower limit) and where to stop (b, the upper limit) on the x-axis.
- The Function (f(x)): This is the curve itself. It represents the height of each of our tiny rectangles.
- dx: This represents the tiny, tiny width of each rectangle.
So, the entire expression $$ \int_{a}^{b} f(x) \,dx $$ means: "Sum up the areas of infinite rectangles of height f(x) and tiny width dx, from x=a to x=b."
Key Takeaway: A definite integral $$ \int_{a}^{b} f(x) \,dx $$ represents the exact area between the curve y = f(x) and the x-axis, from x = a to x = b.
2. The Magic Tool: The Fundamental Theorem of Calculus (FTC)
Adding up an infinite number of rectangles sounds impossible, right? Luckily, mathematicians discovered a breathtakingly simple shortcut. This is the Fundamental Theorem of Calculus (FTC), and it's the most important idea in this chapter!
The Theorem Itself
The FTC connects integration with differentiation. It tells us that to find the definite integral, we just need to:
1. Find the antiderivative (the indefinite integral) of f(x). Let's call it F(x).
2. Plug in the upper and lower limits and subtract!
In mathematical terms:
$$ \int_{a}^{b} f(x) \,dx = F(b) - F(a) $$where F(x) is the antiderivative of f(x) (meaning F'(x) = f(x)).
We often write this using the notation:
$$ \int_{a}^{b} f(x) \,dx = [F(x)]_{a}^{b} = F(b) - F(a) $$Step-by-Step Guide to Using the FTC
Let's find the value of $$ \int_{1}^{3} 2x \,dx $$.
- Find the antiderivative (F(x)): The indefinite integral of 2x is x² + C. For definite integrals, we can ignore the '+ C' because it will always cancel out! ((b² + C) - (a² + C) = b² - a²). So, our F(x) is just x².
- Plug in the upper limit (b=3): F(3) = 3² = 9.
- Plug in the lower limit (a=1): F(1) = 1² = 1.
- Subtract: F(b) - F(a) = 9 - 1 = 8.
That's it! The exact area under the line y=2x from x=1 to x=3 is 8.
Common Mistakes to Avoid
A very common mistake is mixing up the order of subtraction. Always remember it's UPPER limit result minus LOWER limit result. F(b) - F(a), never the other way around!
Key Takeaway: The FTC gives us a simple, powerful way to calculate definite integrals without summing up rectangles. Just find the antiderivative, plug in the limits, and subtract.
3. Properties of Definite Integrals
These are the 'rules of the game' that help simplify problems. Think of them as helpful shortcuts.
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1. Flipping the Limits
$$ \int_{b}^{a} f(x) \,dx = - \int_{a}^{b} f(x) \,dx $$If you integrate 'backwards' (from a larger number to a smaller one), the answer is the negative of integrating 'forwards'.
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2. Zero Width Interval
$$ \int_{a}^{a} f(x) \,dx = 0 $$The area from a point to itself is zero. There's no width!
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3. Splitting the Interval
$$ \int_{a}^{c} f(x) \,dx = \int_{a}^{b} f(x) \,dx + \int_{b}^{c} f(x) \,dx $$The total area from 'a' to 'c' is the same as the area from 'a' to 'b' PLUS the area from 'b' to 'c'. You can split an integral into parts.
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4. Constant Multiples
$$ \int_{a}^{b} kf(x) \,dx = k \int_{a}^{b} f(x) \,dx $$You can pull a constant number out in front of the integral sign, just like with differentiation and indefinite integration.
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5. Sum and Difference
$$ \int_{a}^{b} [f(x) \pm g(x)] \,dx = \int_{a}^{b} f(x) \,dx \pm \int_{a}^{b} g(x) \,dx $$You can integrate a function term by term. This is super useful for polynomials!
Did you know? The "Dummy Variable"
The variable we use for integration doesn't actually matter. 'x' is just a placeholder. The following integrals are all exactly the same!
$$ \int_{1}^{2} x^2 \,dx = \int_{1}^{2} t^2 \,dt = \int_{1}^{2} u^2 \,du $$This is because the final answer will be a number, and the variable will disappear after we substitute the limits.
Key Takeaway: These properties work just like they do for indefinite integrals. Use them to break down complicated-looking integrals into simpler pieces.
4. Finding Definite Integrals in Practice
Integrals of Algebraic and Exponential Functions
Let's use the FTC and our properties with the functions you know.
Example: Find $$ \int_{0}^{1} (3x^2 + e^x) \,dx $$
Step 1: Find the antiderivative.
Using the sum rule, we integrate term by term.
Antiderivative of 3x² is x³.
Antiderivative of eˣ is eˣ.
So, F(x) = x³ + eˣ.
Step 2: Apply the limits using the $$ [F(x)]_{a}^{b} $$ notation.
$$ [x^3 + e^x]_{0}^{1} $$
Step 3: Calculate F(b) - F(a).
F(1) = (1³ + e¹) = 1 + e
F(0) = (0³ + e⁰) = 0 + 1 = 1
So, the answer is (1 + e) - (1) = e.
Definite Integration by Substitution
When using substitution, we have a crucial extra step: we must deal with the limits of integration. There are two ways to do this, but the first method is much faster and less prone to errors!
Example: Find $$ \int_{0}^{2} 2x(x^2 + 1)^3 \,dx $$
Method 1 (Recommended): Change the Limits
- Choose 'u' and find 'du'.
Let $$ u = x^2 + 1 $$.
Then $$ \frac{du}{dx} = 2x $$, so $$ du = 2x \,dx $$. This matches our integral perfectly! - CHANGE THE LIMITS. This is the most important step!
Our old limits were for 'x'. We need new limits for 'u'.
Lower limit: when x = 0, u = (0)² + 1 = 1.
Upper limit: when x = 2, u = (2)² + 1 = 5. - Rewrite the integral entirely in terms of 'u' and the new limits.
$$ \int_{1}^{5} u^3 \,du $$ - Integrate and solve.
$$ [\frac{u^4}{4}]_{1}^{5} = \frac{5^4}{4} - \frac{1^4}{4} = \frac{625}{4} - \frac{1}{4} = \frac{624}{4} = 156 $$
Notice we never had to go back to 'x'!
Common Mistakes to Avoid
The #1 mistake with substitution is forgetting to change the limits and using the original 'x' limits with the 'u' integral. Always change the limits!
Key Takeaway: When using substitution for definite integrals, change the limits from 'x' values to 'u' values. This saves you from having to substitute back at the end.
5. The Main Application: Finding Area
Area Above the x-axis
If the function f(x) is positive (i.e., the curve is above the x-axis) between 'a' and 'b', then the area is simply:
$$ Area = \int_{a}^{b} f(x) \,dx $$Area Below the x-axis
If the function f(x) is negative (below the x-axis), the definite integral will give a negative value. Since area cannot be negative, we must take the absolute value or put a minus sign in front.
$$ Area = - \int_{a}^{b} f(x) \,dx \quad \text{or} \quad Area = \left| \int_{a}^{b} f(x) \,dx \right| $$When the Curve Crosses the x-axis
What if the curve is partly above and partly below the x-axis in our interval? If we just integrate from start to finish, the negative area will cancel out some of the positive area, giving a wrong result for the total area.
Procedure:
- Find where the curve crosses the x-axis by setting f(x) = 0 and solving for x.
- Split the integral at each crossing point.
- Calculate the integral for each part separately.
- Take the absolute value of any negative results.
- Add the positive values together to get the total area.
Example: Find the area bounded by the curve y = x³ from x = -1 to x = 2.
The curve crosses the x-axis at x=0. So we must split the integral.
Area = (Area from -1 to 0) + (Area from 0 to 2)
Area = $$ \left| \int_{-1}^{0} x^3 \,dx \right| + \int_{0}^{2} x^3 \,dx $$
The first part: $$ \int_{-1}^{0} x^3 \,dx = [\frac{x^4}{4}]_{-1}^{0} = \frac{0}{4} - \frac{(-1)^4}{4} = -\frac{1}{4} $$. So the area is $$ |-\frac{1}{4}| = \frac{1}{4} $$.
The second part: $$ \int_{0}^{2} x^3 \,dx = [\frac{x^4}{4}]_{0}^{2} = \frac{2^4}{4} - \frac{0}{4} = \frac{16}{4} = 4 $$.
Total Area = $$ \frac{1}{4} + 4 = 4.25 $$
A Very Important Syllabus Note!
For your HKDSE M1 exam, you are ONLY required to find the area between a curve and the x-axis. You are NOT required to find:
- The area between a curve and the y-axis.
- The area between two different curves.
So, you can focus all your practice on the methods described above!
Key Takeaway: To find the total area, always check if the curve crosses the x-axis in your interval. If it does, split the integral and make sure all parts of the area are counted as positive.
6. Chapter Summary & Quick Review
You've made it! Let's recap the most important points.
- What it is: Definite integration $$ \int_{a}^{b} f(x) \,dx $$ calculates the exact area between f(x) and the x-axis from x=a to x=b.
- The Magic Tool (FTC): The best way to solve it is $$ \int_{a}^{b} f(x) \,dx = [F(x)]_{a}^{b} = F(b) - F(a) $$, where F(x) is the antiderivative.
- Areas Below x-axis: The integral is negative, so the area is $$ -\int ... $$.
- Crossing the x-axis: Find where f(x)=0, split the integral, and add the absolute values of each part.
- Substitution: The golden rule is CHANGE THE LIMITS to be in terms of 'u'.
Definite integration is a cornerstone of calculus. It's a challenging but very rewarding topic. Practice these steps, remember the key ideas, and you'll master it in no time. Keep up the great work!