∫ Definite Integral
Area calculations and the Newton–Leibniz formula
🎯 What Is a Definite Integral?
A definite integral is an integral with limits – a lower limit and an upper limit.
The result is a number (not a function!) that represents the area under the graph.
\(\int_a^b f(x) \, dx\)
\(a\) = lower limit | \(b\) = upper limit
⚖️ Definite vs Indefinite Integral
| Indefinite integral | Definite integral |
|---|---|
| \(\int f(x) \, dx\) | \(\int_a^b f(x) \, dx\) |
| Result: function + C | Result: a number |
| \(\int 2x \, dx = x^2 + C\) | \(\int_1^3 2x \, dx = 8\) |
⭐ Newton–Leibniz Formula (Fundamental Theorem)
\(\int_a^b f(x) \, dx = F(b) - F(a)\)
where \(F(x)\) is an antiderivative of \(f(x)\)
💡 In steps:
- Find the antiderivative \(F(x)\)
- Substitute the upper limit: \(F(b)\)
- Substitute the lower limit: \(F(a)\)
- Subtract: \(F(b) - F(a)\)
📝 Shorthand notation:
\(\int_a^b f(x) \, dx = \Big[ F(x) \Big]_a^b = F(b) - F(a)\)
✏️ Example 1: Basic Calculation
Evaluate: \(\int_1^4 2x \, dx\)
Solution:
Step 1: Find the antiderivative
\(F(x) = x^2\)
Step 2: Substitute the limits
\(\Big[ x^2 \Big]_1^4 = F(4) - F(1)\)
\(= 4^2 - 1^2\)
\(= 16 - 1 = 15\)
Answer: 15
🔍 Note: no +C in the definite integral! (the constant cancels in the subtraction)
📐 Geometric Interpretation: Area
The definite integral \(\int_a^b f(x) \, dx\) represents the area between:
- The graph of \(f(x)\)
- The x-axis
- The lines \(x = a\) and \(x = b\)
⚠️ Area Above and Below the x-Axis
Important:
- Area above the x-axis → contributes a positive value
- Area below the x-axis → contributes a negative value
🔴 Therefore, if you want the true (always positive) area:
\(\text{area} = \int_a^b |f(x)| \, dx\)
or calculate each segment separately and sum the absolute values
✏️ Example 2: Area with a Negative Part
Find the area between \(f(x) = x^2 - 4\) and the x-axis on \([-2, 3]\)
Solution:
Step 1: Find intersection points with the x-axis
\(x^2 - 4 = 0\)
\(x = \pm 2\)
Step 2: Check the sign on each segment
On \([-2, 2]\): the function is negative (below the axis)
On \([2, 3]\): the function is positive (above the axis)
Step 3: Calculate each part separately
Part 1 (negative – take absolute value):
\(\left| \int_{-2}^{2} (x^2-4) \, dx \right| = \left| \Big[ \frac{x^3}{3} - 4x \Big]_{-2}^{2} \right|\)
\(= \left| \left(\frac{8}{3} - 8\right) - \left(\frac{-8}{3} + 8\right) \right|\)
\(= \left| -\frac{16}{3} - \frac{16}{3} \right| = \left| -\frac{32}{3} \right| = \frac{32}{3}\)
Part 2 (positive):
\(\int_{2}^{3} (x^2-4) \, dx = \Big[ \frac{x^3}{3} - 4x \Big]_{2}^{3}\)
\(= \left(9 - 12\right) - \left(\frac{8}{3} - 8\right)\)
\(= -3 + \frac{16}{3} = \frac{7}{3}\)
Step 4: Sum the parts
\(\text{area} = \frac{32}{3} + \frac{7}{3} = \frac{39}{3} = 13\)
Answer: 13
📊 Area Between Two Functions
\(\text{area} = \int_a^b |f(x) - g(x)| \, dx\)
💡 In practice:
\(\text{area} = \int_a^b (\text{upper function} - \text{lower function}) \, dx\)
✏️ Example 3: Area Between Two Functions
Find the area between \(f(x) = x^2\) and \(g(x) = x\)
Solution:
Step 1: Find intersection points
\(x^2 = x\)
\(x^2 - x = 0\)
\(x(x-1) = 0\)
\(x = 0\) or \(x = 1\)
Step 2: Identify which is on top
On \([0, 1]\): check at \(x = 0.5\)
\(f(0.5) = 0.25\), \(g(0.5) = 0.5\)
Therefore \(g(x) = x\) is on top
Step 3: Calculate
\(\text{area} = \int_0^1 (x - x^2) \, dx\)
\(= \Big[ \frac{x^2}{2} - \frac{x^3}{3} \Big]_0^1\)
\(= \left( \frac{1}{2} - \frac{1}{3} \right) - (0)\)
\(= \frac{3-2}{6} = \frac{1}{6}\)
Answer: \(\frac{1}{6}\)
📏 Properties of the Definite Integral
1. Swapping limits:
\(\int_a^b f(x) \, dx = -\int_b^a f(x) \, dx\)
2. Equal limits:
\(\int_a^a f(x) \, dx = 0\)
3. Combining intervals (additivity):
\(\int_a^b f(x) \, dx + \int_b^c f(x) \, dx = \int_a^c f(x) \, dx\)
4. Linearity:
\(\int_a^b [f(x) + g(x)] \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx\)
\(\int_a^b k \cdot f(x) \, dx = k \cdot \int_a^b f(x) \, dx\)
❌ Common Mistakes
❌ Mistake 1: Forgetting that area can be negative
✓ Correct: Check where the function is positive / negative
❌ Mistake 2: Substituting limits in the wrong order
✓ Correct: always \(F(b) - F(a)\) (upper minus lower)
❌ Mistake 3: Writing +C in a definite integral
✓ Correct: no need for C in a definite integral
❌ Mistake 4: Forgetting to check which function is on top when finding area between functions
✓ Correct: always upper minus lower
💡 Tips for the Exam
1️⃣ Sketch!
Always sketch the function to understand the area
2️⃣ Intersection points
Find where the function crosses the x-axis or the second function
3️⃣ Split into segments
If there is a sign change – split the integral
4️⃣ Verify result
Area must be positive! If it comes out negative – something is wrong
📝 Summary
\(\int_a^b f(x) \, dx = F(b) - F(a)\)
Area below the x-axis → negative
Area between functions: upper − lower
The result is a number (not a function)