1) Introduction: From Indefinite to Definite 📚
We’ve mastered indefinite integrals, which give us a family of antiderivatives. Now, let’s explore definite integrals, which calculate a specific value—like the area under a curve—using the Fundamental Theorem of Calculus. This theorem links integration and differentiation in a beautiful way!
In this topic, we’ll cover:
- Definite Integrals: What they represent and how to compute them.
- Fundamental Theorem: The key connection between integrals and derivatives.
- Examples: Applying the theorem step-by-step.
Quick Recap: \( \int f(x) \, dx = F(x) + C \) is the indefinite integral.
2) Definite Integrals: Area and Limits 🎓
A definite integral gives a number, not a function. It represents the net area under the curve of \( f(x) \) from a lower limit \( a \) to an upper limit \( b \). The notation is \( \int_a^b f(x) \, dx \).
Definition 11.1: Definite Integral
The definite integral \( \int_a^b f(x) \, dx \) is the net signed area between \( f(x) \) and the x-axis from \( x = a \) to \( x = b \).
If \( f(x) \) is positive, it’s the area above the x-axis. If negative, it subtracts area below. We’ll use antiderivatives to compute it!
Example 1: Simple Definite Integral
Find \( \int_0^2 3x \, dx \).
- Antiderivative: \( \int 3x \, dx = \frac{3x^2}{2} + C \).
- Evaluate: \( \left[ \frac{3x^2}{2} \right]_0^2 = \frac{3(2)^2}{2} - \frac{3(0)^2}{2} = \frac{12}{2} - 0 = 6 \).
Answer: \( \int_0^2 3x \, dx = 6 \).
Example 2: Negative Area
Find \( \int_0^1 (-2x) \, dx \).
- Antiderivative: \( \int -2x \, dx = -x^2 + C \).
- Evaluate: \( \left[ -x^2 \right]_0^1 = -1^2 - 0^2 = -1 \).
Answer: \( \int_0^1 (-2x) \, dx = -1 \).
3) Fundamental Theorem of Calculus 📐
The Fundamental Theorem connects definite integrals to antiderivatives, making computation straightforward.
Theorem 11.1: Fundamental Theorem of Calculus
If \( F(x) \) is an antiderivative of \( f(x) \) (i.e., \( F'(x) = f(x) \)) and \( f \) is continuous on \([a, b]\), then:
\( \int_a^b f(x) \, dx = F(b) - F(a) \)
This says the definite integral is the change in the antiderivative from \( a \) to \( b \).
Example 3: Applying the Theorem
Find \( \int_1^3 x^2 \, dx \).
- Antiderivative: \( \int x^2 \, dx = \frac{x^3}{3} + C \).
- Evaluate: \( \left[ \frac{x^3}{3} \right]_1^3 = \frac{3^3}{3} - \frac{1^3}{3} = \frac{27}{3} - \frac{1}{3} = 9 - \frac{1}{3} = \frac{26}{3} \).
Answer: \( \int_1^3 x^2 \, dx = \frac{26}{3} \).
Example 4: Mixed Areas
Find \( \int_{-1}^2 (x^2 - 1) \, dx \).
- Antiderivative: \( \int (x^2 - 1) \, dx = \frac{x^3}{3} - x + C \).
- Evaluate: \( \left[ \frac{x^3}{3} - x \right]_{-1}^2 = \left( \frac{2^3}{3} - 2 \right) - \left( \frac{(-1)^3}{3} - (-1) \right) = \left( \frac{8}{3} - 2 \right) - \left( -\frac{1}{3} + 1 \right) \).
- Simplify: \( \frac{8}{3} - \frac{6}{3} - \left( -\frac{1}{3} + \frac{3}{3} \right) = \frac{2}{3} - \left( \frac{2}{3} \right) = 0 \).
Answer: \( \int_{-1}^2 (x^2 - 1) \, dx = 0 \) (areas cancel out).
4) Advanced Examples 🔍
Example 5: Trigonometric Function
Find \( \int_0^{\pi/2} \sin(x) \, dx \).
- Antiderivative: \( \int \sin(x) \, dx = -\cos(x) + C \).
- Evaluate: \( \left[ -\cos(x) \right]_0^{\pi/2} = -\cos(\pi/2) - (-\cos(0)) = 0 - (-1) = 1 \).
Answer: \( \int_0^{\pi/2} \sin(x) \, dx = 1 \).
Example 6: Piecewise Behavior
Find \( \int_{-2}^2 |x| \, dx \).
- Split at \( x = 0 \): \( \int_{-2}^0 (-x) \, dx + \int_0^2 x \, dx \).
- First part: \( \int_{-2}^0 (-x) \, dx = \left[ -\frac{x^2}{2} \right]_{-2}^0 = 0 - \left( -\frac{(-2)^2}{2} \right) = 2 \).
- Second part: \( \int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = 2 - 0 = 2 \).
- Total: \( 2 + 2 = 4 \).
Answer: \( \int_{-2}^2 |x| \, dx = 4 \) (total area, ignoring sign).
5) Practice Questions 🎯
Fundamental Practice Questions 🌱
Instructions: Compute the definite integral using the Fundamental Theorem of Calculus. 📚
\( \int_0^1 2x \, dx \)
\( \int_1^3 x^2 \, dx \)
\( \int_0^2 (3x + 1) \, dx \)
\( \int_{-1}^1 x^3 \, dx \)
\( \int_0^{\pi} \sin(x) \, dx \)
\( \int_0^4 \sqrt{x} \, dx \)
\( \int_{-2}^2 (x^2 - 4) \, dx \)
\( \int_0^1 \cos(x) \, dx \)
\( \int_1^2 (2x^2 - x) \, dx \)
\( \int_{-\pi/2}^{\pi/2} 2\cos(x) \, dx \)
\( \int_0^3 (x^3 - 2x) \, dx \)
Challenging Practice Questions 🌟
Instructions: These require deeper analysis or special considerations. 🧠
Find \( \int_0^2 |x - 1| \, dx \) and explain the result geometrically.
Compute \( \int_{-1}^1 (x^4 - x^2) \, dx \) and interpret the sign of the answer.
Evaluate \( \int_0^{\pi/2} (2\sin(x) - \cos(x)) \, dx \) and verify.
Determine \( \int_{-3}^3 \sqrt{9 - x^2} \, dx \) and relate it to a geometric shape.
Find \( \int_0^1 x e^{x^2} \, dx \) using substitution and the Fundamental Theorem.
6) Summary & Cheat Sheet 📋
6.1) Definite Integral
\( \int_a^b f(x) \, dx \) is the net area from \( a \) to \( b \).
6.2) Fundamental Theorem
\( \int_a^b f(x) \, dx = F(b) - F(a) \), where \( F'(x) = f(x) \).
6.3) Tip
Split integrals at points where \( f(x) \) changes sign for absolute value functions.
You’ve connected integrals to areas! Next, we’ll calculate specific areas. 🎉