1) Introduction: Beyond Net Area 📚
Definite integrals gave us net area, but what if we want the actual area under a curve or between two curves, even when they cross the x-axis? This topic shows how to use integrals to calculate these areas, opening the door to real-world applications like calculating land areas or fluid volumes!
We’ll explore:
- Area Under a Curve: Using definite integrals for positive regions.
- Area Between Curves: Handling two functions with overlaps.
- Examples: Step-by-step area calculations.
Quick Recap: \( \int_a^b f(x) \, dx = F(b) - F(a) \) gives net area.
2) Area Under a Curve 🎓
The area under a curve from \( a \) to \( b \) is the definite integral of \( f(x) \) when \( f(x) \geq 0 \). If \( f(x) \) dips below the x-axis, we take the absolute value or split the integral.
Definition 12.1: Area Under a Curve
The area under \( f(x) \) from \( a \) to \( b \) (where \( f(x) \geq 0 \)) is \( \int_a^b f(x) \, dx \). For negative regions, use \( \int_a^b |f(x)| \, dx \).
Example 1: Simple Area
Find the area under \( f(x) = x \) from 0 to 2.
- Since \( x \geq 0 \), \( \int_0^2 x \, dx \).
- Antiderivative: \( \frac{x^2}{2} \).
- Evaluate: \( \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - 0 = 2 \).
Answer: Area = 2 square units.
Example 2: With Negative Part
Find the area under \( f(x) = x^2 - 1 \) from -1 to 2.
- Split at \( x = 0 \) (where \( f(x) = -1 < 0 \)): \( \int_{-1}^0 (1 - x^2) \, dx + \int_0^2 (x^2 - 1) \, dx \).
- First part: \( \left[ x - \frac{x^3}{3} \right]_{-1}^0 = 0 - \left( -1 + \frac{1}{3} \right) = \frac{2}{3} \).
- Second part: \( \left[ \frac{x^3}{3} - x \right]_0^2 = \left( \frac{8}{3} - 2 \right) - 0 = \frac{2}{3} \).
- Total: \( \frac{2}{3} + \frac{2}{3} = \frac{4}{3} \).
Answer: Area = \( \frac{4}{3} \) square units.
3) Area Between Two Curves 📐
To find the area between two curves \( y = f(x) \) and \( y = g(x) \) from \( a \) to \( b \), integrate the difference \( |f(x) - g(x)| \) over the interval, adjusting for which curve is on top.
Definition 12.2: Area Between Curves
The area between \( f(x) \) and \( g(x) \) from \( a \) to \( b \) is \( \int_a^b |f(x) - g(x)| \, dx \), or \( \int_a^b [f(x) - g(x)] \, dx \) if \( f(x) \geq g(x) \).
Example 3: Basic Case
Find the area between \( y = x \) and \( y = x^2 \) from 0 to 1.
- Check: \( x \geq x^2 \) for \( x \in [0, 1] \).
- Integrate: \( \int_0^1 (x - x^2) \, dx \).
- Antiderivative: \( \frac{x^2}{2} - \frac{x^3}{3} \).
- Evaluate: \( \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \left( \frac{1}{2} - \frac{1}{3} \right) - 0 = \frac{1}{6} \).
Answer: Area = \( \frac{1}{6} \) square units.
Example 4: Crossing Curves
Find the area between \( y = x^2 \) and \( y = x \) from -1 to 2.
- Find intersection: \( x^2 = x \) gives \( x = 0, 1 \).
- Split: \( \int_{-1}^0 (x - x^2) \, dx + \int_0^1 (x - x^2) \, dx + \int_1^2 (x^2 - x) \, dx \).
- First part: \( \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^0 = 0 - \left( \frac{1}{2} - \frac{1}{3} \right) = -\frac{1}{6} \).
- Second part: \( \left[ \frac{x^2}{2} - \frac{x^3}{3} \right]_0^1 = \frac{1}{6} \).
- Third part: \( \left[ \frac{x^3}{3} - \frac{x^2}{2} \right]_1^2 = \left( \frac{8}{3} - 2 \right) - \left( \frac{1}{3} - \frac{1}{2} \right) = \frac{7}{6} \).
- Total: \( \frac{1}{6} + \frac{1}{6} + \frac{7}{6} = \frac{9}{6} = 1.5 \).
Answer: Area = 1.5 square units.
4) Advanced Examples 🔍
Example 5: Trigonometric Curves
Find the area between \( y = \sin(x) \) and \( y = \cos(x) \) from 0 to \( \pi/4 \).
- Check: \( \sin(x) \leq \cos(x) \) from 0 to \( \pi/4 \).
- Integrate: \( \int_0^{\pi/4} (\cos(x) - \sin(x)) \, dx \).
- Antiderivative: \( \sin(x) + \cos(x) \).
- Evaluate: \( \left[ \sin(x) + \cos(x) \right]_0^{\pi/4} = (\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}) - (0 + 1) = \sqrt{2} - 1 \).
Answer: Area = \( \sqrt{2} - 1 \) square units.
Example 6: Multiple Intersections
Find the area between \( y = x^2 - 4 \) and \( y = 2x - 1 \) from -2 to 3.
- Intersections: \( x^2 - 4 = 2x - 1 \) gives \( x = 1, 3 \).
- Split: \( \int_{-2}^1 (2x - 1 - (x^2 - 4)) \, dx + \int_1^3 ((x^2 - 4) - (2x - 1)) \, dx \).
- First part: \( \int_{-2}^1 (2x - 1 - x^2 + 4) \, dx = \int_{-2}^1 (-x^2 + 2x + 3) \, dx \).
- Antiderivative: \( -\frac{x^3}{3} + x^2 + 3x \).
- Evaluate: \( \left[ -\frac{x^3}{3} + x^2 + 3x \right]_{-2}^1 = (-\frac{1}{3} + 1 + 3) - (-\frac{-8}{3} + 4 - 6) = \frac{10}{3} \).
- Second part: \( \int_1^3 (x^2 - 2x - 3) \, dx \).
- Evaluate: \( \left[ \frac{x^3}{3} - x^2 - 3x \right]_1^3 = (9 - 9 - 9) - (\frac{1}{3} - 1 - 3) = -9 + \frac{10}{3} = -\frac{17}{3} \).
- Total: \( \frac{10}{3} - \frac{17}{3} = -2 \) (adjust for absolute area if needed).
Answer: Area = 7 square units (absolute value adjustment).
5) Practice Questions 🎯
Fundamental Practice Questions 🌱
Instructions: Compute the area under the curve or between the given curves. 📚
\( \int_0^2 x \, dx \)
\( \int_0^1 x^2 \, dx \)
\( \int_1^3 (x - 1) \, dx \)
\( \int_{-1}^2 (2 - x) \, dx \)
\( \int_0^{\pi/2} \cos(x) \, dx \)
\( \int_0^4 \sqrt{x} \, dx \)
\( \int_{-1}^1 |x| \, dx \)
\( \int_0^2 (x^2 + 1) \, dx \)
\( \int_0^1 (2x - x^2) \, dx \)
\( \int_{-\pi}^{\pi} \sin(x) \, dx \)
\( \int_1^2 (x^3 - x) \, dx \)
Challenging Practice Questions 🌟
Instructions: These require advanced analysis or multiple steps. 🧠
Find the area under \( y = x^2 - 4 \) from -2 to 2.
Compute the area between \( y = x^2 \) and \( y = x + 2 \) from -1 to 2.
Determine the area between \( y = \sin(x) \) and \( y = 0 \) from 0 to \( 2\pi \).
Evaluate the area between \( y = x^3 \) and \( y = x \) from -1 to 1.
Find the area between \( y = e^x \) and \( y = e^{-x} \) from 0 to 1.
6) Summary & Cheat Sheet 📋
6.1) Area Under a Curve
\( \int_a^b f(x) \, dx \) for \( f(x) \geq 0 \); use \( |f(x)| \) for total area.
6.2) Area Between Curves
\( \int_a^b [f(x) - g(x)] \, dx \) if \( f(x) \geq g(x) \); split at intersections.
6.3) Tip
Check for sign changes and intersections to split integrals accurately.
You’ve mastered area calculations! Next, we’ll explore more applications. 🎉