Calculating Area Using Integrals
After learning how to integrate algebraic and trigonometric functions, we will now study the applications of integrals. One important application is for calculating the area bounded by several curves.
To understand this material, you need to master:
- Integration techniques
- Graphing functions (especially linear and quadratic)
- Using derivatives to sketch curves (for functions other than linear and quadratic)
Area with Boundaries on the X-Axis
Area bounded by curve \( y = f(x) \), the X-axis, lines \( x = a \) and \( x = b \), with \( f(x) \geq 0 \) on interval \([a, b]\):
Area bounded by curve \( y = g(x) \), the X-axis, lines \( x = c \) and \( x = d \), with \( g(x) \leq 0 \) on interval \([c, d]\):
The negative sign is used because the function values are negative, but area must be positive.
Area bounded by two curves \( y_1 = f(x) \) and \( y_2 = g(x) \) on interval \([a, b]\):
Upper curve minus lower curve.
Example Problems and Solutions
Calculate the area bounded by the curve \( y = 4x - x^2 \), \( x = 1 \), \( x = 3 \), and the X-axis.
Solution:
Area = \( \int_1^3 (4x - x^2) \, dx \)
= \( \left[ 2x^2 - \frac{1}{3}x^3 \right]_1^3 \)
= \( \left( 2(3)^2 - \frac{1}{3}(3)^3 \right) - \left( 2(1)^2 - \frac{1}{3}(1)^3 \right) \)
= \( (18 - 9) - (2 - \frac{1}{3}) \)
= \( 9 - \frac{5}{3} = \frac{22}{3} = 7\frac{1}{3} \)
Answer: Area = \( 7\frac{1}{3} \) square units.
Determine the area bounded by the curve \( y = x^2 - 5x + 4 \), the X-axis, on interval [0, 4].
Solution:
The region consists of two parts: L₁ above the X-axis (0 ≤ x ≤ 1) and L₂ below the X-axis (1 ≤ x ≤ 4).
Total area = \( L_1 + (-L_2) = L_1 - L_2 \)
= \( \int_0^1 (x^2 - 5x + 4) \, dx - \int_1^4 (x^2 - 5x + 4) \, dx \)
= \( \left[ \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x \right]_0^1 - \left[ \frac{1}{3}x^3 - \frac{5}{2}x^2 + 4x \right]_1^4 \)
= \( \left( \frac{1}{3} - \frac{5}{2} + 4 \right) - \left( \frac{64}{3} - 40 + 16 \right) + \left( \frac{1}{3} - \frac{5}{2} + 4 \right) \)
= \( \frac{19}{3} = 6\frac{1}{3} \)
Answer: Area = \( 6\frac{1}{3} \) square units.
Calculate the area bounded by the curves \( y = x^2 - 2x \) and \( y = 6x - x^2 \).
Solution:
1. Find intersection points of the curves:
\( x^2 - 2x = 6x - x^2 \)
\( 2x^2 - 8x = 0 \)
\( 2x(x - 4) = 0 \)
\( x = 0 \) or \( x = 4 \)
2. On interval [0, 4], the upper curve is \( y = 6x - x^2 \) and the lower curve is \( y = x^2 - 2x \).
3. Area = \( \int_0^4 [(6x - x^2) - (x^2 - 2x)] \, dx \)
= \( \int_0^4 (8x - 2x^2) \, dx \)
= \( \left[ 4x^2 - \frac{2}{3}x^3 \right]_0^4 \)
= \( 4(16) - \frac{2}{3}(64) = 64 - \frac{128}{3} = \frac{192 - 128}{3} = \frac{64}{3} = 21\frac{1}{3} \)
Answer: Area = \( 21\frac{1}{3} \) square units.
Calculate the area bounded by the curve \( y = 4 - x^2 \), the Y-axis, and the line \( y = 1 \) in quadrant I.
Solution (using X-axis boundaries):
1. Find intersection points: \( 4 - x^2 = 1 \) → \( x^2 = 3 \) → \( x = \pm \sqrt{3} \)
In quadrant I, \( x = \sqrt{3} \)
2. Area = \( \int_0^{\sqrt{3}} [(4 - x^2) - 1] \, dx \)
= \( \int_0^{\sqrt{3}} (3 - x^2) \, dx \)
= \( \left[ 3x - \frac{1}{3}x^3 \right]_0^{\sqrt{3}} \)
= \( 3\sqrt{3} - \frac{1}{3}(\sqrt{3})^3 = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3} \)
Solution (using Y-axis boundaries):
Transform the function: \( y = 4 - x^2 \) → \( x = \sqrt{4 - y} \)
Area = \( \int_1^4 \sqrt{4 - y} \, dy \)
= \( \left[ -\frac{2}{3}(4 - y)^{3/2} \right]_1^4 \)
= \( 0 - \left( -\frac{2}{3}(3)^{3/2} \right) = \frac{2}{3} \cdot 3\sqrt{3} = 2\sqrt{3} \)
Answer: Area = \( 2\sqrt{3} \) square units.
Area with Boundaries on the Y-Axis
The formulas and calculation methods for area with boundaries on the Y-axis are almost the same as for boundaries on the X-axis. The differences are:
- Functions must be transformed to the form \( x = f(y) \)
- For area between two curves: right curve minus left curve
- Formula: \( \text{Area} = \int_{c}^{d} [f_{\text{right}}(y) - f_{\text{left}}(y)] \, dy \)
Conclusion
1. Every integral calculation requires:
- Each curve's function
- A clearly defined shaded region
- Integration boundaries (X-axis or Y-axis)
2. Choice of integration boundaries (X or Y axis) should be adapted to the problem and given functions.
3. For some types of problems, area can be calculated without sketching the curve first.
• Always sketch the curve and shade the region if possible
• Pay attention to whether the region is above or below the X-axis
• For area between two curves, identify which curve is above and which is below
• For trigonometric functions, consider the given interval carefully