AP Calculus BC 6.6 Applying Properties of Definite Integrals Study Notes - New Syllabus
AP Calculus BC 6.6 Applying Properties of Definite Integrals Study Notes- New syllabus
AP Calculus BC 6.6 Applying Properties of Definite Integrals Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Recognizing opportunities to apply knowledge of geometry and mathematical rules can simplify integration.
Key Concepts:
- Applying Properties of Definite Integrals
Applying Properties of Definite Integrals
Applying Properties of Definite Integrals
Definite integrals have several algebraic and geometric properties that can be used to simplify and compute integrals without evaluating them directly. These properties are often useful in understanding symmetry, interval changes, or breaking down complex integrals.
Properties of Definite Integrals:
| Property | Explanation |
|---|---|
| \(\displaystyle \int_a^a f(x) \, dx = 0\) | Integral over an interval of zero length is zero |
| \(\displaystyle \int_a^b f(x) \, dx = -\displaystyle \int_b^a f(x) \, dx\) | Reversing limits changes the sign of the integral |
| \(\displaystyle \int_a^b [f(x) + g(x)] \, dx = \displaystyle \int_a^b f(x)\, dx + \displaystyle \int_a^b g(x)\, dx\) | Integral of a sum is the sum of the integrals |
| \(\displaystyle \int_a^b c \cdot f(x) \, dx = c \cdot \displaystyle \int_a^b f(x) \, dx\) | Constants can be factored out of the integral |
| \(\displaystyle \int_a^b f(x) \, dx = \displaystyle \int_a^c f(x) \, dx + \displaystyle \int_c^b f(x) \, dx\) | Integrals can be split at a point in the interval |
Note on Symmetry:
- Even functions: \( f(-x) = f(x) \), so \( \displaystyle \int_{-a}^a f(x)\,dx = 2\displaystyle \int_0^a f(x)\,dx \)
- Odd functions: \( f(-x) = -f(x) \), so \( \displaystyle \int_{-a}^a f(x)\,dx = 0 \)
Example:
Given \( \displaystyle \int_1^5 f(x) \, dx = 7 \), find:
a. \( \displaystyle \int_5^1 f(x) \, dx \) b. \( \displaystyle \int_1^5 3f(x) \, dx \)
▶️Answer/Explanation
a. By reversing limits: \( \displaystyle \int_5^1 f(x) \, dx = -\displaystyle \int_1^5 f(x) \, dx = -7 \)
b. By constant multiple rule: \( \displaystyle \int_1^5 3f(x) \, dx = 3 \cdot \displaystyle \int_1^5 f(x) \, dx = 3 \cdot 7 = 21 \)
Example:
Let \( f(x) \) be an odd function. Evaluate \( \displaystyle \int_{-3}^{3} f(x) \, dx \).
▶️Answer/Explanation
Since \( f(x) \) is odd: \( f(-x) = -f(x) \)
Then: \( \displaystyle \int_{-3}^{3} f(x) \, dx = 0 \)
Example:
Given:
\( \displaystyle \int_0^2 f(x) \, dx = 4 \)
\( \displaystyle \int_2^5 f(x) \, dx = -1 \),
Find: \( \displaystyle \int_0^5 [2f(x)] \, dx \)
▶️Answer/Explanation
Split the integral and apply constant rule:
\( \displaystyle \int_0^5 2f(x) \, dx = 2\displaystyle \int_0^2 f(x)\, dx + 2\displaystyle \int_2^5 f(x)\, dx \)
\( = 2(4) + 2(-1) = 8 – 2 = 6 \)
Evaluating Definite Integrals Using Geometry
In many cases, a definite integral can be evaluated without antiderivatives, by interpreting the area under the curve geometrically. This is especially useful when the graph forms basic geometric shapes (like rectangles, triangles, or semicircles).
Concept: The definite integral from \( a \) to \( b \) of a function \( f(x) \) represents the net area between the curve and the x-axis.
- Areas above the x-axis are counted as positive
- Areas below the x-axis are counted as negative
Example :
Evaluate \( \displaystyle \int_{-2}^3 f(x)\,dx \), where \( f(x) \) represents a straight line passing through the origin with slope 1, so that the graph is a straight line from \( (-2, -2) \) to \( (3, 3) \).
▶️Answer/Explanation
We are integrating a straight line \( f(x) = x \) from \( x = -2 \) to \( x = 3 \).
We split the interval at \( x = 0 \):
- From \( -2 \) to \( 0 \), the graph is below the x-axis (triangle with base 2 and height 2).
- From \( 0 \) to \( 3 \), the graph is above the x-axis (triangle with base 3 and height 3).
Area below x-axis (negative): \( A_1 = -\dfrac{1}{2} \times 2 \times 2 = -2 \)
Area above x-axis (positive): \( A_2 = \dfrac{1}{2} \times 3 \times 3 = 4.5 \)
Net area (value of the integral): \( \displaystyle \int_{-2}^3 f(x)\,dx = -2 + 4.5 = 2.5 \)
Definite Integrals with Discontinuities
While the definition of the definite integral is based on limits of Riemann sums, it can be extended to functions with certain types of discontinuities as long as the total area can still be calculated.
Allowed Discontinuities:
- Removable discontinuity: A hole at a single point does not affect the area.
- Jump discontinuity: A finite jump integral can still be computed if the jump occurs at a finite number of points.
As long as there are only a finite number of discontinuities in the interval \([a, b]\), the definite integral can still be evaluated.
Example:
Let \( f(x) = \begin{cases} 1 & x \ne 2 \\ 5 & x = 2 \end{cases} \).
Evaluate \( \displaystyle \int_0^4 f(x)\, dx \)
▶️Answer/Explanation
Even though \( f \) has a removable discontinuity at \( x = 2 \), the value at a single point doesn’t affect the area.
So \( f(x) = 1 \) almost everywhere from 0 to 4.
Area under the curve is a rectangle: \( \displaystyle \int_0^4 f(x) \, dx = 1 \times 4 = 4 \)
Example:
Let \( f(x) = \begin{cases} 2 & \text{for } 0 \leq x < 3 \\ 5 & \text{for } 3 \leq x \leq 5 \end{cases} \).
Find \( \displaystyle \int_0^5 f(x)\, dx \)
▶️Answer/Explanation
Split the integral at the jump:
\( \displaystyle \int_0^5 f(x)\,dx = \displaystyle \int_0^3 2\,dx + \displaystyle \int_3^5 5\,dx \)
= \( 2(3) + 5(2) = 6 + 10 = 16 \)