AP Calculus BC 6.14 Selecting Techniques for Antidifferentiation Study Notes - New Syllabus
AP Calculus BC 6.14 Selecting Techniques for Antidifferentiation Study Notes- New syllabus
AP Calculus BC 6.14 Selecting Techniques for Antidifferentiation 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:
- Selecting Techniques for Antidifferentiation
Selecting Techniques for Antidifferentiation
Selecting Techniques for Antidifferentiation
When solving indefinite or definite integrals, there is no single method that works for all functions. Instead, we must choose an integration technique based on the form of the integrand. The table below outlines common forms and the most suitable techniques.
| Form of Integrand | Recommended Technique |
|---|---|
| \( f(g(x)) \cdot g'(x) \) | Substitution (u-substitution) |
| \( P_n(x) \) (polynomial) | Direct integration using power rule |
| Product of two functions \( u(x) \cdot v(x) \) | Integration by parts |
| Rational functions \( \dfrac{P_m(x)}{P_n(x)} \) | Polynomial long division and/or partial fractions |
| Square roots of quadratic expressions | Trigonometric substitution or completing the square |
| Trigonometric functions | Trigonometric identities, u-substitution, or parts |
| Improper integrals | Rewrite using limits, then apply above techniques |
Example :
Evaluate \( \displaystyle \int x \cos(x^2) \, dx \).
▶️Answer/Explanation
We notice the integrand has the form \( f(g(x)) \cdot g'(x) \), which suggests u-substitution.
Let \( u = x^2 \), then \( du = 2x \, dx \) or \( \dfrac{du}{2} = x \, dx \).
\(\displaystyle\int x \cos(x^2) \, dx = \dfrac{1}{2} \int \cos(u) \, du = \dfrac{1}{2} \sin(u) + C \).
Substitute back \( u = x^2 \):
\( \boxed{\dfrac{1}{2} \sin(x^2) + C} \).
Example :
Evaluate \( \displaystyle \int x^3 e^x \, dx \).
▶️Answer/Explanation
The integrand is a product of two functions \( x^3 \) and \( e^x \), suggesting integration by parts.
We choose \( u = x^3 \), \( dv = e^x dx \).
Then \( du = 3x^2 dx \) and \( v = e^x \).
Applying integration by parts:
\(\displaystyle \int x^3 e^x dx = x^3 e^x – \int 3x^2 e^x dx \).
We repeat the process until we reach \( \int e^x dx \).
Final answer:
\( e^x(x^3 – 3x^2 + 6x – 6) + C \).
Example:
Evaluate \( \displaystyle \int \sin^3 x \cos^2 x \, dx \).
▶️Answer/Explanation
We can use the identity \( \sin^2 x = 1 – \cos^2 x \).
Write \( \sin^3 x = \sin x (1 – \cos^2 x) \).
Substitute \( u = \cos x \), so \( du = -\sin x \, dx \).
The integral becomes:
\( – \displaystyle\int (1 – u^2) u^2 \, du = -\displaystyle\int (u^2 – u^4) \, du \)
\( = -\left( \dfrac{u^3}{3} – \dfrac{u^5}{5} \right) + C \)
Substitute back \( u = \cos x \):
\( -\dfrac{\cos^3 x}{3} + \dfrac{\cos^5 x}{5} + C \)
Example:
Evaluate \( \displaystyle \int \dfrac{dx}{x^2 + 6x + 13} \).
▶️Answer/Explanation
Complete the square: \( x^2 + 6x + 13 = (x+3)^2 + 4 \).
Let \( u = x+3 \), so \( du = dx \).
The integral becomes:
\( \displaystyle\int \dfrac{du}{u^2 + 2^2} = \dfrac{1}{2} \arctan\left( \dfrac{u}{2} \right) + C \)
Substitute back \( u = x+3 \):
\( \dfrac{1}{2} \arctan\left( \dfrac{x+3}{2} \right) + C \)
Example:
Evaluate \( \displaystyle \int \dfrac{3x+5}{x^2 – x – 6} \, dx \).
▶️Answer/Explanation
Factor the denominator: \( x^2 – x – 6 = (x-3)(x+2) \).
Write:
\( \dfrac{3x+5}{(x-3)(x+2)} = \dfrac{A}{x-3} + \dfrac{B}{x+2} \)
Multiply through by \((x-3)(x+2)\):
\( 3x + 5 = A(x+2) + B(x-3) \)
Set \( x = 3 \): \( 9 + 5 = 5A \) → \( A = \dfrac{14}{5} \).
Set \( x = -2 \): \( -6 + 5 = -5B \) → \( B = \dfrac{1}{5} \).
Integral becomes:
\( \dfrac{14}{5} \int \dfrac{dx}{x-3} + \dfrac{1}{5} \int \dfrac{dx}{x+2} \)
\( = \dfrac{14}{5} \ln|x-3| + \dfrac{1}{5} \ln|x+2| + C \)
Example:
Evaluate \( \displaystyle \int x e^{2x} \, dx \).
▶️Answer/Explanation
Use the formula \( \int u \, dv = uv – \int v \, du \).
Let \( u = x \) → \( du = dx \).
Let \( dv = e^{2x} dx \) → \( v = \dfrac{e^{2x}}{2} \).
Then:
\( \displaystyle\int x e^{2x} \, dx = \dfrac{x e^{2x}}{2} – \int \dfrac{e^{2x}}{2} \, dx \)
\( = \dfrac{x e^{2x}}{2} – \dfrac{e^{2x}}{4} + C \)