AP Calculus BC 6.11 Integrating Using Integration by Parts Study Notes - New Syllabus
AP Calculus BC 6.11 Integrating Using Integration by Parts by Parts Study Notes- New syllabus
AP Calculus BC 6.11 Integrating Using Integration by Parts by Parts 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:
- Integrating Using Integration by Parts
Integrating Using Integration by Parts
Integrating Using Integration by Parts
For two functions \( u(x) \) and \( v(x) \), the integration by parts formula is:
\(\displaystyle \int u \, dv = uv -\displaystyle \int v \, du \)
This is derived from the product rule for derivatives: \( \dfrac{d}{dx}(uv) = u \, \dfrac{dv}{dx} + v \, \dfrac{du}{dx} \).
Strategy to Choose \( u \) and \( dv \):
Use the ILATE rule as a general guide to pick \( u \):
| Priority | Function Type (Choose as \( u \)) |
|---|---|
| 1 | Inverse trig (e.g. \( \tan^{-1}x \)) |
| 2 | Logarithmic (e.g. \( \ln x \)) |
| 3 | Algebraic (e.g. \( x^2, x \)) |
| 4 | Trigonometric (e.g. \( \sin x, \cos x \)) |
| 5 | Exponential (e.g. \( e^x \)) |
Example:
Let \(\displaystyle \int x \ln x \, dx \) be evaluated using integration by parts. Which of the following is the correct indefinite integral?
(A) \( \dfrac{x^2}{2} \ln x – \dfrac{x^2}{4} + C \)
(B) \( x \ln x – x + C \)
(C) \( \dfrac{x^2}{2} \ln x – \dfrac{x}{2} + C \)
(D) \( x \ln x + \dfrac{x^2}{2} + C \)
(E) \( x \ln x – \dfrac{x^2}{2} + C \)
▶️Answer/Explanation
We apply integration by parts:
Let \( u = \ln x \), \( dv = x\,dx \). Then:
- \( du = \dfrac{1}{x}\,dx \)
- \( v = \dfrac{x^2}{2} \)
Using integration by parts: \(\displaystyle \int u\,dv = uv -\displaystyle \int v\,du \)
\(\displaystyle \int x \ln x\, dx = \dfrac{x^2}{2} \ln x -\displaystyle \int \dfrac{x^2}{2} \cdot \dfrac{1}{x} \, dx \)
\( = \dfrac{x^2}{2} \ln x -\displaystyle \int \dfrac{x}{2} \, dx = \dfrac{x^2}{2} \ln x – \dfrac{x^2}{4} + C \)
Correct answer: (A)
Example:
Evaluate \(\displaystyle \int x e^x \, dx \)
▶️Answer/Explanation
Use integration by parts. Choose:
- \( u = x \) ⇒ \( du = dx \)
- \( dv = e^x dx \) ⇒ \( v = e^x \)
Now apply:
\(\displaystyle \int x e^x dx = uv -\displaystyle \int v \, du = x e^x -\displaystyle \int e^x dx = x e^x – e^x + C \)
Final Answer: \( e^x(x – 1) + C \)
Example:
Evaluate \(\displaystyle \int \ln x \, dx \)
▶️Answer/Explanation
Let:
- \( u = \ln x \) ⇒ \( du = \dfrac{1}{x} dx \)
- \( dv = dx \) ⇒ \( v = x \)
\(\displaystyle \int \ln x \, dx = x \ln x -\displaystyle \int x \cdot \dfrac{1}{x} dx = x \ln x -\displaystyle \int 1 \, dx = x \ln x – x + C \)
Final Answer: \( x \ln x – x + C \)
Example:
Evaluate \(\displaystyle \int_0^1 x \ln(x+1) \, dx \)
▶️Answer/Explanation
Let:
- \( u = \ln(x+1) \) ⇒ \( du = \dfrac{1}{x+1} dx \)
- \( dv = x dx \) ⇒ \( v = \dfrac{1}{2}x^2 \)
Then:
\(\displaystyle \int_0^1 x \ln(x+1) dx = \left[\dfrac{1}{2} x^2 \ln(x+1) \right]_0^1 -\displaystyle \int_0^1 \dfrac{1}{2}x^2 \cdot \dfrac{1}{x+1} dx \)
Evaluate the boundary term:
\( \left[\dfrac{1}{2} x^2 \ln(x+1) \right]_0^1 = \dfrac{1}{2} (1)^2 \ln(2) – 0 = \dfrac{1}{2} \ln 2 \)
Now evaluate the remaining integral:
\(\displaystyle \int_0^1 \dfrac{x^2}{x+1} dx \) ← Use polynomial division (or long division):
\( \dfrac{x^2}{x+1} = x – 1 + \dfrac{1}{x+1} \)
So:
\(\displaystyle \int_0^1 \dfrac{x^2}{x+1} dx =\displaystyle \int_0^1 (x – 1 + \dfrac{1}{x+1}) dx = \left[ \dfrac{1}{2}x^2 – x + \ln(x+1) \right]_0^1 \)
= \( \left( \dfrac{1}{2} – 1 + \ln 2 \right) – (0 – 0 + \ln 1) = -\dfrac{1}{2} + \ln 2 \)
So the full answer becomes:
\( \dfrac{1}{2} \ln 2 – \dfrac{1}{2}(-\dfrac{1}{2} + \ln 2) = \dfrac{1}{2} \ln 2 + \dfrac{1}{4} – \dfrac{1}{2} \ln 2 = \dfrac{1}{4} \)
Final Answer: \( \dfrac{1}{4} \)
Example:
Evaluate the definite integral: \(\displaystyle \int_1^e \ln x \, dx \)
(A) \( 1 \)
(B) \( e – 1 \)
(C) \( e \ln e – e + 1 \)
(D) \( e – 1 – \ln e \)
(E) \( e – 1 \)
▶️Answer/Explanation
Use integration by parts:
Let \( u = \ln x \), so \( du = \dfrac{1}{x} dx \)
Let \( dv = dx \), so \( v = x \)
Using the formula: \(\displaystyle \int u\,dv = uv -\displaystyle \int v\,du \)
\(\displaystyle \int \ln x \, dx = x \ln x -\displaystyle \int x \cdot \dfrac{1}{x} dx = x \ln x -\displaystyle \int 1 \, dx = x \ln x – x \)
Now apply the limits from 1 to \( e \):
\(\displaystyle \int_1^e \ln x \, dx = \left[ x \ln x – x \right]_1^e \)
At \( x = e \): \( e \ln e – e = e – e = 0 \)
At \( x = 1 \): \( 1 \ln 1 – 1 = 0 – 1 = -1 \)
So, \(\displaystyle \int_1^e \ln x \, dx = 0 – (-1) = 1 \)
Correct answer: (A)