Question

Use integration by parts to find the integrals in Exercise.
∫(4x-12)e-8x dx.

Answer 1

`\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (e^(-8x))/(2) \bigg( (23)/(8) - x \bigg) + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• [Indefinite Integrals] Integration Constant C

Integration Property [Multiplied Constant]:
`\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx`

U-Substitution

Integration by Parts:
`\displaystyle \int {u} \, dv = uv - \int {v} \, du`

• [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Explanation:

Step 1: Define

Identify

`\displaystyle \int {(4x - 12)e^(-8x)} \, dx`

Step 2: Integrate Pt. 1

1. [Integrand] Rewrite [Factor]:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = \int {4(x - 3)e^(-8x)} \, dx`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = 4 \int {(x - 3)e^(-8x)} \, dx`

Step 3: Integrate Pt. 2

Identify variables for integration by parts using LIPET.

1. Set u:
   `\displaystyle u = x - 3`
2. [u] Basic Power Rule [Derivative Property - Addition/Subtraction]:
   `\displaystyle du = dx`
3. Set dv:
   `\displaystyle dv = e^(-8x) \ dx`
4. [dv] Exponential Integration [U-Substitution]:
   `\displaystyle v = (-e^(-8x))/(8)`

Step 4: Integrate Pt. 3

1. [Integral] Integration by Parts:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = 4 \bigg( (-(x - 3)e^(-8x))/(8) - \int {(-e^(-8x))/(8)} \, dx \bigg)`
2. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = 4 \bigg( (-(x - 3)e^(-8x))/(8) + (1)/(8) \int {e^(-8x)} \, dx \bigg)`
3. Factor:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (1)/(2) \bigg( -(x - 3)e^(-8x) + \int {e^(-8x)} \, dx \bigg)`

Step 5: Integrate Pt. 4

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = -8x`
2. [u] Basic Power Rule [Derivative Rule - Multiplied Constant]:
   `\displaystyle du = -8 \ dx`

Step 6: Integrate Pt. 5

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (1)/(2) \bigg( -(x - 3)e^(-8x) - (1)/(8) \int {-8e^(-8x)} \, dx \bigg)`
2. [Integral] U-Substitution:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (1)/(2) \bigg( (x - 3)e^(-8x) - (1)/(8) \int {e^u} \, dx \bigg)`
3. [Integral] Exponential Integration:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (1)/(2) \bigg( (x - 3)e^(-8x) - (e^u)/(8) \bigg) + C`
4. [u] Back-Substitute:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (1)/(2) \bigg( (x - 3)e^(-8x) - (e^(-8x))/(8) \bigg) + C`
5. Factor:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (e^(-8x))/(2) \bigg( -(x - 3) - (1)/(8) \bigg) + C`
6. Simplify:
   `\displaystyle \int {(4x - 12)e^(-8x)} \, dx = (e^(-8x))/(2) \bigg( (23)/(8) - x \bigg) + C`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration