Answer:
[tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}[/tex]
General Formulas and Concepts:
Pre-Algebra
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
Algebra I
- Terms/Coefficients
- Factoring
- Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
Calculus
Derivatives
Derivative Notation
Basic Power Rule:
- f(x) = cxⁿ
- f’(x) = c·nxⁿ⁻¹
Integrals
Integration Constant C
Integration Rule [Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]
Integration Property [Multiplied Constant]: [tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]
U-Substitution
eˣ Integration: [tex]\displaystyle \int {e^u} \, dx = e^u + C[/tex]
Step-by-step explanation:
Step 1: Define
[tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx[/tex]
Step 2: Integrate Pt. 1
Identify variables for u-substitution.
- Set: [tex]\displaystyle u = 4x^{-2}[/tex]
- [u] Differentiate [Derivative Rule - Basic Power Rule]: [tex]\displaystyle du = -8x^{-3} \ dx[/tex]
- [du] Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle du = \frac{-8}{x^3} \ dx[/tex]
Step 3: Integrate Pt. 2
- [Integral] Rewrite [Integration Property - Multiplied Constant]: [tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^6_4 {\frac{-8}{x^3}e^{4x^{-2}}} \, dx[/tex]
- [Integral] U-Substitution: [tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{1}{9}}_{\frac{1}{4}} {e^u} \, dx[/tex]
- [Integral] eˣ Integration: [tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}(e^u) \bigg| \limits^{\frac{1}{9}}_{\frac{1}{4}}[/tex]
- Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: [tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8} \bigg[ -e^\bigg{\frac{1}{9}} \bigg( e^\bigg{\frac{5}{36}} - 1 \bigg) \bigg][/tex]
- Simplify: [tex]\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}[/tex]
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Integration
Book: College Calculus 10e
