Question

IF ANYONE HELPS ME WITH THIS I WILL GIVE U BRAINLIEST AND 100 POINTS BTW ITS INTEGRALS FOR CALCULUS

Answer 1

Answer:

$\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Terms/Coefficients
• Factoring
• Exponential Rule [Rewrite]:                                                                           $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integrals

• Definite Integrals

Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]:                                     $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

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

U-Substitution

eˣ Integration:                                                                                                         $\displaystyle \int {e^u} \, dx = e^u + C$

Step-by-step explanation:

Step 1: Define

$\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

1. Set:                                                                                                                 $\displaystyle u = 4x^{-2}$
2. [u] Differentiate [Derivative Rule - Basic Power Rule]:                               $\displaystyle du = -8x^{-3} \ dx$
3. [du] Rewrite [Exponential Rule - Rewrite]:                                                   $\displaystyle du = \frac{-8}{x^3} \ dx$

Step 3: Integrate Pt. 2

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^6_4 {\frac{-8}{x^3}e^{4x^{-2}}} \, dx$
2. [Integral] U-Substitution:                                                                                $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{1}{9}}_{\frac{1}{4}} {e^u} \, dx$
3. [Integral] eˣ Integration:                                                                                $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8}(e^u) \bigg| \limits^{\frac{1}{9}}_{\frac{1}{4}}$
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:          $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{-1}{8} \bigg[ -e^\bigg{\frac{1}{9}} \bigg( e^\bigg{\frac{5}{36}} - 1 \bigg) \bigg]$
5. Simplify:                                                                                                         $\displaystyle \int\limits^6_4 {\frac{1}{x^3}e^{4x^{-2}}} \, dx = \frac{e^\bigg{\frac{1}{4}}}{8} - \frac{e^\bigg{\frac{1}{9}}}{8}$

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

Unit: Integration

Book: College Calculus 10e