Question

If anyone knows about definite integrals for calculus then please I request help! I

Answer 1

Answer:

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:                                                           $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

1. Set u:                                                                                                             $\displaystyle u = 4x^{-2}$
2. [u] Differentiate [Basic Power Rule, Derivative Properties]:                       $\displaystyle du = \frac{-8}{x^3} \ dx$
3. [Bounds] Switch:                                                                                           $\displaystyle \left \{ {{x = 9 ,\ u = 4(9)^{-2} = \frac{4}{81}} \atop {x = 5 ,\ u = 4(5)^{-2} = \frac{4}{25}}} \right.$

Step 3: Integrate Pt. 2

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:                 $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^9_5 {\frac{-8}{x^3}e^\big{4x^{-2}}} \, dx$
2. [Integral] U-Substitution:                                                                              $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}\int\limits^{\frac{4}{81}}_{\frac{4}{25}} {e^\big{u}} \, du$
3. [Integral] Exponential Integration:                                                               $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8}(e^\big{u}) \bigg| \limits^{\frac{4}{81}}_{\frac{4}{25}}$
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:           $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{-1}{8} \bigg( e^\Big{\frac{4}{81}} - e^\Big{\frac{4}{25}} \bigg)$
5. Simplify:                                                                                                         $\displaystyle \int\limits^9_5 {\frac{1}{x^3}e^\big{4x^{-2}}} \, dx = \frac{1}{8} \bigg( e^\Big{\frac{4}{25}} - e^\Big{\frac{4}{81}} \bigg)$

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

Unit: Integration