Question

PLEASE HELP!! CALCULUS QUESTION

How to do this?

Answer 1

Answer:

$\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = 28ln(7) - 24ln(6) - 4$

General Formulas and Concepts:

Algebra I

• Exponential Rule [Rewrite]:                                                                           $\displaystyle b^{-m} = \frac{1}{b^m}$
• Exponential Rule [Root Rewrite]:                                                                 $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Logarithmic Derivative:                                                                                         $\displaystyle \frac{d}{dx} [lnu] = \frac{u'}{u}$

Integrals

• Definite Integrals

Integration Rule [Reverse Power Rule]:                                                               $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

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

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

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

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy$

Step 2: Integrate Pt. 1

Identify and find variables for Integration by Parts.

1. [LIPET] Set:                                                                                                     $\displaystyle u = ln(y)$
2. [LIPET] Set:                                                                                                     $\displaystyle dv = \frac{1}{\sqrt{y}} \ dy$
3. [u] Differentiate [Logarithmic Derivative]:                                                   $\displaystyle \frac{du}{dy} = \frac{1}{y}$
4. [u] Rewrite:                                                                                                     $\displaystyle du = \frac{1}{y} \ dy$
5. [dv] Integrate [Integration Rule - Reverse Power Rule]:                             $\displaystyle v = 2\sqrt{y}$

Step 3: Integrate Pt. 2

1. [Integral] Integration by Parts:                                                                       $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = \bigg[ 2ln(y)\sqrt{y} \bigg] \bigg| \limits^{49}_{36} - \int\limits^{49}_{36} {\frac{2\sqrt{y}}{y}} \, dy$
2. [Integral] Simplify/Rewrite [Integration Property - Multiplied Constant]:   $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = \bigg[ 2ln(y)\sqrt{y} \bigg] \bigg| \limits^{49}_{36} - 2\int\limits^{49}_{36} {\frac{1}{\sqrt{y}}} \, dy$
3. [Integral] Reverse Power Rule:                                                                     $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = \bigg[ 2ln(y)\sqrt{y} \bigg] \bigg| \limits^{49}_{36} - 2 \bigg[ 2\sqrt{y} \bigg] \bigg| \limits^{49}_{36}$
4. Evaluate [Integration Rule - FTC 1]:                                                             $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = 28ln(7) - 24ln(6) - 2(2)$
5. Simplify:                                                                                                         $\displaystyle \int\limits^{49}_{36} {\frac{ln(y)}{\sqrt{y}}} \, dy = 28ln(7) - 24ln(6) - 4$

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

Unit: Integration

Book: College Calculus 10e