Mathematics, 04.01.2020 13:31 hilton894

Calculus question

how to do this?

Calculus question how to do this?

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Answer from: 20alondra04

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

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) = cxⁿf’(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.

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

Step 3: Integrate Pt. 2

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

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