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
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
