Mathematics, 14.01.2020 21:31 epicchicken420

Use integration by parts to find the integrals in exercise.
∫^0_1 ln 3x dx.

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Answer from: makiasdillard0925

$\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = \ln(3) - 1$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

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

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]: $\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)$

Integration

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

$\displaystyle \int\limits^1_0 {\ln(3x)} \, dx$

Step 2: Integrate Pt. 1

Identify variables for integration by parts using LIPET.

Set u: $\displaystyle u = \ln(3x)$ [u] Logarithmic Differentiation [Derivative Rule - Chain Rule]: $\displaystyle du = \frac{(3x)'}{3x} \ dx$ [du] Basic Power Rule [Derivative Property - Multiplied Constant]: $\displaystyle du = \frac{3}{3x} \ dx$ [du] Simplify: $\displaystyle du = \frac{1}{x} \ dx$ Set dv: $\displaystyle dv = 1 \ dx$ [dv] Integration Rule [Reverse Power Rule]: $\displaystyle v = x$

Step 3: Integrate Pt. 2

[Integral] Integration by Parts: $\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = x \ln(3x) \bigg| \limits^1_0 - \int\limits^1_0 {1} \, dx$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = x \ln(3x) \bigg| \limits^1_0 - \int\limits^1_0 {} \, dx$ [Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = x \ln(3x) \bigg| \limits^1_0 - x \bigg| \limits^1_0$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^1_0 {\ln(3x)} \, dx = \ln(3) - 1$

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

Unit: Integration

Answer from: Quest

That is right sorry i was gonna answer but i already saw his

Answer from: Quest

The answer is 9.7 cm

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Use integration by parts to find the integrals in exercise.
∫^0_1 ln 3x dx....