Mathematics, 14.01.2020 21:31 bossdde

Use integration by parts to find the integrals in exercise.
∫^2_1 ln 5x dx.

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Answer from: InfinityVicky

$\displaystyle \int\limits^2_1 {\ln 5x} \, dx = \ln(20) - 1$

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

$\displaystyle \int\limits^2_1 {\ln 5x} \, dx$

Step 2: Integrate Pt. 1

Identify variables for integration by parts using LIPET.

Set u: $\displaystyle u = \ln 5x$ [u] Logarithmic Differentiation [Derivative Rule - Chain Rule]: $\displaystyle du = \frac{(5x)'}{5x} \ dx$ [du] Basic Power Rule [Derivative Rule - Multiplied Constant]: $\displaystyle du = \frac{5}{5x} \ dx$ [du] Simplify: $\displaystyle du = \frac{1}{x} \ dx$ Set dv: $\displaystyle dv = dx$ [dv] Integration Rule [Reverse Power Rule]: $\displaystyle v = x$

Step 3: Integrate Pt. 2

[Integral] Integration by Parts: $\displaystyle \int\limits^2_1 {\ln 5x} \, dx = x \ln(5x) \bigg| \limits^2_1 - \int\limits^2_1 {} \, dx$ [Integral] Integration Rule [Reverse Power Rule]: $\displaystyle \int\limits^2_1 {\ln 5x} \, dx = x \ln(5x) \bigg| \limits^2_1 - x \bigg| \limits^2_1$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^2_1 {\ln 5x} \, dx = \ln(20) - 1$

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

Unit: Integration

Answer from: miahbaby2003p2af1b

$\int^2_1 ln 5x dx=1.99$

Step-by-step explanation:

we have the integral:

$\int^2_1 ln 5x dx$

The formula to integrate by parts is:

$\int udv=uv-\int vdu$

in this case we will use u=ln5x and dv=dx .

so v=x

and using logarithm derivation $d(lnw)=\frac{dw}{w}$

$du=\frac{5}{5x} =\frac{1}{x}$

thus:

$\int ln 5x dx=(ln5x)(x)- \int x(\frac{1}{x})dx$

$\int ln 5x dx=xln5x - \int dx$

$\int ln 5x dx=xln5x - x$

evaluating the limits:

$\int^2_1 ln 5x dx=[xln5x - x]^2_1\\=[2ln(5*2)- 2]-[1ln(5*1) - 1]\\=[2ln10-2]-[ln5-1]\\=[2(2.3)-2]-[1.61-1]\\=[4.6-2]-0.61\\=2.6-0.61\\=1.99$

Answer from: Quest

yes the data was not prsented

step-by-step explanation:

Answer from: Quest

The answer is d, none of above hope i

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Use integration by parts to find the integrals in exercise.
∫^2_1 ln 5x dx....

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