Mathematics, 14.01.2020 21:31 RiceLord9562

Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on integration.
∫x^2/2x^3+1 dx.

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Answer from: nyandog50

$\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln \big| 2x^3 + 1 \big|}{6} + C$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

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

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals\ Integration Constant C

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

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 2x^3 + 1$ [u] Basic Power Rules [Derivative Properties]: $\displaystyle du = 6x^2 \ dx$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \int {\frac{6x^2}{2x^3 + 1}} \, dx$ [Integral] U-Substitution: $\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{1}{6} \int {\frac{1}{u}} \, du$ [Integral] Logarithmic Integration: $\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln \big| u \big|}{6} + C$ [u] Back-Substitute: $\displaystyle \int {\frac{x^2}{2x^3 + 1}} \, dx = \frac{\ln \big| 2x^3 + 1 \big|}{6} + C$

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

Unit: Integration

Answer from: Quest

a) one student is chosen from classroom a, then a student is chosen from classroom b.

step-by-step explanation:

independent events are ones in which the probability of one event does not affect the probability of the second event.

one student is chosen from classroom a; this does not affect the chances of choosing a student from classroom b. this makes these independent events.

when one card is chosen from a standard deck and then set aside before a second card is chosen, the probability of the second card being chosen changes. this makes them dependent events.

when one student is chosen from classroom a and then a second student is chosen from classroom a, the probability of choosing the second student changes. this makes them dependent events.

when a letter is chosen from the alphabet and then a second letter is chosen, the probability of choosing the second letter changes. this makes them dependent events.

Answer from: Quest

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )

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