Mathematics, 06.09.2020 14:01 PlaneGamer5678

Integrate the following (3x+4)^2

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Answer from: zacharynugent

$\displaystyle \int {(3x + 4)^2} \, dx = \frac{(3x + 4)^3}{9} + C$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

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

IntegralsIndefinite IntegralsIntegration Constant C

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {(3x + 4)^2} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

Set u: $\displaystyle u = 3x + 4$ [u] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

Step 3: Integrate Pt. 2

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {(3x + 4)^2} \, dx = \frac{1}{3}\int {3(3x + 4)^2} \, dx$ [Integral] U-Substitution: $\displaystyle \int {(3x + 4)^2} \, dx = \frac{1}{3}\int {u^2} \, du$ [Integral] Reverse Power Rule: $\displaystyle \int {(3x + 4)^2} \, dx = \frac{1}{3} \bigg( \frac{u^3}{3} \bigg) + C$ Simplify: $\displaystyle \int {(3x + 4)^2} \, dx = \frac{u^3}{9} + C$ Back-Substitute: $\displaystyle \int {(3x + 4)^2} \, dx = \frac{(3x + 4)^3}{9} + C$

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

Unit: Integration

Book: College Calculus 10e

Answer from: Quest

step-by-step explanation:

Answer from: Quest

$7 {}^{6} 7 {}^{3} =$

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