Mathematics, 14.01.2020 21:31 annafellows

Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on integration.
∫x^2√x+4 dx.

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Answer from: cheergirl2133

$\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \frac{2(x + 4)^\Big{\frac{3}{2}}(15x^2 - 48x + 128)}{105} + C$

General Formulas and Concepts:

Calculus

Differentiation

DerivativesDerivative Notation

Basic Power Rule:

f(x) = cxⁿf’(x) = c·nxⁿ⁻¹

Integration

Integrals[Indefinite Integrals] Integration Constant C

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

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

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

U-Substitution

U-Solve

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int {x^2\sqrt{x + 4}} \, dx$

Step 2: Integrate Pt. 1

Identify variables for u-solve.

Set u: $\displaystyle u = x + 4$ [u] Rewrite: $\displaystyle x = u - 4$ [u] Manipulate: $\displaystyle x^2 = (u - 4)^2$ [u] Basic Power Rule: $\displaystyle du = dx$

Step 3: Integrate Pt. 2

[Integral] U-Solve: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \int {(u - 4)^2\sqrt{u}} \, dx$ [Integrand] Rewrite: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \int {u^\Big{\frac{5}{2}} - 8u^\Big{\frac{3}{2}} + 16u^\Big{\frac{1}{2}}} \, dx$ [Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \int {u^\Big{\frac{5}{2}}} \, dx - \int {8u^\Big{\frac{3}{2}}} \, dx + \int {16u^\Big{\frac{1}{2}}} \, dx$ [Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \int {u^\Big{\frac{5}{2}}} \, dx - 8 \int {u^\Big{\frac{3}{2}}} \, dx + 16 \int {u^\Big{\frac{1}{2}}} \, dx$ [Integrals] Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \frac{2u^\Big{\frac{7}{2}}}{7} - 8 \Bigg( \frac{2u^\Big{\frac{5}{2}}}{5} \Bigg) + 16 \Bigg( \frac{2u^\Big{\frac{3}{2}}}{3} \Bigg) + C$ Simplify: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \frac{2u^\Big{\frac{7}{2}}}{7} - \frac{16u^\Big{\frac{5}{2}}}{5} + \frac{32u^\Big{\frac{3}{2}}}{3} + C$ [u] Back-Substitute: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \frac{2(x + 4)^\Big{\frac{7}{2}}}{7} - \frac{16(x + 4)^\Big{\frac{5}{2}}}{5} + \frac{32(x + 4)^\Big{\frac{3}{2}}}{3} + C$ Rewrite: $\displaystyle \int {x^2\sqrt{x + 4}} \, dx = \frac{2(x + 4)^\Big{\frac{3}{2}}(15x^2 - 48x + 128)}{105} + C$

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

Unit: Integration

Answer from: Quest

give more details

step-by-step explanation:

Answer from: Quest

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step-by-step explanation:

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