Mathematics, 17.09.2019 16:00 alinegonzalez0027

Evaluate the definite integral (s is the integration symbols with the indicated values)
1
s 4x(6sqrt(1+x^2) dx
0

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Answer from: ethanhose05

$\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 8(2\sqrt{2} - 1)$

General Formulas and Concepts:

Algebra I

Terms/Coefficients

Expanding/Factoring

Exponential Property [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$

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

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]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Simplify: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = \int\limits^1_0 {24x\sqrt{1 + x^2}} \, dx$ [Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 24\int\limits^1_0 {x\sqrt{1 + x^2}} \, dx$

Step 3: integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 1 + x^2$ [u] Differentiate [Basic Power Rule, Derivative Properties]: $\displaystyle du = 2x \ dx$ [Bounds] Switch: $\displaystyle \left \{ {{x = 1 ,\ u = 1 + 1^2 = 2} \atop {x = 0 ,\ u = 1 + 0^2 = 1}} \right.$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12\int\limits^1_0 {2x\sqrt{1 + x^2}} \, dx$ [Integral] U-Substitution: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12\int\limits^2_1 {\sqrt{u}} \, du$ [Integrand] Rewrite [Exponential Property - Root Rewrite]: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12\int\limits^2_1 {u^\big{\frac{1}{2}}} \, du$ [Integral] Reverse Power Rule: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12 \bigg( \frac{2u^\big{\frac{3}{2}}}{3} \bigg) \bigg| \limits^2_1$ Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 12 \bigg( \frac{4\sqrt{2}}{3} - \frac{2}{3} \bigg)$ Factor: $\displaystyle \int\limits^1_0 {4x(6\sqrt{1 + x^2})} \, dx = 8(2\sqrt{2} - 1)$

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

Unit: Integration

Answer from: Quest

4th option is correct xy=-10

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Answer from: Quest

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Evaluate the definite integral (s is the integration symbols with the indicated values)
1

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