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Find the integral of x(4x² + 1) from 0 to 2. a. 18 c. 22 b. 16 d. 20

User Postelrich
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1 Answer

5 votes

Answer:

a. 18

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Distributive Property

Algebra I

  • Terms/Coefficients

Calculus

Integrals

  • Definite Integrals

Integration Rule [Reverse Power Rule]:
\displaystyle \int {x^n} \, dx = (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 Property [Addition/Subtraction]:
\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx

Explanation:

Step 1: Define

Identify


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

Step 2: Integrate

  1. [Integrand] Distribute x [Distributive Property]:
    \displaystyle \int\limits^2_0 {(4x^3 + x)} \, dx
  2. Rewrite Integral [Integration Property - Addition/Subtraction]:
    \displaystyle \int\limits^2_0 {4x^3} \, dx + \int\limits^2_0 {x} \, dx
  3. Rewrite 1st Integral [Integration Property - Multiplied Constant]:
    \displaystyle 4\int\limits^2_0 {x^3} \, dx + \int\limits^2_0 {x} \, dx
  4. [Integrals] Reverse Power Rule:
    \displaystyle 4((x^4)/(4)) \bigg| \limits^2_0 + ((x^2)/(2)) \bigg| \limits^2_0
  5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
    \displaystyle 4(4) + 2
  6. Evaluate:
    \displaystyle 18

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

Unit: Integration

Book: College Calculus 10e

User David Underwood
by
5.3k points
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