Question

Exercise is mixed —some require integration by parts, while others can be integrated by using techniques discussed in the chapter on Integration.

∫^5_0 x √x^2+2 dx.

Answer 1

`\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx = 27√(3) - (2√(2))/(3)`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• 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`

U-Substitution

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx`

Step 2: Integrate Pt. 1

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = x^2 + 2`
2. [u] Basic Power Rule [Derivative Property - Addition/Subtraction]:
   `\displaystyle du = 2x \ dx`
3. [Limits] Swap:
   `\displaystyle \left \{ {{x = 5 ,\ u = 5^2 + 2 = 27} \atop {x = 0 ,\ u = 0^2 + 2 = 2}} \right.`

Step 3: Integrate Pt. 2

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx = (1)/(2) \int\limits^5_0 {2x√(x^2 + 2)} \, dx`
2. [Integral] U-Substitution:
   `\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx = (1)/(2) \int\limits^(27)_2 {√(u)} \, du`
3. [Integral] Integration Rule [Reverse Power Rule]:
   `\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx = (1)/(2) \Bigg( \frac{2x^\bigg{(3)/(2)}}{3} \Bigg) \Bigg| \limits^(27)_2`
4. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx = (1)/(2) \bigg( 54√(3) - (4√(2))/(3) \bigg)`
5. Simplify:
   `\displaystyle \int\limits^5_0 {x√(x^2 + 2)} \, dx = 27√(3) - (2√(2))/(3)`

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

Unit: Integration