Question

This is not really a question because I already know the answer to it but I will give 34 points for someone that can answer this. What is the definite integral from 1 to ∞ for the function e^-x (derivative with respect to x).

Answer 1

`\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = (1)/(e)`

General Formulas and Concepts:

Calculus

Limits

Limit Property [Addition/Subtraction]:
`\displaystyle \lim_(x \to c) [f(x) \pm g(x)] = \lim_(x \to c) f(x) \pm \lim_(x \to c) g(x)`

Limit Property [Multiplied Constant]:
`\displaystyle \lim_(x \to c) bf(x) = b \lim_(x \to c) f(x)`

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals

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^(\infty)_1 {e^(-x)} \, dx`

Step 2: Integrate Pt. 1

1. [Integral] Rewrite [Improper Integral]:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = \lim_(b \to \infty) \int\limits^b_1 {e^(-x)} \, dx`

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

1. Set u:
   `\displaystyle u = -x`
2. [u] Differentiate [Basic Power Rule, Derivative Properties]:
   `\displaystyle du = -dx`
3. [Bounds] Switch:
   `\displaystyle \left \{ {{x = b ,\ u = -b} \atop {x = 1 ,\ u = -1}} \right.`

Step 4: Integrate Pt. 3

1. [Integral] Rewrite [Integration Property - Multiplied Constant]:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = \lim_(b \to \infty) -\int\limits^b_1 {-e^(-x)} \, dx`
2. [Integral] U-Substitution:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = \lim_(b \to \infty) -\int\limits^(-b)_(-1) {e^u} \, du`
3. [Integral] Rewrite [Limit Property - Multiplied Constant]:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = -\lim_(b \to \infty) \int\limits^(-b)_(-1) {e^u} \, du`
4. [Integral] Exponential Integration:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = -\lim_(b \to \infty) e^u \bigg| \limits^(-b)_(-1)`
5. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = -\lim_(b \to \infty) \big( e^(-b) - e^(-1) \big)`
6. Rewrite:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = -\lim_(b \to \infty) \bigg( (1)/(e^b) - (1)/(e) \bigg)`
7. Evaluate limit:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = -\bigg( 0 - (1)/(e) \bigg)`
8. Simplify:
   `\displaystyle \int\limits^(\infty)_1 {e^(-x)} \, dx = (1)/(e)`

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

Unit: Integration