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

Answer:

$\displaystyle \int\limits^{\infty}_1 {e^{-x}} \, dx = \frac{1}{e}$

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]:                                                           $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

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

Step-by-step 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:                                                                                                         $\displaystyle \int\limits^{\infty}_1 {e^{-x}} \, dx = -\lim_{b \to \infty} \bigg( \frac{1}{e^b} - \frac{1}{e} \bigg)$
7. Evaluate limit:                                                                                                 $\displaystyle \int\limits^{\infty}_1 {e^{-x}} \, dx = -\bigg( 0 - \frac{1}{e} \bigg)$
8. Simplify:                                                                                                         $\displaystyle \int\limits^{\infty}_1 {e^{-x}} \, dx = \frac{1}{e}$

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

Unit: Integration