Mathematics, 12.12.2019 19:31, Gladistshiala267

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).

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answered: Jackpumpkin

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

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

DerivativesDerivative Notation

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

Basic Power Rule:

f(x) = cxⁿf’(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

[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.

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

Step 4: Integrate Pt. 3

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

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

Unit: Integration

answered: Guest

h who knows

step-by-step explanation:

answered: Guest

Sorry i didn’t know i had