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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Answer from: 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

Answer from: Quest

a. we have two

lines: y = 4-x and

y = 8-x^-1

given two simultaneous equations that are both to be

true, then the solution is the points where the lines cross. the intersection is

where the two equations are equal. therefore the solution that works for both

equations is when

4-x = 8-x^-1

this is where the two lines will cross and that is the

common point that satisfies both equations.

b. 4-x = 8-x^-1

x 4-x 8-x^-1

-3 7

8.33

-2 6 8.5

-1 5 9

0 4

1 3

2 2

7.5

3 1 7.67

the table shows that none of the x values from -3 to 3 is

the solution because in no case does

4-x = 8-x^-1

to find the solution we need to rearrange the equation to

find for x:

4-x = 8-x^-1

multiply both sides with x:

4x-x^2 = 8x-1

x^2+4x-1=0

x= -4.236, 0.236

therefore there are two points that satisfies the

equation.

find y:

x=-4.236

y = 4-x = 4 – (-4.236)

= 8.236

y = 8-x^-1 = .236)^-1

= 8.236

x=0.236

y = 4-x = 4 – (0.236) = 3.764

y = 8-x^-1 = 8-(0.236)^-1

= 3.764

thus the two lines cross at 2 points:

(-4.236, 8.236) & (0.236, 3.764)

c. to solve

graphically the equation 4-x = 8-x^-1

we would graph both lines: y = 4-x and y

= 8-x^-1

the point on the graph where the lines cross is the

solution to the system of equations.

just graph the points on part b on a cartesian coordinate

system and extend the two lines. the

solution is, as stated, the point where the two lines cross on the graph.

step-by-step explanation:

mark brainliest!

Answer from: Quest

there is nothing below so i cannot

step-by-step explanation:

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