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Determine whether the improper integral converges or diverges, and find the value of each that converges.∫^0_-[infinity] 5e^60x dx

Answers

Answer 1

Answer:

Answer:

The improper integral converges.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

General Formulas and Concepts:
Calculus

Limit

Limit Rule [Variable Direct Substitution]: $\displaystyle \lim_(x \to c) x = c$

Differentiation

Derivatives
Derivative Notation

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

Derivative Rule [Basic Power Rule]:

f(x) = cxⁿ
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]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Method: U-Substitution

Improper Integral: $\displaystyle \int\limits^(\infty)_a {f(x)} \, dx = \lim_(b \to \infty) \int\limits^b_a {f(x)} \, dx$

Step-by-step explanation:

Step 1: Define

Identify.

$\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = 5 \int\limits^0_(- \infty) {e^(60x)} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) 5 \int\limits^0_(a) {e^(60x)} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 60x$
[u] Differentiate [Derivative Properties and Rules]: $\displaystyle du = 60 \ dx$
[Bounds] Swap: $\displaystyle \left \{ {{x = 0 \rightarrow u = 0} \atop {x = a \rightarrow u = 60a}} \right.$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(a) {60e^(60x)} \, dx$
[Integral] Apply Integration Method [U-Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) \int\limits^0_(60a) {e^(u)} \, du$
[Integral] Apply Exponential Integration: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1)/(12) e^u \bigg| \limits^0_(60a)$
Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = \lim_(a \to - \infty) (1 - e^(60a))/(12)$
[Limit] Evaluate [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1 - e^(60(-\infty)))/(12)$
Rewrite: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12) - (1)/(12e^(60(\infty)))$
Simplify: $\displaystyle \int\limits^0_(- \infty) {5e^(60x)} \, dx = (1)/(12)$

∴ the improper integral equals $\displaystyle \bold{(1)/(12)}$ and is convergent.

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Learn more about improper integrals: brainly.com/question/14413972

Learn more about calculus: brainly.com/question/23558817

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Topic: AP Calculus BC (Calculus I + II)

Unit: Integration

Answer 2

Answer:

Answer:

$\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)$

Step-by-step explanation:

Assuming this integral:

$\int_(-\infty)^0 5 e^(60x) dx$

We can do this as the first step:

$5 \int_(-\infty)^0 e^(60x) dx$

Now we can solve the integral and we got:

$5 (e^(60x))/(60) \Big|_(-\infty)^0$

$\int_(-\infty)^0 5 e^(60x) dx = (e^(60x))/(12)\Big|_(-\infty)^0 = (1)/(12) [e^(60*0) -e^(-\infty)]$

$\int_(-\infty)^0 5 e^(60x) dx = (1)/(12)[e^0 -0]= (1)/(12)$

So then we see that the integral on this case converges amd the values is 1/12 on this case.

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Answers

Answer: the value of A is 2

Step-by-step explanation:

Ac and bd are perpendicular bisectors of each other. adc. Find eab

Answers

Let ∠ ADC = 2β

Ac and BD are perpendicular bisectors of each other ⇒⇒ (Given information)

∴ BD bisects the angle ADC
∴ ∠ADE = 0.5 ∠ADC = β

And in ΔADE:
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