Caculate the monthly payment for a 5 year car loan of $23,570 at 10.48% interest, compounded monthly

Calculus 2. Please help

Anarel [89]

Answer:

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}}} \, dx = \infty$

General Formulas and Concepts:

Algebra I

Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Limits

Right-Side Limit: $\displaystyle \lim_{x \to c^+} f(x)$

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

Derivatives

Derivative Notation

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integrals

Definite Integrals

Integration Constant C

Integration Rule [Fundamental Theorem of Calculus 1]: $\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

U-Substitution

U-Solve

Improper Integrals

Exponential Integral Function: $\displaystyle \int {\frac{e^x}{x}} \, dx = Ei(x) + C$

Step-by-step explanation:

Step 1: Define

Identify

$\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$

Step 2: Integrate Pt. 1

[Integral] Rewrite [Exponential Rule - Rewrite]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \int\limits^1_0 {\frac{e^{-x^2}}{x} \, dx$
[Integral] Rewrite [Improper Integral]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \int\limits^1_a {\frac{e^{-x^2}}{x} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set: $\displaystyle u = -x^2$
Differentiate [Basic Power Rule]: $\displaystyle \frac{du}{dx} = -2x$
[Derivative] Rewrite: $\displaystyle du = -2x \ dx$

Rewrite u-substitution to format u-solve.

Rewrite du: $\displaystyle dx = \frac{-1}{2x} \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {-\frac{e^{-x^2}}{x} \, dx$
[Integral] Substitute in variables: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} -\int\limits^1_a {\frac{e^{u}}{-2u} \, du$
[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}\int\limits^1_a {\frac{e^{u}}{u} \, du$
[Integral] Substitute [Exponential Integral Function]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(u)] \bigg| \limits^1_a$
Back-Substitute: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-x^2)] \bigg| \limits^1_a$
Evaluate [Integration Rule - FTC 1]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{1}{2}[Ei(-1) - Ei(a)]$
Simplify: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \lim_{a \to 0^+} \frac{Ei(-1) - Ei(a)}{2}$
Evaluate limit [Limit Rule - Variable Direct Substitution]: $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx = \infty$

∴ $\displaystyle \int\limits^1_0 {\frac{1}{xe^{x^2}} \, dx$ diverges.

Topic: Multivariable Calculus

7 0

3 years ago

Please simplify picture attached

Mekhanik [1.2K]

Answer:

It is the first bubble.

Step-by-step explanation:

multiply the fraction $\frac{\sqrt{5} }{\sqrt{} 5}$

multiply the numerators and denominators seperatly

calculate the product

multiply the numbers

solution

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3 years ago

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astra-53 [7]

2/5 at the top nd bottom, but 0/5 on the sides. So 2/5 + 0/5= 1/5

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3 years ago

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Morgarella [4.7K]

You are converting the numbers to a word problem.

1. 5+x = the number five plus x

2. 5-x= the number 5 minus x

3. 5x= five times x

4. 5/x= five divided by x

5. three-fourths of x plus 2

6. two minus three-fourths of x

7. 3 times x plus 10

8. ten times x plus three

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4 years ago

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svlad2 [7]

Answer:

S: (-2,7)

T: (-5,7)

U: (-1,8)

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3 years ago

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