Mathematics, 27.06.2021 01:20 eelebron0905

Calculus 2. Please help

Answers: 2

Show answers

Answers

Answer from: weberalycia

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

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) = cxⁿf’(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

Answer from: Quest

mark branliest

step-by-step explanation:

.6 is 1/2 of 12

the answer is actually d. i just took the test

Answer from: Quest

The answer that i got was 8/3