Respuesta :

Nayefx

Answer:

[tex] \rm \displaystyle \int x ^{2} \cdot {e}^{x} \: dx = {x}^{2} {e}^{x}- 2x {e}^{x} + 2 e^{x} + \rm C[/tex]

Step-by-step explanation:

we want to integrate the following integration:

[tex] \displaystyle \int {x}^{2} \cdot {e}^{x} dx[/tex]

notice that, the integrand is in the multiplication of two different function in that case we can consider integration by parts given by:

[tex] \rm \displaystyle \int u \cdot v \: dx = u \int vdx - \int u' \left( \int v dx \right)dx[/tex]

where u' can be defined by the differentiation of u

now we have to choose our u and v

which can be chosen by using the guideline:ILATE Inverse trig. , Logarithm, Algebraic expression, Trigonometry, Exponent.

since x comes first thus,

[tex] \displaystyle u = {x}^{2} \quad\text{and} \quad v = {e}^{x} [/tex]

by figuring out the defferentiation of u,we acquire:

[tex] \displaystyle u' = 2x[/tex]

altogether we get:

[tex] \rm \displaystyle \int {x}^{2} \cdot {e}^{x} \: dx = {x}^{2} \int {e}^{x} dx - \int 2x \left( \int {e}^{x} dx \right)dx[/tex]

by figuring out the parentheses integration

we get:

[tex] \rm \displaystyle \int x^2 \cdot {e}^{x} \: dx = {x}^{2} \int {e}^{x} dx - \int2x \cdot {e}^{x} dx[/tex]

now we again have integration of two different functions so let's choose our u and v once again which can be done using the guideline

[tex] \displaystyle u = 2x \quad\text{and} \quad v = {e}^{x} [/tex]

by figuring out the defferentiation of u

we acquire:

[tex] \displaystyle \: u' = 2[/tex]

altogether we get:

[tex] \rm \displaystyle \int x ^{2} \cdot {e}^{x} \: dx = {x}^{2} \int {e}^{x} dx - \left( 2x \int{e}^{x} dx- \int2 \left( \int{e}^{x}dx \right) dx \right)[/tex]

by solving parentheses we get

[tex] \rm \displaystyle \int x ^{2} \cdot {e}^{x} \: dx = {x}^{2} \int {e}^{x} dx - \left( 2x {e}^{x} - 2 e^{x} dx \right)[/tex]

by figuring out the integral we get:

[tex] \rm \displaystyle \int x ^{2} \cdot {e}^{x} \: dx = {x}^{2} {e}^{x}- \left( 2x {e}^{x} - 2 e^{x} \right)[/tex]

remove parentheses:

[tex] \rm \displaystyle \int x ^{2} \cdot {e}^{x} \: dx = {x}^{2} {e}^{x}- 2x {e}^{x} + 2 e^{x} [/tex]

at last we of course have to add constant of integration:

[tex] \rm \displaystyle \int x ^{2} \cdot {e}^{x} \: dx = {x}^{2} {e}^{x}- 2x {e}^{x} + 2 e^{x} + \rm C[/tex]

and we are done!

Space

Answer:

[tex]\displaystyle \int {x^2e^x} \, dx = e^x(x^2 - 2x + 2) + C[/tex]

General Formulas and Concepts:

Calculus

Differentiation

  • Derivatives
  • Derivative Notation

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

Basic Power Rule:

  1. f(x) = cxⁿ
  2. f’(x) = c·nxⁿ⁻¹

Integration

  • Integrals
  • Indefinite Integrals
  • Integration Constant C

Integration by Parts:                                                                                               [tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]

  • [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig
  • Tabular Integration

Step-by-step explanation:

*Note:

The answer below me is correct, but there is a simpler method of obtaining the answer.

Step 1: Define

Identify

[tex]\displaystyle \int {x^2e^x} \, dx[/tex]

Step 2: Integrate

Use tabular integration.

  1. [Integrand] Differentiate/Integrate [Tabular Integration]:                           [tex]\displaystyle \\\begin{center}\begin{tabular}{ c | c }\line{u & dv} \\\cline{1 - 2} x^2 & e^x \\\ 2x & e^x \\\ 2 & e^x \\\ 0 & e^x\end{tabular}\end{center}[/tex]
  2. Write out expansion [Tabular Integration]:                                                 [tex]\displaystyle \int {x^2e^x} \, dx = x^2e^x - 2xe^x + 2e^x + C[/tex]
  3. Factor:                                                                                                           [tex]\displaystyle \int {x^2e^x} \, dx = e^x(x^2 - 2x + 2) + C[/tex]

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

Unit: Integration

Book: College Calculus 10e