Respuesta :
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!
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:
- f(x) = cxⁿ
- 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.
- [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]
- Write out expansion [Tabular Integration]: [tex]\displaystyle \int {x^2e^x} \, dx = x^2e^x - 2xe^x + 2e^x + C[/tex]
- 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