Question

Help please!!

Integrate the following:

`\displaystyle \int {x}^(2) {e}^(x) dx`
​

Answer 1

`\rm \displaystyle \int x ^(2) \cdot {e}^(x) \: dx = {x}^(2) {e}^(x)- 2x {e}^(x) + 2 e^(x) + \rm C`

Explanation:

we want to integrate the following integration:

`\displaystyle \int {x}^(2) \cdot {e}^(x) dx`

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

`\rm \displaystyle \int u \cdot v \: dx = u \int vdx - \int u' \left( \int v dx \right)dx`

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,

`\displaystyle u = {x}^(2) \quad\text{and} \quad v = {e}^(x)`

by figuring out the defferentiation of u,we acquire:

`\displaystyle u' = 2x`

altogether we get:

`\rm \displaystyle \int {x}^(2) \cdot {e}^(x) \: dx = {x}^(2) \int {e}^(x) dx - \int 2x \left( \int {e}^(x) dx \right)dx`

by figuring out the parentheses integration

we get:

`\rm \displaystyle \int x^2 \cdot {e}^(x) \: dx = {x}^(2) \int {e}^(x) dx - \int2x \cdot {e}^(x) dx`

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

`\displaystyle u = 2x \quad\text{and} \quad v = {e}^(x)`

by figuring out the defferentiation of u

we acquire:

`\displaystyle \: u' = 2`

altogether we get:

`\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)`

by solving parentheses we get

`\rm \displaystyle \int x ^(2) \cdot {e}^(x) \: dx = {x}^(2) \int {e}^(x) dx - \left( 2x {e}^(x) - 2 e^(x) dx \right)`

by figuring out the integral we get:

`\rm \displaystyle \int x ^(2) \cdot {e}^(x) \: dx = {x}^(2) {e}^(x)- \left( 2x {e}^(x) - 2 e^(x) \right)`

remove parentheses:

`\rm \displaystyle \int x ^(2) \cdot {e}^(x) \: dx = {x}^(2) {e}^(x)- 2x {e}^(x) + 2 e^(x)`

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

`\rm \displaystyle \int x ^(2) \cdot {e}^(x) \: dx = {x}^(2) {e}^(x)- 2x {e}^(x) + 2 e^(x) + \rm C`

and we are done!

Answer 2

`\displaystyle \int {x^2e^x} \, dx = e^x(x^2 - 2x + 2) + C`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

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

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Integration

• Integrals
• Indefinite Integrals
• Integration Constant C

Integration by Parts:
`\displaystyle \int {u} \, dv = uv - \int {v} \, du`

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

Explanation:

*Note:

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

Step 1: Define

Identify

`\displaystyle \int {x^2e^x} \, dx`

Step 2: Integrate

Use tabular integration.

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

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

Unit: Integration

Book: College Calculus 10e