Chapter 7 · Integrals

Exercise 7.6 · Q3: $\int x^2e^x\,dx$

Exercise 7.6Integration by parts · Section 7.6

The question

Integrate the functions in Exercises 1 to 22.

3. $x^2e^x$

Exam answer

Let $I = \displaystyle\int x^2e^x\,dx$ ⁠.

Taking $x^2$ as the first function, by parts:

\begin{aligned} I &= x^2e^x - \int 2x\cdot e^x\,dx \\ &= x^2e^x - 2\int xe^x\,dx \end{aligned}

Again by parts, taking $x$ as the first function:

\begin{aligned} \int xe^x\,dx &= xe^x - \int 1\cdot e^x\,dx \\ &= xe^x - e^x \end{aligned}

\therefore\ I = x^2e^x - 2\left(xe^x - e^x\right) + C

$\therefore\ I = e^x\left(x^2 - 2x + 2\right) + C$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

Each use of integration by parts with $u$ = the power of $x$ lowers that power by one, while $e^x$ stays $e^x$ when integrated. Starting from $x^2$ ⁠, two uses bring the power down to $x^0 = 1$ ⁠, and then $\int e^x\,dx$ is standard.

Given, to find, and the plan

Given: The function $x^2e^x$ ⁠, defined and continuous for every real $x$ ⁠.
To find: $\int x^2e^x\,dx$ ⁠.
Plan: (1) Name $u$ and $\frac{dv}{dx}$ ⁠. (2) First integration by parts, which leaves $\int xe^x\,dx$ ⁠. (3) Second integration by parts for $\int xe^x\,dx$ ⁠. (4) Put the pieces together. (5) Differentiate back.

Basics you need

1Integration by parts

Section 7.6, equation (1):

\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx,

with $u$ the first function (differentiated) and $\frac{dv}{dx}$ the second (integrated to give $v$ ⁠). Lesson: integration by parts

2Which factor should be $u$ ⁠?

Remark (iii) after Example 17: a power of $x$ is usually the first function. In the memory aid ILATE (Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential; not in the book), Algebraic comes before Exponential.

3The exponential function

$\int e^x\,dx = e^x + C$ (standard formula (xi), Section 7.2) and $\frac{d}{dx}e^x = e^x$ ⁠. Basics: integration undoes differentiation

4Power rule for derivatives

$\frac{d}{dx}x^2 = 2x$ ⁠, $\frac{d}{dx}x = 1$ ⁠. Basics: powers of x

Solution

Name $u$ and $\frac{dv}{dx}$
$\begin{aligned} u &= x^2 \quad (\text{first}), \\ \frac{dv}{dx} &= e^x \quad (\text{second}). \end{aligned}$
Then $\frac{du}{dx} = 2x$ and $v = \int e^x\,dx = e^x$ ⁠.

ILATE: $x^2$ is Algebraic, $e^x$ is Exponential; A comes first. Differentiating $x^2$ lowers the power to $x$ ⁠; integrating $e^x$ gives $e^x$ again, no harder. The opposite choice would raise the power to $\frac{x^3}{3}$ ⁠. No constant is added to $v$ (Remark (ii) after Example 17).

Remark (iii) after Example 17, Section 7.6; standard formula (xi). Lesson: integration by parts
First integration by parts
$\begin{aligned} \int x^2e^x\,dx &= uv - \int v\,\frac{du}{dx}\,dx \\ &= x^2e^x - \int e^x\cdot 2x\,dx \\ &= x^2e^x - 2\int xe^x\,dx. \end{aligned}$
The new integral $\int xe^x\,dx$ has $x$ instead of $x^2$ ⁠: one step simpler, but still a product. So we use integration by parts once more.

Section 7.6, equation (1); constant multiple rule (Section 7.2.1). Lesson: integration by parts
Second integration by parts: $\int xe^x\,dx$

Now take $u = x$ (first) and $\frac{dv}{dx} = e^x$ (second), so $\frac{du}{dx} = 1$ and $v = e^x$ ⁠:
$\begin{aligned} \int xe^x\,dx &= xe^x - \int e^x\cdot 1\,dx \\ &= xe^x - e^x. \end{aligned}$
The same choice for the same reason: differentiating $x$ leaves 1, and $\int e^x\,dx = e^x$ is standard. This is exactly the book's Example 19. We add the constant only once, at the end of Step 4.

Example 19, Section 7.6. Example 19
Put the pieces together
$\begin{aligned} \int x^2e^x\,dx &= x^2e^x - 2\left(xe^x - e^x\right) + C \\ &= x^2e^x - 2xe^x + 2e^x + C \\ &= e^x\left(x^2 - 2x + 2\right) + C. \end{aligned}$
Substitute the result of Step 3 into Step 2, open the bracket (the $-2$ multiplies both terms, so $-2\cdot(-e^x) = +2e^x$ ⁠), and take out the common factor $e^x$ ⁠.

Distributive law; common factor. Lesson: integration by parts
Check: differentiate the answer

By the product rule, with $\frac{d}{dx}(x^2 - 2x + 2) = 2x - 2$ ⁠:
$\begin{aligned} &\frac{d}{dx}\Big[e^x\left(x^2 - 2x + 2\right)\Big] \\ &= e^x\left(x^2 - 2x + 2\right) + e^x(2x - 2) \\ &= e^x\left(x^2 - 2x + 2 + 2x - 2\right) \\ &= x^2e^x. \end{aligned}$
The $-2x$ and $+2x$ cancel, and so do the $+2$ and $-2$ ⁠: only $x^2e^x$ is left, for every real $x$ ⁠.

Product rule; $\frac{d}{dx}e^x = e^x$ ⁠. Basics: integration undoes differentiation

Result

\int x^2e^x\,dx = e^x(x^2 - 2x + 2) + C,

valid for every real $x$ ⁠. (Textbook answer: $e^x(x^2 - 2x + 2) + C$ ⁠, the same.)

Check it with numbers

With $F(x) = e^x(x^2 - 2x + 2)$ ⁠, on $[0, 1]$ ⁠:

\begin{aligned} F(1) - F(0) &= e(1 - 2 + 2) - 1\cdot 2 \\ &= e - 2 = 0.7182818285\ldots \end{aligned}

Numerical integration of $x^2e^x$ over $[0, 1]$ gives $0.7182818285\ldots$ ⁠, the same.

Common mistakes

Dropping the 2. $\frac{d}{dx}x^2 = 2x$ ⁠, so the first use of the formula leaves $2\int xe^x\,dx$ ⁠, not $\int xe^x\,dx$ ⁠.
Sign of the last term. $-2(xe^x - e^x) = -2xe^x + 2e^x$ ⁠: the last term is $+2e^x$ ⁠.
Choosing $u = e^x$ ⁠. Then $v = \frac{x^3}{3}$ and the power of $x$ goes up at every step.

What to take away

For $x^n e^x$ ⁠, use integration by parts $n$ times, each time with the power of $x$ as $u$ ⁠.
The pattern behind the answer: $e^x$ times (the polynomial, minus its derivative, plus its second derivative, and so on). For $x^2$ this is $x^2 - 2x + 2$ ⁠.
Add the constant $C$ once, at the very end.

Every fact used, and where it comes from

Fact	Used in	Source
Integration by parts	Steps 2, 3	Section 7.6, equation (1) (Lesson: integration by parts)
Choice of the first function (ILATE)	Steps 1, 3	Remark (iii) after Example 17, Section 7.6
No constant needed in $v$	Steps 1, 3	Remark (ii) after Example 17, Section 7.6
$\int xe^x\,dx = xe^x - e^x + C$	Step 3	Example 19, Section 7.6
$\int e^x\,dx = e^x + C$	Steps 1, 3	Standard formula (xi), Section 7.2
Constant multiple rule	Step 2	Property (IV), Section 7.2.1
Product rule	Step 5	Chapter 5, Section 5.3

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Exercise 7.6 · Q3: ∫x2ex dx\int x^2e^x\,dx∫x2exdx

Basics you need

Solution

Name uuu and dvdx\frac{dv}{dx}dxdv​

First integration by parts

Second integration by parts: ∫xex dx\int xe^x\,dx∫xexdx

Put the pieces together

Check: differentiate the answer

Every fact used, and where it comes from

Exercise 7.6 · Q3: $\int x^2e^x\,dx$

Name $u$ and $\frac{dv}{dx}$

Second integration by parts: $\int xe^x\,dx$