Chapter 7 · Integrals

Miscellaneous · Q33: $\int_0^1 x\,e^x\,dx=1$

Miscellaneous ExerciseDefinite integrals by parts · Miscellaneous Exercise

The question

Prove that

\int_0^1 x\,e^x\,dx = 1

Exam answer

Let $I = \displaystyle\int_0^1 x\,e^x\,dx$ ⁠.

Taking $x$ as the first function and $e^x$ as the second, by parts:

\begin{aligned} I &= \Big[x\,e^x\Big]_0^1 - \int_0^1 1\cdot e^x\,dx \\ &= \Big[x\,e^x - e^x\Big]_0^1 = \Big[(x - 1)e^x\Big]_0^1 \\ &= (1 - 1)e^1 - (0 - 1)e^0 \\ &= 0 - (-1) \end{aligned}

$\therefore\ I = 1 = \text{RHS}$ ⁠. Hence proved.

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

The one idea this problem is built on

The integrand is a product of a polynomial $x$ and an exponential $e^x$ ⁠. Integration by parts with $x$ as the first function removes the $x$ (its derivative is 1) while $e^x$ integrates to itself. What is left, $\int e^x\,dx$ ⁠, is immediate. The antiderivative $(x - 1)e^x$ happens to vanish at the upper limit, which is why the answer is so clean.

Given, to show, and the plan

Given: $I = \displaystyle\int_0^1 x\,e^x\,dx$ ⁠.
To show: $I = 1$ ⁠.
Plan: (1) Continuity. (2) Choose the first and second functions. (3) Integrate by parts. (4) Check by differentiating. (5) Evaluate from 0 to 1.

Basics you need

1Integration by parts (Section 7.6)

$\int u\,dv = uv - \int v\,du$ ⁠: "first function times integral of the second, minus the integral of (derivative of the first times integral of the second)". Lesson: integration by parts

2The exponential

$\frac{d}{dx}e^x = e^x$ and $\int e^x\,dx = e^x + C$ (standard formula (xi), Section 7.2); $e^0 = 1$ ⁠. Basics: integration undoes differentiation

3Theorem 2, Section 7.8.3

$\int_a^b f(x)\,dx = F(b) - F(a)$ for continuous $f$ with antiderivative $F$ ⁠. Lesson: definite integrals

Solution

The integrand is continuous on $[0, 1]$

$x$ and $e^x$ are continuous on all of $\R$ ⁠, so their product is continuous on $[0, 1]$ ⁠, and Theorem 2 applies.

Every definite integral starts with this check (Remark (iv) after Theorem 2). Here it is immediate.

Lesson: definite integrals and the fundamental theorem
Choose the first function: $u = x$
$\begin{aligned} u &= x, & du &= dx, \\ dv &= e^x\,dx, & v &= e^x . \end{aligned}$
Remark (iii) of Section 7.6: when one factor is a polynomial and the other is not inverse trigonometric or logarithmic, take the polynomial as the first function. Differentiating $x$ gives 1, so the new integral has no $x$ left in it. The other choice, $u = e^x$ ⁠, $dv = x\,dx$ ⁠, would produce $\int\frac{x^2}{2}e^x\,dx$ ⁠, which is worse. (The book makes the same choice in Example 19.)

Lesson: integration by parts
Integrate by parts
$\begin{aligned} \int x\,e^x\,dx &= x\,e^x - \int e^x\,dx \\ &= x\,e^x - e^x + C \\ &= (x - 1)e^x + C . \end{aligned}$
Call $F(x) = (x - 1)e^x$ ⁠. It is valid for every real $x$ ⁠.

This is $uv - \int v\,du$ with $uv = xe^x$ and $\int v\,du = \int e^x\,dx = e^x$ ⁠. No constant is needed when finding $v$ (Remark (ii), Section 7.6).

Section 7.6, formula (1); standard formula (xi), Section 7.2. Lesson: integration by parts
Check: differentiate $F$ back
$F'(x) = 1\cdot e^x + (x - 1)e^x = x\,e^x \ ✓$
The product rule gives two terms; the $e^x$ and $-e^x$ cancel, leaving the integrand.

Product rule (Chapter 5); $\frac{d}{dx}e^x = e^x$ ⁠. Basics: integration undoes differentiation
Evaluate between 0 and 1

$F(1) = (1 - 1)e^1 = 0$ ⁠, and $F(0) = (0 - 1)e^0 = -1$ ⁠.
$I = F(1) - F(0) = 0 - (-1) = 1 .$
This is the required result.

Theorem 2. The upper value is 0 because of the factor $x - 1$ ⁠; the lower value is $-1$ because $e^0 = 1$ ⁠. Size check: $xe^x$ rises from 0 to $e \approx 2.718$ on $[0, 1]$ and bends upwards, so the area under it is well below the triangle-like estimate $\frac12\cdot 2.718 \approx 1.36$ ⁠. The value 1 fits.

Basics: area as a sum of thin strips

Result

$\begin{aligned} \int_0^1 x\,e^x\,dx &= \Big[(x - 1)e^x\Big]_0^1 \\ &= 0 - (-1) = 1, \end{aligned}$ as was to be proved.

Check it with numbers

Numerical integration (adaptive quadrature) of $xe^x$ over $[0, 1]$ gives 1.0000000000 ✓.

Derivative spot check at $x = 0.5$ ⁠: $F(0.501) \approx -0.82353504$ and $F(0.499) \approx -0.82518376$ ⁠, so the slope is about $\frac{0.00164872}{0.002} = 0.82436$ (that is, $F(0.501) - F(0.499)$ over 0.002), and $0.5\,e^{0.5} \approx 0.8243606$ ✓.

Common mistakes

Choosing $u = e^x$ : the power of $x$ goes up instead of down.
Sign of the second term: it is $xe^x - \int e^x\,dx = xe^x - e^x$ ⁠, not $xe^x + e^x$ ⁠.
Taking $F(0) = 0$ because "everything has an $x$ in it". $F(0) = (0 - 1)e^0 = -1$ ⁠.

What to take away

$\int x\,e^x\,dx = (x - 1)e^x + C$ ⁠. More generally, by parts with the polynomial first, $\int p(x)e^x\,dx$ is (polynomial) $\cdot e^x$ ⁠.
This is also an example of $\int e^x\big(f(x) + f'(x)\big)dx = e^x f(x) + C$ (Section 7.6.1) with $f(x) = x - 1$ ⁠: $f + f' = x - 1 + 1 = x$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
A definite integral needs $f$ continuous on $[a, b]$	Step 1	Theorem 2 and Remark (iv), Section 7.8.3
Integration by parts; choice of first function	Steps 2, 3	Section 7.6, formula (1), Remarks (ii), (iii); Example 19
$\int e^x\,dx = e^x + C$	Step 3	Standard formula (xi), Section 7.2
Product rule	Step 4	Chapter 5, Section 5.3
$F(b) - F(a)$	Step 5	Theorem 2, Section 7.8.3
$\int e^x(f + f')\,dx = e^x f + C$	Take-away	Section 7.6.1

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Miscellaneous · Q33: ∫01x ex dx=1\int_0^1 x\,e^x\,dx=1∫01​xexdx=1

Basics you need

Solution

The integrand is continuous on [0,1][0, 1][0,1]

Choose the first function: u=xu = xu=x

Integrate by parts

Check: differentiate FFF back

Evaluate between 0 and 1

Every fact used, and where it comes from

Miscellaneous · Q33: $\int_0^1 x\,e^x\,dx=1$

The integrand is continuous on $[0, 1]$

Choose the first function: $u = x$

Check: differentiate $F$ back