Chapter 7 · Integrals

Miscellaneous · Q40 (MCQ): $\int_a^b x f(x)\,dx$ when $f(a+b-x)=f(x)$

Miscellaneous ExerciseProperties of definite integrals (P3) · Miscellaneous Exercise

The question

Choose the correct answer.

If $f(a + b - x) = f(x)$ ⁠, then $\displaystyle\int_a^b x\,f(x)\,dx$ is equal to

(A) $\dfrac{a+b}{2}\displaystyle\int_a^b f(b - x)\,dx$ (B) $\dfrac{a+b}{2}\displaystyle\int_a^b f(b + x)\,dx$
(C) $\dfrac{b-a}{2}\displaystyle\int_a^b f(x)\,dx$ (D) $\dfrac{a+b}{2}\displaystyle\int_a^b f(x)\,dx$

Exam answer

Let $I = \displaystyle\int_a^b x\,f(x)\,dx$ ⁠. Using $\displaystyle\int_a^b g(x)\,dx$ $= \displaystyle\int_a^b g(a + b - x)\,dx$ and $f(a + b - x) = f(x)$ ⁠,

\begin{aligned} I &= \int_a^b (a + b - x)f(x)\,dx \\ &= (a + b)\int_a^b f(x)\,dx - I \\ \Rightarrow\ 2I &= (a + b)\int_a^b f(x)\,dx \end{aligned}

Answer: (D) $\dfrac{a + b}{2}\displaystyle\int_a^b f(x)\,dx$

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

The one idea this problem is built on

Property P3 says an integral over $[a, b]$ is unchanged when $x$ is replaced by $a + b - x$ (reading the interval backwards). Doing that to $I = \int_a^b x f(x)\,dx$ changes the factor $x$ into $a + b - x$ but, by the given symmetry, leaves $f$ alone. So $I$ $= \int_a^b (a + b - x)f(x)\,dx$ $= (a + b)\int_a^b f - I$ ⁠, and solving gives option (D).

Given, to find, and the plan

Given: A function $f$ ⁠, continuous on $[a, b]$ (needed for the integrals to make sense), with $f(a + b - x) = f(x)$ for all $x$ in $[a, b]$ ⁠.
To find: Which option equals $I = \int_a^b x\,f(x)\,dx$ ⁠.
Plan: (1) Read the condition as a symmetry. (2) Apply P3. (3) Solve for $I$ ⁠. (4) Test every option on a concrete example.

Basics you need

1Property P3 (Section 7.10)

$\int_a^b g(x)\,dx = \int_a^b g(a + b - x)\,dx$ for any $g$ continuous on $[a, b]$ ⁠. It is proved with the substitution $t = a + b - x$ ⁠. Lesson: properties of definite integrals

2Mirror symmetry

$x$ and $a + b - x$ are mirror images in the midpoint $\frac{a+b}{2}$ of $[a, b]$ ⁠. So $f(a + b - x) = f(x)$ says the graph of $f$ is symmetric about the line $x = \frac{a+b}{2}$ ⁠. Basics: symmetry and mirror images

3Linearity

$\int_a^b (c - x)f(x)\,dx$ $= c\int_a^b f(x)\,dx - \int_a^b x f(x)\,dx$ for a constant $c$ ⁠. Lesson: definite integrals

Solution

What the condition means

The midpoint of $x$ and $a + b - x$ is $\frac{x + (a + b - x)}{2} = \frac{a+b}{2}$ ⁠, the midpoint of $[a, b]$ ⁠. So the condition $f(a + b - x) = f(x)$ says that $f$ takes equal values at points placed symmetrically about $\frac{a+b}{2}$ ⁠.

Seeing the condition as a symmetry suggests the answer: a symmetric weight $f$ balances at its centre $\frac{a+b}{2}$ ⁠, so " $x$ weighted by $f$ " should average to $\frac{a+b}{2}$ ⁠. Option (D) says exactly that: $\int x f = \frac{a+b}{2}\int f$ ⁠. The next steps prove it.

Basics: symmetry and mirror images
Apply P3 to $g(x) = x\,f(x)$
$\begin{aligned} I &= \int_a^b x\,f(x)\,dx \\ &= \int_a^b (a + b - x)\,f(a + b - x)\,dx \\ &= \int_a^b (a + b - x)\,f(x)\,dx . \end{aligned}$
The second line is P3: every $x$ in $g(x) = x f(x)$ is replaced by $a + b - x$ ⁠, including the one inside $f$ ⁠. The third line uses the given condition $f(a + b - x) = f(x)$ ⁠. P3 applies because $g$ is continuous on $[a, b]$ (a product of continuous functions).

P3, Section 7.10. Lesson: properties of definite integrals
Split and solve for $I$
$\begin{aligned} I &= (a + b)\int_a^b f(x)\,dx \\ &\quad - \int_a^b x\,f(x)\,dx \\ &= (a + b)\int_a^b f(x)\,dx - I . \end{aligned}$
Adding $I$ to both sides, $2I = (a + b)\int_a^b f(x)\,dx$ ⁠, so
$\int_a^b x\,f(x)\,dx = \frac{a + b}{2}\int_a^b f(x)\,dx .$
This is option (D).

The original integral reappears on the right with a minus sign, which lets us solve for it like an unknown in an equation (the same trick as in Examples 30 and 32 of Section 7.10). $I$ is a finite number because $x f(x)$ is continuous on $[a, b]$ ⁠.

Linearity of the definite integral. Lesson: definite integrals
Examine every option on a concrete example

Take $a = 1$ ⁠, $b = 3$ and $f(x) = (x - 2)^2$ ⁠. The condition holds: $f(4 - x) = (2 - x)^2 = (x - 2)^2 = f(x)$ ⁠. Then
$\begin{aligned} I &= \int_1^3 x(x - 2)^2\,dx \\ &= \int_{-1}^{1}(u + 2)u^2\,du \\ &= 0 + 2\cdot\tfrac23 = \tfrac43 , \end{aligned}$
with $u = x - 2$ (the $u^3$ part is odd and gives 0 by P7 (ii)). Also $\int_1^3 (x - 2)^2\,dx = \frac23$ and $\frac{a+b}{2} = 2$ ⁠, $\frac{b-a}{2} = 1$ ⁠.
- (A) $2\int_1^3 f(3 - x)\,dx$ $= 2\int_1^3 (1 - x)^2\,dx$ $= 2\cdot\frac83$ $= \frac{16}{3}$ $\ne \frac43$ ⁠. Wrong.
- (B) $2\int_1^3 f(3 + x)\,dx$ $= 2\int_1^3 (x + 1)^2\,dx$ $= 2\cdot\frac{56}{3}$ $= \frac{112}{3}$ $\ne \frac43$ ⁠. Wrong.
- (C) $1\cdot\frac23 = \frac23 \ne \frac43$ ⁠. Wrong. (It has the length $b - a$ where the midpoint sum $a + b$ belongs.)
- (D) $2\cdot\frac23 = \frac43$ ✓. Correct, as Step 3 proves in general.
One example where an option gives the wrong number is enough to reject it, because the option claims to hold for every such $f$ ⁠. A constant $f$ would not do as the test: with $f = 1$ ⁠, (A) and (B) also give the right value, since then $f(b - x) = f(b + x) = f(x)$ ⁠; only (C) would be rejected. A non-constant symmetric $f$ and $a \ne 0$ are needed to separate (A) from (D). (When $a = 0$ ⁠, P4 makes (A) the same as (D).)

$\int_1^3(1 - x)^2dx$ $= \big[\frac{(x-1)^3}{3}\big]_1^3$ $= \frac83$ ⁠; $\int_1^3(x+1)^2dx = \frac{64 - 8}{3} = \frac{56}{3}$ ⁠.

Result

If $f(a + b - x) = f(x)$ ⁠, then $\int_a^b x\,f(x)\,dx = \frac{a + b}{2}\int_a^b f(x)\,dx .$ The correct option is (D). (Textbook answer: D.)

Check it with numbers

For $f(x) = (x - 2)^2$ on $[1, 3]$ ⁠, numerical integration (adaptive quadrature) gives $\int_1^3 x f(x)\,dx \approx 1.3333333333$ and $\int_1^3 f(x)\,dx \approx 0.6666666667$ ⁠; twice the second is $\frac43 \approx 1.3333333333$ ⁠, the first ✓. The options compute to (A) 5.3333333333, (B) 37.3333333333, (C) 0.6666666667, (D) 1.3333333333.

A second example, $f(x) = \sin x$ on $[0, \pi]$ (where $\sin(\pi - x) = \sin x$ ⁠): $\int_0^\pi x\sin x\,dx = \pi \approx 3.1415927$ ⁠, and $\frac{\pi}{2}\int_0^\pi\sin x\,dx = \frac{\pi}{2}\cdot 2 = \pi$ ✓.

Common mistakes

Applying P3 only to $f$ and forgetting to replace the factor $x$ by $a + b - x$ ⁠.
Confusing $a + b$ with $b - a$ : the symmetry is about the midpoint $\frac{a+b}{2}$ ⁠, not the half-length $\frac{b-a}{2}$ ⁠.
Testing with a constant $f$ : it cannot tell (A) or (B) from (D).

What to take away

If $f$ is symmetric about the midpoint of $[a, b]$ ⁠, then $\int_a^b x f(x)\,dx = \frac{a+b}{2}\int_a^b f(x)\,dx$ ⁠.
The special case $a = 0$ ⁠, $b = \pi$ ⁠, $f(\pi - x) = f(x)$ gives $\int_0^\pi x f(x)\,dx = \frac{\pi}{2}\int_0^\pi f(x)\,dx$ ⁠, used in Example 30.

Every fact used, and where it comes from

Fact	Used in	Source
P3: $\int_a^b g(x)\,dx = \int_a^b g(a + b - x)\,dx$	Step 2	Section 7.10, property P3 (Lesson: properties)
Linearity of the definite integral	Step 3	Properties (III), (IV), Section 7.2.1, with Theorem 2, Section 7.8.3
Solving for the integral that reappears	Step 3	Examples 30 and 32, Section 7.10
P7 (ii); P4	Step 4	Section 7.10
All values	Step 4, Check	Recomputed (exact sympy and mpmath quadrature) in the site's verification script

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Miscellaneous · Q40 (MCQ): ∫abxf(x) dx\int_a^b x f(x)\,dx∫ab​xf(x)dx when f(a+b−x)=f(x)f(a+b-x)=f(x)f(a+b−x)=f(x)

Basics you need

Solution

What the condition means

Apply P3 to g(x)=x f(x)g(x) = x\,f(x)g(x)=xf(x)

Split and solve for III

Examine every option on a concrete example

Every fact used, and where it comes from

Miscellaneous · Q40 (MCQ): $\int_a^b x f(x)\,dx$ when $f(a+b-x)=f(x)$

Apply P3 to $g(x) = x\,f(x)$

Split and solve for $I$