Sample 2025-26 Class 12 Maths Q11 Solution: ∫ from a to b of x f(x) dx when f(a + b − x) = f(x)

The question

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

(A) $\frac{a + b}{2}\int_a^b f(b - x)\,dx$

(B) $\frac{a + b}{2}\int_a^b f(a - x)\,dx$

(C) $\frac{b - a}{2}\int_a^b f(x)\,dx$

(D) $\frac{a + b}{2}\int_a^b f(x)\,dx$

Exam answer

Let $I = \int_a^b x\,f(x)\,dx$ ⁠.

Using $\int_a^b g(x)\,dx = \int_a^b g(a + b - x)\,dx$ ⁠,

Why?

This property holds for every integrable $g$ ⁠: putting $x = a + b - t$ maps the interval $[a, b]$ onto itself, so the value does not change.

Here $g(x) = x\,f(x)$ ⁠, so

\begin{aligned} &g(a + b - x) \\ &= (a + b - x)\,f(a + b - x) \end{aligned}

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

(using $f(a + b - x) = f(x)$ ⁠)

Why?

The question gives $f(a + b - x) = f(x)$ ⁠, so the $f$ part goes back to $f(x)$ ⁠. Only the factor $x$ has changed, to $a + b - x$ ⁠.

\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}

Why?

Split the bracket:

\begin{aligned} &(a + b - x)\,f(x) \\ &= (a + b)\,f(x) - x\,f(x) \end{aligned}

$a + b$ is a constant, so it comes outside the integral. The last integral is $I$ itself.

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

Why?

Add $I$ to both sides, then divide by 2.

Watch out: the factor is $\frac{a + b}{2}$ ⁠, the midpoint of $[a, b]$ ⁠, not $\frac{b - a}{2}$ as in (C).

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

Where this comes from

Chapter 7, Integrals: Definite integrals: the f(a + b − x) property.

Question type: Limits from 0 to $a$

How to spot it: click to see

Type 12. Limits from 0 to $a$ ⁠: the $f(a - x)$ trick

Spot it: replacing $x$ by $a - x$ swaps the parts of the question, like sin and cos when $a$ is $\frac{\pi}{2}$ ⁠.

Here: Replacing $x$ by $a + b - x$ leaves $f$ alone, since $f(a + b - x) = f(x)$ ⁠, but turns the factor $x$ into $a + b - x$ ⁠. So $I = (a + b)\int_a^b f(x)\,dx - I$ ⁠, and halving $2I$ gives option (D).

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Q11: ∫abx f(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)

Q11: $\int_a^b x\,f(x)\,dx$ when $f(a + b - x) = f(x)$