Board 2024 Class 12 Maths Q6 Solution: ∫ₐᵇ f(x) dx = ∫ₐᵇ f(a + b − x) dx

The question

$\displaystyle\int_a^b f(x)\,dx$ is equal to :

(A) $\displaystyle\int_a^b f(a - x)\,dx$

(B) $\displaystyle\int_a^b f(a + b - x)\,dx$

(C) $\displaystyle\int_a^b f(x - (a + b))\,dx$

(D) $\displaystyle\int_a^b f((a - x) + (b - x))\,dx$

Exam answer

Put $x = a + b - t$ ⁠. Then $dx = -dt$ ⁠.

When $x = a$ ⁠, $t = b$ ⁠; when $x = b$ ⁠, $t = a$ ⁠.

Why?

This substitution reflects the interval $[a, b]$ end to end, so it turns $f(x)$ into $f(a + b - t)$ ⁠, the form in option (B).

Differentiating $x = a + b - t$ with respect to $t$ gives $dx/dt = -1$ ⁠, because $a$ and $b$ are constants.

For the limits, solve $t = a + b - x$ ⁠: $x = a$ gives $t = b$ ⁠, and $x = b$ gives $t = a$ ⁠. The limits swap.

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

Why?

Swapping the limits of a definite integral changes its sign:

\int_b^a g(t)\,dt = -\int_a^b g(t)\,dt

This cancels the minus sign from $dx = -dt$ ⁠.

In the last line, $t$ is renamed $x$ ⁠. A definite integral does not depend on the letter used for the variable.

Check with $f(x) = x$ ⁠, $a = 0$ ⁠, $b = 1$ ⁠: the left side is $\int_0^1 x\,dx = \frac12$ ⁠.

\begin{aligned} \text{(A)}: \int_0^1 (-x)\,dx &= -\tfrac12 \\ \text{(B)}: \int_0^1 (1 - x)\,dx &= \tfrac12 \\ \text{(C)}: \int_0^1 (x - 1)\,dx &= -\tfrac12 \\ \text{(D)}: \int_0^1 (1 - 2x)\,dx &= 0 \end{aligned}

Why?

With $a = 0$ and $b = 1$ ⁠, the four integrands are $f(0 - x) = -x$ ⁠, $f(1 - x) = 1 - x$ ⁠, $f(x - 1) = x - 1$ and $f(1 - 2x) = 1 - 2x$ ⁠.

For (D), the inside is $(0 - x) + (1 - x) = 1 - 2x$ ⁠.

Each integral uses $\int x\,dx = \frac{x^2}{2}$ and $\int 1\,dx = x$ ⁠. Put in the top limit 1, then take away the bottom limit 0:

\begin{aligned} &\int_0^1 x\,dx = \Big[\tfrac{x^2}{2}\Big]_0^1 \\ &= \tfrac{(1)^2}{2} - \tfrac{(0)^2}{2} = \tfrac12 \\[4pt] &\text{(A)}: \int_0^1 (-x)\,dx \\ &= -\Big[\tfrac{x^2}{2}\Big]_0^1 \\ &= -\big(\tfrac12 - 0\big) = -\tfrac12 \\[4pt] &\text{(B)}: \int_0^1 (1 - x)\,dx \\ &= \Big[x - \tfrac{x^2}{2}\Big]_0^1 \\ &= \big(1 - \tfrac12\big) - 0 = \tfrac12 \\[4pt] &\text{(C)}: \int_0^1 (x - 1)\,dx \\ &= \Big[\tfrac{x^2}{2} - x\Big]_0^1 \\ &= \big(\tfrac12 - 1\big) - 0 = -\tfrac12 \\[4pt] &\text{(D)}: \int_0^1 (1 - 2x)\,dx \\ &= \Big[x - x^2\Big]_0^1 \\ &= (1 - 1) - 0 = 0 \end{aligned}

A single example where an option fails is enough to rule that option out. Only (B) matches here, as the property says it must.

Answer: (B) $\displaystyle\int_a^b f(a + b - x)\,dx$

Where this comes from

Chapter 7, Integrals: Definite integrals: properties.

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Q6: ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx∫ab​f(x)dx=∫ab​f(a+b−x)dx