Board 2025 Class 12 Maths Q3 Solution: Value of ∫ from −1 to 1 of |x|/x dx

The question

$\displaystyle\int_{-1}^{1} \frac{|x|}{x}\,dx$ ⁠, $x \ne 0$ is equal to

(A) $-1$

(B) $0$

(C) $1$

(D) $2$

Exam answer

For $x < 0$ ⁠, $|x| = -x$ ⁠; for $x > 0$ ⁠, $|x| = x$ ⁠.

So $\frac{|x|}{x} = -1$ for $x < 0$ ⁠, and $\frac{|x|}{x} = 1$ for $x > 0$ ⁠.

Why?

The modulus has a different formula on each side of 0:

|x| = \begin{cases} x, & x \ge 0 \\ -x, & x < 0 \end{cases}

Dividing by x gives $\frac{-x}{x} = -1$ on the left of 0 and $\frac{x}{x} = 1$ on the right.

\begin{aligned} &\int_{-1}^{1} \frac{|x|}{x}\,dx \\ &= \int_{-1}^{0} (-1)\,dx + \int_{0}^{1} 1\,dx \\ &= \Big[-x\Big]_{-1}^{0} + \Big[x\Big]_{0}^{1} \\ &= (0 - 1) + (1 - 0) \\ &= -1 + 1 = 0 \end{aligned}

Why?

The formula changes at $x = 0$ ⁠, so split the integral there, using

\begin{aligned} \int_a^b f\,dx &= \int_a^c f\,dx \\ &\quad + \int_c^b f\,dx \end{aligned}

On each piece the integrand is a constant. At $x = 0$ the integrand is not defined, but a single point does not change the integral.

For $\Big[-x\Big]_{-1}^{0}$ ⁠: the value at 0 is 0, and the value at $-1$ is $-(-1) = 1$ ⁠.

Check: $f(x) = \frac{|x|}{x}$ is odd, since $f(-x) = -f(x)$ ⁠. The integral of an odd function from $-a$ to $a$ is 0.

Watch out: forgetting the minus sign on the left half gives $1 + 1 = 2$ ⁠, option (D).

Answer: (B) $0$

Where this comes from

Chapter 7, Integrals: Definite integral of |x|/x.

Question type: Limits from $-a$ to $a$

How to spot it: click to see

Type 11. Limits from $-a$ to $a$ ⁠: check odd or even

Spot it: the two limits are the same number with opposite signs.

Here: The limits are $-1$ and $1$ ⁠, and $\frac{|x|}{x}$ is $-1$ left of 0 and $1$ right of it, so $f(-x) = -f(x)$ ⁠. It is odd, so the answer is 0; splitting at 0 gives the same $-1 + 1$ ⁠.

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Q3: Value of $\int_{-1}^{1} \frac{|x|}{x}\,dx$