$\int_{-1}^{1} (1-|x|)\, dx$ is equal to: (A) $2 \int_{0}^{1} (1+x)\, dx$ (B) $2 \int_{-1}^{0} (1+x)\, dx$ (C) $0$ (D) $2 \int_{-1}^{0} (1-x)\, dx$

> [!TLDR] > The integral $\int_{-1}^{1} (1-|x|)\, dx$ evaluates to $1$. Using symmetry, the integrand is even, so the integral equals $2\int_{0}^{1} (1-x)\, dx$, which matches option **(B)** after a variable substitution. The key here is the absolute value function $|x|$. It makes the integrand $1-|x|$ an **even function** — symmetric about the $y$-axis. For any even function $f(x)$, we have the property: $$ \int_{-a}^{a} f(x)\, dx = 2 \int_{0}^{a} f(x)\, dx $$ This is because the area from $-a$ to $0$ is a mirror image of the area from $0$ to $a$. So instead of dealing with the absolute value directly, we can exploit this symmetry to simplify the integral. Let’s work through it step by step. 1. **Identify the symmetry.** The function $f(x) = 1 - |x|$ is even because $|x|$ is even, and subtracting an even function from a constant keeps it even. So: $$ \int_{-1}^{1} (1-|x|)\, dx = 2 \int_{0}^{1} (1-|x|)\, dx $$ But for $x \ge 0$, $|x| = x$. Therefore: $$ \int_{-1}^{1} (1-|x|)\, dx = 2 \int_{0}^{1} (1-x)\, dx $$ 2. **Evaluate the integral directly (to know the target value).** Compute: $$ 2 \int_{0}^{1} (1-x)\, dx = 2 \left[ x - \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1 $$ So the integral equals $1$. Now we check which option also gives $1$. 3. **Examine each option.** - **(A)** $2 \int_{0}^{1} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 + \frac{1}{2} \right) = 3$ — not equal to $1$. - **(B)** $2 \int_{-1}^{0} (1+x)\, dx$ — let’s evaluate this carefully. For $x$ from $-1$ to $0$, $1+x$ is positive? Actually at $x=-1$, $1+(-1)=0$; at $x=0$, $1+0=1$. So: $$ 2 \int_{-1}^{0} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{-1}^{0} = 2 \left( 0 - \left( -1 + \frac{1}{2} \right) \right) = 2 \left( 0 - \left( -\frac{1}{2} \right) \right) = 2 \cdot \frac{1}{2} = 1 $$ This matches! - **(C)** $0$ — clearly not $1$. - **(D)**

∫-1¹ (1-|x|) dx is equal to: (A) 2 ∫₀¹ (1+x) dx (B) 2 ∫-1⁰ (1+x) dx (C) 0 (D) 2 ∫-1⁰ (1-x) dx

Question

\int_{-1}^{1} (1-|x|)\, dx

CBSECBSE Class XII Board 2026MCQ· 1mImportance★★★★★

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Concept understanding — Definite Integral Symmetry

Definite Integral Symmetry: The Shortcut That Saves You Work

Asked to find the area under $f(x) = x^3$ from $x = -2$ to $x = 2$ ? You could integrate directly — but there's a much faster way if you notice the symmetry of the graph.

The Intuition: What Does "Symmetry" Mean Here?

A function can be symmetric about the y-axis (like $x^2$ or $\cos x$ ) or about the origin (like $x^3$ or $\sin x$ ). When you integrate over a symmetric interval — from $-a$ to $a$ — these symmetries create a predictable cancellation or doubling.

Even functions (symmetric about the y-axis): $f(-x) = f(x)$ — think $x^2$ , $x^4$ , $\cos x$ , $|x|$ . The left side mirrors the right, so the area from $-a$ to $0$ equals the area from $0$ to $a$ . The total is double the area on one side.

Odd functions (symmetric about the origin): $f(-x) = -f(x)$ — think $x^3$ , $x^5$ , $\sin x$ , $\tan x$ . The left side is the negative mirror of the right, so every positive area on the right is cancelled by an equal negative area on the left. The total is zero.

Note

This only works when the limits are symmetric about zero — from $-a$ to $a$ . For $[0, a]$ or $[1, 3]$ , symmetry doesn't help directly.

The Precise Statement

Let $f$ be continuous on $[-a, a]$ .

If $f$ is even ( $f(-x) = f(x)$ ), then $\;\int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx$
If $f$ is odd ( $f(-x) = -f(x)$ ), then $\;\int_{-a}^{a} f(x) \, dx = 0$

\int_{-a}^{a} f(x) \, dx = \begin{cases} 2 \int_{0}^{a} f(x) \, dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}

Why Does This Work? (A Quick Proof)

Split at zero:

\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx

For the first term substitute $u = -x$ ( $dx = -du$ ; $x=-a \to u=a$ , $x=0 \to u=0$ ):

\int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(-u) \, du

Now use symmetry:

If $f$ is even, $f(-u) = f(u)$ , so this becomes $\int_{0}^{a} f(u) \, du$ ; adding the second term gives $2 \int_{0}^{a} f(x) \, dx$ .
If $f$ is odd, $f(-u) = -f(u)$ , so this becomes $-\int_{0}^{a} f(u) \, du$ ; adding the second term gives $0$ .

Common Mistakes to Avoid

Watch out

Don't assume symmetry without checking. A "balanced"-looking function need not be even or odd — e.g. $f(x) = x^2 + x$ is neither, so these formulas don't apply. Always verify $f(-x) = f(x)$ or $f(-x) = -f(x)$ for all $x$ .

Watch out

The interval must be $[-a, a]$ . If the limits are $[-2, 3]$ , symmetry doesn't apply directly — split the integral or shift variables.

Examples to Cement the Idea

Example 1: $\int_{-3}^{3} x^4 \, dx$ — $x^4$ is even, so

$\int_{-3}^{3} x^4 \, dx = 2 \int_{0}^{3} x^4 \, dx = 2 \left[ \frac{x^5}{5} \right]_{0}^{3} = 2 \cdot \frac{243}{5} = \frac{486}{5}$ …

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The integral $\int_{-1}^{1} (1-|x|)\, dx$ evaluates to $1$ . Using symmetry, the integrand is even, so the integral equals $2\int_{0}^{1} (1-x)\, dx$ , which matches option (B) after a variable substitution.

The key here is the absolute value function $|x|$ . It makes the integrand $1-|x|$ an even function — symmetric about the $y$ -axis. For any even function $f(x)$ , we have the property:

\int_{-a}^{a} f(x)\, dx = 2 \int_{0}^{a} f(x)\, dx

This is because the area from $-a$ to $0$ is a mirror image of the area from $0$ to $a$ . So instead of dealing with the absolute value directly, we can exploit this symmetry to simplify the integral.

Let’s work through it step by step.

Identify the symmetry. The function $f(x) = 1 - |x|$ is even because $|x|$ is even, and subtracting an even function from a constant keeps it even. So:

\int_{-1}^{1} (1-|x|)\, dx = 2 \int_{0}^{1} (1-|x|)\, dx

But for $x \ge 0$ , $|x| = x$ . Therefore:

\int_{-1}^{1} (1-|x|)\, dx = 2 \int_{0}^{1} (1-x)\, dx

Evaluate the integral directly (to know the target value). Compute:

2 \int_{0}^{1} (1-x)\, dx = 2 \left[ x - \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1

So the integral equals $1$ . Now we check which option also gives $1$ .

Examine each option.
- (A) $2 \int_{0}^{1} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 + \frac{1}{2} \right) = 3$ — not equal to $1$ .
- (B) $2 \int_{-1}^{0} (1+x)\, dx$ — let’s evaluate this carefully. For $x$ from $-1$ to $0$ , $1+x$ is positive? Actually at $x=-1$ , $1+(-1)=0$ ; at $x=0$ , $1+0=1$ . So:

2 \int_{-1}^{0} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{-1}^{0} = 2 \left( 0 - \left( -1 + \frac{1}{2} \right) \right) = 2 \left( 0 - \left( -\frac{1}{2} \right) \right) = 2 \cdot \frac{1}{2} = 1

 This matches!

(C) $0$ — clearly not $1$ .
(D) $2 \int_{-1}^{0} (1-x)\, dx$ — for $x$ from $-1$ to $0$ , $1-x$ ranges from $2$ to $1$ , so: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 45 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$\displaystyle\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx$
›Reveal solutionSolution
The integrand is an odd function (satisfies $f(-x) = -f(x)$ ), so its integral over the symmetric interval $[-1, 1]$ vanishes. The value is $0$ .

When you see an integral over a symmetric interval like $[-1, 1]$ , the first instinct should be to check whether the integrand has any special symmetry. Two types matter:

Even function: $f(-x) = f(x)$ for all $x$ . Then $\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$ .

Odd function: $f(-x) = -f(x)$ for all $x$ . Then $\int_{-a}^{a} f(x)\,dx = 0$ .

The second property is powerful: if you can show the integrand is odd, the integral is zero immediately, no computation needed. The geometric reason is that the area under the curve from $-a$ to $0$ exactly cancels the area from $0$ to $a$ .

Let's check our integrand $f(x) = \frac{x^3}{x^2 + 2|x| + 1}$ .

Step-by-step verification

Compute $f(-x)$ .

Substitute $-x$ for $x$ :

$f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} = \frac{-x^3}{x^2 + 2|x| + 1}$
Notice that $(-x)^3 = -x^3$ , $(-x)^2 = x^2$ , and crucially $|-x| = |x|$ (the absolute value kills the sign).

Compare with $f(x)$ .

We have

$f(-x) = \frac{-x^3}{x^2 + 2|x| + 1} = -\frac{x^3}{x^2 + 2|x| + 1} = -f(x)$
This is exactly the definition of an odd function.

Apply the symmetry property.

Since $f(x)$ is odd and the interval $[-1, 1]$ is symmetric about the origin, …
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CBSE 2026Set 65/3/11 markMCQ
$\int_{-1}^{1} (1-|x|)\, dx$
›Reveal solutionSolution
The integral $\int_{-1}^{1} (1-|x|)\, dx$ evaluates to $1$ . Using symmetry, the integrand is even, so the integral equals $2\int_{0}^{1} (1-x)\, dx$ , which matches option (B) after a variable substitution.

The key here is the absolute value function $|x|$ . It makes the integrand $1-|x|$ an even function — symmetric about the $y$ -axis. For any even function $f(x)$ , we have the property:
$\int_{-a}^{a} f(x)\, dx = 2 \int_{0}^{a} f(x)\, dx$
This is because the area from $-a$ to $0$ is a mirror image of the area from $0$ to $a$ . So instead of dealing with the absolute value directly, we can exploit this symmetry to simplify the integral.

Let’s work through it step by step.

Identify the symmetry. The function $f(x) = 1 - |x|$ is even because $|x|$ is even, and subtracting an even function from a constant keeps it even. So:

$\int_{-1}^{1} (1-|x|)\, dx = 2 \int_{0}^{1} (1-|x|)\, dx$
But for $x \ge 0$ , $|x| = x$ . Therefore:
$\int_{-1}^{1} (1-|x|)\, dx = 2 \int_{0}^{1} (1-x)\, dx$

Evaluate the integral directly (to know the target value). Compute:

$2 \int_{0}^{1} (1-x)\, dx = 2 \left[ x - \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1$
So the integral equals $1$ . Now we check which option also gives $1$ .

Examine each option.

(A) $2 \int_{0}^{1} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 + \frac{1}{2} \right) = 3$ — not equal to $1$ .

(B) $2 \int_{-1}^{0} (1+x)\, dx$ — let’s evaluate this carefully. For $x$ from $-1$ to $0$ , $1+x$ is positive? Actually at $x=-1$ , $1+(-1)=0$ ; at $x=0$ , $1+0=1$ . So:

$2 \int_{-1}^{0} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{-1}^{0} = 2 \left( 0 - \left( -1 + \frac{1}{2} \right) \right) = 2 \left( 0 - \left( -\frac{1}{2} \right) \right) = 2 \cdot \frac{1}{2} = 1$
This matches!

(C) $0$ — clearly not $1$ .

(D) $2 \int_{-1}^{0} (1-x)\, dx$ — for $x$ from $-1$ to $0$ , $1-x$ ranges from $2$ to $1$ , so: …
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CBSE 2026Set V11 markMCQ
$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x\, dx$
(a) $1$
(b) $0$
(c) $-1$
(d) $7$
›Reveal solutionSolution

An odd function integrated over symmetric limits gives $0$ ; answer (b).

Let $g(x)=\sin^7 x$ . Then
$g(-x)=\sin^7(-x)=(-\sin x)^7=-\sin^7 x=-g(x),$
so $g$ is odd. For any odd function, …
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CBSE 2026Set A1 markMCQ
$\int_{-\pi}^{\pi}\sin^5 x\,dx =$
(a) $\frac{3\pi}{4}$
(b) $2\pi$
(c) $\frac{5\pi}{6}$
(d) $0$
›Reveal solutionSolution

Odd function over a symmetric interval integrates to $0$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int_{-a}^{a}\{f(x)-f(-x)\}\,dx$
(a) 0
(b) $2f(a)$
(c) $2f(-a)$
(d) $2f(0)$
›Reveal solutionSolution

$g(x)=f(x)-f(-x)$ is an odd function, so $\displaystyle\int_{-a}^{a}g(x)\,dx=0$ .

Let $g(x)=f(x)-f(-x)$ . Check whether $g$ is odd or even by evaluating $g(-x)$ :

$g(-x)=f(-x)-f(-(-x))=f(-x)-f(x)=-\big[f(x)-f(-x)\big]=-g(x)$

Since $g(-x)=-g(x)$ , $g$ is an odd function.

…

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CBSE 2026Set ANNUAL1 mark
$\displaystyle\int_{-1}^1 \sin^5 x \cos^4 x\, dx$
›Reveal solutionSolution

Recognise that the integrand is an odd function; the integral of any odd function over a symmetric interval $[-a,a]$ is always zero.

Let $f(x) = \sin^5x\cos^4x$ .

Test for odd/even symmetry:

$f(-x) = \sin^5(-x)\cos^4(-x) = (-\sin x)^5(\cos x)^4 = -\sin^5x\cos^4x = -f(x)$

(using $\sin(-x)=-\sin x$ so $\sin^5(-x)=-\sin^5x$ , and $\cos(-x)=\cos x$ so $\cos^4(-x)=\cos^4x$ .)

…

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CBSE 2025Set 65/2/11 markMCQ
$f(2a - x) = f(x)$
›Reveal solutionSolution
The condition $f(2a-x) = f(x)$ implies that the function $f(x)$ is symmetric about the line $x=a$ , which simplifies the definite integral $\int_{0}^{2a} f(x)\,dx$ to $2\int_{0}^{a} f(x)\,dx$ .

This problem tests your understanding of a fundamental property of definite integrals related to symmetry. The condition $f(2a-x) = f(x)$ is key here. It tells us something profound about the function's behaviour over the interval $[0, 2a]$ .

Concept and Intuition: Symmetry in Definite Integrals

Imagine the interval $[0, 2a]$ on the x-axis. The midpoint of this interval is $x=a$ .

The condition $f(2a-x) = f(x)$ means that the value of the function at any point $x$ is the same as its value at the point $2a-x$ .

Let's pick a point $x_1$ in the interval $[0, a]$ . Its symmetric counterpart with respect to $x=a$ is $x_2 = 2a - x_1$ .

For example, if $a=5$ and $x_1=2$ , then $2a-x_1 = 10-2=8$ . The condition $f(2)=f(8)$ means the function has the same height at $x=2$ and $x=8$ . Both points are $3$ units away from $x=5$ .

This implies that the graph of $f(x)$ is symmetric about the vertical line $x=a$ .

When a function is symmetric about $x=a$ over the interval $[0, 2a]$ , the area under the curve from $0$ to $a$ must be exactly equal to the area under the curve from $a$ to $2a$ .

Therefore, the total area from $0$ to $2a$ is simply twice the area from $0$ to $a$ . This is the intuition behind the property we are about to derive.
$f(2a-x) = f(x)$
Let's prove this property step-by-step.

Split the integral: We can split the given integral into two parts at the midpoint $a$ :

$\int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{2a} f(x)\,dx$
Let's call the second integral $I_2 = \int_{a}^{2a} f(x)\,dx$.

2. Apply substitution to the second integral:

To make use of the given condition $f(2a-x) = f(x)$ , we perform a substitution in $I_2$ .

Let $t = 2a - x$ .

Then, differentiating with respect to $x$ , we get $dt = -dx$ .

We also need to change the limits of integration:

* When $x = a$ , $t = 2a - a = a$ .

* When $x = 2a$ , $t = 2a - 2a = 0$ .

Substituting these into $I_2$:
$I_2 = \int_{a}^{0} f(2a-t)(-dt)$

Simplify the substituted integral: Using the property $\int_{b}^{a} g(t)\,dt = -\int_{a}^{b} g(t)\,dt$ , we can reverse the limits and remove the negative sign:

$I_2 = -\int_{a}^{0} f(2a-t)\,dt = \int_{0}^{a} f(2a-t)\,dt$

Apply the given condition: We are given that $f(2a-x) = f(x)$ . Since $t$ is just a dummy variable, this also means $f(2a-t) = f(t)$ . …
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CBSE 2025Set E1 markMCQ
$\int_{-1}^{1} \sin^7 x \cos^{13} x\,dx =$
(a) $0$
(b) $1$
(c) $20$
(d) $6$
›Reveal solutionSolution

An odd function integrated over a symmetric interval gives $0$ .

Let $f(x)=\sin^7x\cos^{13}x$ . Since $\sin^7(-x)=-\sin^7x$ (odd power of an odd function) and $\cos^{13}(-x)=\cos^{13}x$ (even function),

$f(-x)=-\sin^7x\cos^{13}x=-f(x),$ …

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CBSE 2025Set E1 markMCQ
$\int_{\alpha}^{\beta} \phi(x)\,dx + \int_{\beta}^{\alpha} \phi(x)\,dx =$
(a) $2$
(b) $1$
(c) $0$
(d) $2\int_{\alpha}^{\beta} \phi(x)\,dx$
›Reveal solutionSolution

By the property $\int_b^a\phi=-\int_a^b\phi$ , the two integrals cancel to $0$ .

A basic property of definite integrals is $\int_{\beta}^{\alpha}\phi(x)\,dx=-\int_{\alpha}^{\beta}\phi(x)\,dx$ . Therefore …

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CBSE 2025Set E1 markMCQ
$\int_{-1}^{1} \sin^{13}x \cos^{12}x\,dx =$
(a) $0$
(b) $1$
(c) $\frac{1}{2}$
(d) $2$
›Reveal solutionSolution

$\sin^{13}x\cos^{12}x$ is odd, so its integral over $[-1,1]$ is $0$ .

Let $f(x) = \sin^{13}x\cos^{12}x$ . Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$ :

$f(-x) = (-\sin x)^{13}(\cos x)^{12} = -\sin^{13}x\cos^{12}x = -f(x),$ …

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CBSE 2025Set E1 markMCQ
$\int_{-1}^{1} \sin x \cdot \cos^3 x\,dx =$
(a) $2$
(b) $1$
(c) $0$
(d) $-1$
›Reveal solutionSolution

$\sin x\cos^3 x$ is odd (odd $\times$ even), so its integral over $[-1,1]$ is $0$ .

Check parity: $\sin x$ is odd and $\cos^3 x$ is even, so their product $\sin x\cos^3 x$ is odd. The integral of an odd function over a sym …

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CBSE 2025Set ANNUAL1 markMCQ
$\displaystyle\int_{-a}^{a} f(x)\, dx = 0$
(a) $f(x)$ is an even function
(b) $f(x)$ is an odd function
(c) $f(2a-x) = f(x)$
(d) $f(2a-x) = -f(x)$
›Reveal solutionSolution

This is the standard symmetry property of definite integrals over [-a, a].

Property: $\displaystyle\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\int_0^a f(x)\,dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}$

…

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Q.∫−11(1−∣x∣) dx\int_{-1}^{1} (1-|x|)\, dx∫−11​(1−∣x∣)dx is equal to: (A) 2∫01(1+x) dx2 \int_{0}^{1} (1+x)\, dx2∫01​(1+x)dx (B) 2∫−10(1+x) dx2 \int_{-1}^{0} (1+x)\, dx2∫−10​(1+x)dx (C) 000 (D) 2∫−10(1−x) dx2 \int_{-1}^{0} (1-x)\, dx2∫−10​(1−x)dx

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Q. $\int_{-1}^{1} (1-|x|)\, dx$ is equal to: (A) $2 \int_{0}^{1} (1+x)\, dx$ (B) $2 \int_{-1}^{0} (1+x)\, dx$ (C) $0$ (D) $2 \int_{-1}^{0} (1-x)\, dx$

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