Evaluate : $$\int_{-\pi/2}^{\pi/2} |\sin x| \, dx$$

Evaluate : ∫-π/2π/2 |sin x| dx

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\int_{-\pi/2}^{\pi/2} |\sin x| \, dx

Meghalaya MboseMBOSE Meghalaya Intermediate Board 2019Subjective· 2mImportance★★★★★

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Concept understanding — Even Function Property

The Even Function Property: A Mirror in Mathematics

Stand in front of a mirror: the distance from your nose to the mirror equals the distance from the mirror to your reflection. That's the core idea of an even function — it's symmetric about the vertical axis (the y-axis).

The Intuition

Take $f(x) = x^2$ . At $x = 3$ , $f(3) = 9$ ; at $x = -3$ , $f(-3) = 9$ as well. The output is identical for a number and its negative — and this happens for every single $x$ in the domain.

Graphically, if you fold the paper along the y-axis, the left half of the graph lands exactly on top of the right half. The curve is a perfect mirror image of itself.

The Precise Statement

f(-x) = f(x) \quad \text{for all } x \text{ in the domain}

One equation — but it must hold for every $x$ where the function is defined, not just for a few nice numbers.

What This Means in Practice

If you know the value at $x = 5$ , you automatically know the value at $x = -5$ — they're the same. This property lets you halve your work when analyzing the function.

Examples that satisfy the property:

$f(x) = x^2$ (check: $(-x)^2 = x^2$ )
$f(x) = \cos x$ (check: $\cos(-x) = \cos x$ )
$f(x) = |x|$ (check: $|-x| = |x|$ )
$f(x) = x^4 - 3x^2 + 1$ (only even powers of $x$ )

Watch out

A common mistake: thinking $f(x) = (x+1)^2$ is even because it has a square. Check: $f(-x) = (-x+1)^2 = (1-x)^2$ , which is not equal to $(x+1)^2$ for most $x$ . Only functions with only even powers of $x$ (and constants) are even — unless the function is defined piecewise.

Why "Even"?

The name comes from even powers: $x^2$ , $x^4$ , $x^6$ all satisfy $(-x)^n = x^n$ when $n$ is even. Odd powers like $x^3$ give $(-x)^3 = -x^3$ , which is a different property (odd functions).

A Quick Test

Replace every $x$ with $-x$ in the formula.
Simplify.
If you get back exactly the original expression, it's even. …

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$|\sin x|$ is an even function of $x$ , so the symmetric integral doubles the integral over $[0,\pi/2]$ , where $\sin x\ge0$ and the absolute value can be dropped.

Step 1 — Check evenness.

$|\sin(-x)|=|-\sin x|=|\sin x|$ , so $|\sin x|$ is even. For an even function, $\displaystyle\int_{-a}^{a}g(x)\,dx=2\int_0^a g(x)\,dx$ . Here $a=\dfrac{\pi}{2}$ .

$\int_{-\pi/2}^{\pi/2}|\sin x|\,dx = 2\int_0^{\pi/2}|\sin x|\,dx.$

Step 2 — Drop the absolute value on $[0,\pi/2]$ .

On $\left[0,\dfrac{\pi}{2}\right]$ , $\sin x\ge 0$ , so $|\sin x|=\sin x$ : …

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Q.Evaluate : ∫−π/2π/2∣sin⁡x∣ dx\int_{-\pi/2}^{\pi/2} |\sin x| \, dx∫−π/2π/2​∣sinx∣dx

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Q.Evaluate : $\int_{-\pi/2}^{\pi/2} |\sin x| \, dx$

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