Geometric approach to evaluate ∫₀¹ √(1-x²) dx as area of quarter circle.

Evaluate ∫₀¹ √(1-x²) dx — Area Under Semicircle

Question

Evaluate the definite integral:

\int_0^1 \sqrt{1 - x^2} \, dx

Solution — Step by Step

Ask yourself: what does $y = \sqrt{1 - x^2}$ look like? Squaring both sides gives $x^2 + y^2 = 1$ , and since $y \geq 0$ , this is the upper semicircle of radius 1 centered at the origin.

The limits $x = 0$ to $x = 1$ don’t cover the full semicircle — they cover only the first quadrant. So the integral represents the area bounded by the curve, the x-axis, and the lines $x = 0$ and $x = 1$ , which is exactly a quarter circle of radius 1.

Area of a full circle with radius $r = 1$ is $\pi r^2 = \pi$ . A quarter of that is:

\text{Area} = \frac{\pi r^2}{4} = \frac{\pi (1)^2}{4} = \frac{\pi}{4}

Substitute $x = \sin\theta$ , so $dx = \cos\theta \, d\theta$ . When $x = 0$ , $\theta = 0$ ; when $x = 1$ , $\theta = \pi/2$ .

\int_0^{\pi/2} \sqrt{1 - \sin^2\theta} \cdot \cos\theta \, d\theta = \int_0^{\pi/2} \cos^2\theta \, d\theta

Using the identity $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ :

= \frac{1}{2}\left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\pi/2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}

Both methods agree. ✓

Answer: $\dfrac{\pi}{4}$

Why This Works

The fundamental insight here is that a definite integral $\int_a^b f(x)\,dx$ gives the area under the curve $y = f(x)$ between $x = a$ and $x = b$ . When the curve is a familiar geometric shape, we can bypass all the algebra.

The function $y = \sqrt{1 - x^2}$ is non-negative on $[0, 1]$ , so there’s no signed-area complication. The curve is exactly the arc of a unit circle in the first quadrant, and the region below it down to the x-axis is a quarter-disk with area $\pi/4$ .

This geometric shortcut is not a trick — it’s the actual definition of the integral at work. JEE Advanced regularly tests whether students can see the geometry instead of grinding through substitutions.

Alternative Method

Use the standard reduction formula directly. We know:

\int_0^{\pi/2} \cos^n\theta \, d\theta = \frac{(n-1)!!}{n!!} \cdot \begin{cases} \pi/2 & \text{if } n \text{ even} \\ 1 & \text{if } n \text{ odd} \end{cases}

After the substitution $x = \sin\theta$ , we get $\int_0^{\pi/2} \cos^2\theta \, d\theta$ , which has $n = 2$ (even):

= \frac{1!!}{2!!} \cdot \frac{\pi}{2} = \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4}

Same answer. This reduction formula is worth memorising for NEET and JEE Main where speed matters.

Any time you see $\sqrt{r^2 - x^2}$ under an integral from $0$ to $r$ , the answer is $\frac{\pi r^2}{4}$ — no calculation needed. This appeared in JEE Advanced 2023 essentially to test whether students waste time on trigonometric substitution or see the quarter circle immediately.

Common Mistake

A very common error is trying to integrate $\sqrt{1 - x^2}$ by treating it like a power: writing it as $(1 - x^2)^{1/2}$ and applying the power rule as $\frac{(1-x^2)^{3/2}}{3/2}$ . This is wrong — the power rule $\int u^n \, du = \frac{u^{n+1}}{n+1}$ only works when $du$ matches the derivative of the inner function. Here, $d(1 - x^2) = -2x\,dx$ , but we only have $dx$ , not $x\,dx$ . The composite function rule doesn’t apply. Always check whether the “chain rule reverse” condition is satisfied before applying it.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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