The area of the region {(x, y) : |x - 1| ≤ y ≤ √(5 - x²)} is equal to:

Q: Why do we split the integral into two parts?

The absolute value function $|x - 1|$ is defined differently for $x < 1$ and $x \ge 1$. While you could split the integral at $x = 1$, it is often much faster to treat the area under the V-shape as two simple right-angled triangles using geometry, which avoids the need for complex piecewise integration.

Q: Can we use geometry for the circle part instead of the integral formula?

Yes, you absolutely can! The area under the semi-circle is a circular sector. However, using the standard integral formula $\int \sqrt{a^2 - x^2} dx$ is a robust, universal method that works regardless of the limits, making it a reliable tool for your JEE toolkit.

Q: How did you get $\frac{\pi}{2}$ for the sum of the inverse sines?

This is a beautiful application of the identity $\sin^{-1}(\theta) + \cos^{-1}(\theta) = \frac{\pi}{2}$. If you let $\alpha = \sin^{-1}(\frac{2}{\sqrt{5}})$, then $\sin(\alpha) = \frac{2}{\sqrt{5}}$. This implies $\cos(\alpha) = \frac{1}{\sqrt{5}}$, which means $\alpha = \cos^{-1}(\frac{1}{\sqrt{5}})$. Therefore, $\sin^{-1}(\frac{2}{\sqrt{5}}) + \sin^{-1}(\frac{1}{\sqrt{5}}) = \cos^{-1}(\frac{1}{\sqrt{5}}) + \sin^{-1}(\frac{1}{\sqrt{5}}) = \frac{\pi}{2}$.

Question

The area of the region $\{(x, y) : |x - 1| \le y \le \sqrt{5 - x^2}\}$ is equal to:

Options:

(A) $\frac{5}{2} \sin^{-1}(\frac{3}{5}) - \frac{1}{2}$

(B) $\frac{5\pi}{4} - \frac{3}{2}$

(C) $\frac{3\pi}{4} + \frac{3}{2}$

(D) $\frac{5\pi}{4} - \frac{1}{2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two distinct mathematical entities. One is a smooth, elegant semi-circle defined by $y = \sqrt{5 - x^2}$, which is the upper half of the circle $x^2 + y^2 = 5$.

The other is a sharp, angular V-shape defined by $y = |x - 1|$, with its vertex anchored firmly at $(1, 0)$. The problem asks us to find the area of the region trapped between these two curves.

### The Hunt for Intersections

Before we can calculate the area, we must determine where these two curves meet. We set the upper boundary equal to the lower boundary:
$$\sqrt{5 - x^2} = |x - 1|$$

To solve this, we square both sides, which removes both the square root and the absolute value:
$$(x - 1)^2 = 5 - x^2$$

Expanding the left side gives us $x^2 - 2x + 1 = 5 - x^2$. Rearranging this into a standard quadratic form, we get $2x^2 - 2x - 4 = 0$. Dividing by $2$ simplifies our equation to:
$$x^2 - x - 2 = 0$$

Factoring this quadratic, we find $(x - 2)(x + 1) = 0$. Our intersection points are $x = -1$ and $x = 2$. These are the boundaries of our region; everything we need to calculate lies between these two values.

### Setting the Stage for Integration

The area $A$ is defined by the integral of the upper curve minus the lower curve:
$$A = \int_{-1}^{2} (\sqrt{5 - x^2} - |x - 1|) dx$$

We can split this into two separate integrals:
$$A = \int_{-1}^{2} \sqrt{5 - x^2} dx - \int_{-1}^{2} |x - 1| dx$$

The first part, the integral of the semi-circle, utilizes the standard integral formula:
$$\int \sqrt{a^2 - x^2} dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}ight)$$

Here, $a^2 = 5$, so $a = \sqrt{5}$. Applying this formula and evaluating from $-1$ to $2$ requires careful substitution.

### The Beauty of Geometry

For the second part, the integral of the absolute value function $\int_{-1}^{2} |x - 1| dx$, we can use geometric intuition. If you visualize the graph of $y = |x - 1|$, you will see two right-angled triangles.

The first triangle, from $x = -1$ to $x = 1$, has a base of $2$ and a height of $2$, giving an area of:
$$\frac{1}{2} 	imes 2 	imes 2 = 2$$

The second triangle, from $x = 1$ to $x = 2$, has a base of $1$ and a height of $1$, giving an area of:
$$\frac{1}{2} 	imes 1 	imes 1 = \frac{1}{2}$$

The total area under the V-shape is $2 + \frac{1}{2} = \frac{5}{2}$.

### The Grand Finale

When we evaluate the semi-circle integral, we encounter a sum of inverse sine functions. Using the trigonometric identity $\sin^{-1}(\frac{2}{\sqrt{5}}) + \sin^{-1}(\frac{1}{\sqrt{5}}) = \frac{\pi}{2}$, the expression simplifies beautifully.

Combining the results, the area under the semi-circle becomes $2 + \frac{5\pi}{4}$. Finally, we subtract the area of the V-shape:
$$A = \left(2 + \frac{5\pi}{4}ight) - \frac{5}{2} = \frac{5\pi}{4} - \frac{1}{2}$$

The final resulting area is **$\frac{5\pi}{4} - \frac{1}{2}$**.

Animated Solution for Mathematics - Definite Integration: The area of the region ${(x, y) : ∣ x - 1∣ \leq y \leq 5 - x^{2}}$ is equal to:

Select Answer:

Visualized Solution

Visualizing the Boundaries

Finding Intersection Points

Solving for $x$

Setting up the Area Integral

Integrating the Semi-circle

Evaluating the Limits

Area of the V-shape Region

Using Inverse Trig Identity

Final Calculation

The Sigma Insight: Area Bounded by Curves

Analyzing the Setup

The Hunt for Intersections

Setting the Stage for Integration

The Beauty of Geometry

The Grand Finale

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Animated Solution for Mathematics - Definite Integration: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The area of the region {(x,y):∣x−1∣≤y≤5−x2​} is equal to: