Question

The area bounded by the curves y = |x² – 1| and y = 1 is

• A. \(\frac{2}{3}\left(\sqrt2+1 \right)\)
• B. \(\frac{4}{3}\left(\sqrt2-1\right)\)
• C. \(2\left(\sqrt2-1\right)\)
• D. \(\frac{8}{3}\left(\sqrt2-1\right)\)

Answer 1

The Correct Option is D

 To find the area bounded by the curves \(y = |x^2 - 1|\) and \(y = 1\), we need to consider the point of intersections and the areas where these curves overlap.

The function \(y = |x^2 - 1|\) can be piecewise defined as:

• \(y = x^2 - 1\) when \(x^2 - 1 \geq 0\) (i.e., \(|x| \geq 1\))
• \(y = -(x^2 - 1)\) when \(x^2 - 1 < 0\) (i.e., \(|x| < 1\))

The line \(y = 1\) intersects with \(y = x^2 - 1\) when:

\(x^2 - 1 = 1 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}\)

Therefore, the area can be divided into two parts:

1. Area I: From \(x = -\sqrt{2}\) to \(x = -1\) under \(y = 1 - (x^2 - 1)\)
2. Area II: From \(x = -1\) to \(x = 1\) under \(y = 1 - (-(x^2 - 1))\)
3. Area III: From \(x = 1\) to \(x = \sqrt{2}\) under \(y = 1 - (x^2 - 1)\)

Let's calculate each area:

• For Area I and III:
• For Area II:

Integration gives us:

• Calculating for Area I:
• Calculating for Area II:
• Calculating for Area III (similar to Area I):

Adding the areas together gives us:

\[\text{Total Area} = \text{Area I} + \text{Area II} + \text{Area III} = 2 \left( \frac{8\sqrt{2}}{3} - \frac{4}{3} \right) + \frac{2}{3} \] \[ = \frac{8}{3}\left(\sqrt{2}-1\right)\]

Therefore, the correct answer is \( \frac{8}{3}\left(\sqrt{2}-1\right) \).