Using Integration, find the area of the region \(x,y):0≤ y≤√(4-x²),\ x≥2-y\.

MathsCBSE Class 12 · Sample paper, session 2026-27 · 5 marks · Long answer

CBSE Class 12 Mathematics sample paper 2026-27: question 34(A)

From the official CBSE sample question paper for 2026-27, section D, question 34(A), for 5 marks. Try it first; the hints and our step-by-step solution are below.

Q34(A) · 5 marks · Long answerApplication of Integrals

Using Integration, find the area of the region ${(x, y) : 0 \leq y \leq 4 - x^{2}, x \geq 2 - y}$ .

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

Identify the curve: $y = 4 - x^{2}$ is part of a circle. Then decide which side of the line $x + y = 2$ the region lies on.

Hint 2

Find where the line meets the circle. The region lies between the upper curve and the lower curve over that interval.

Hint 3

Use $\int a^{2} - x^{2} d x = \frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}$ and subtract the area under the line.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

Figure (described): $y = 4 - x^{2}$ is the upper half of the circle $x^{2} + y^{2} = 4$ , with centre $(0, 0)$ and radius $2$ . The line $x + y = 2$ cuts both axes at $2$ . The condition $x \geq 2 - y$ means the point lies on or above the line, i.e. on the side away from the origin. The required region is the small segment in the first quadrant between the line and the arc, from $x = 0$ to $x = 2$ .
1 mark
Points of intersection: substitute $y = 2 - x$ in $x^{2} + y^{2} = 4$ to get $x^{2} + (2 - x)^{2} = 4 \Rightarrow 2 x^{2} - 4 x = 0 \Rightarrow x = 0$ or $x = 2$ . So the curves meet at $(0, 2)$ and $(2, 0)$ .
1 mark
Required area $= \int_{0}^{2} (4 - x^{2} - (2 - x)) d x$ , because the arc lies above the line on $[0, 2]$ .
1 mark
Integrating, the area is $[\frac{x}{2} 4 - x^{2} + \frac{4}{2} sin^{- 1} \frac{x}{2} - 2 x + \frac{x ^{2}}{2}]_{0}^{2}$ .
1 mark
Substituting the limits gives $(0 + 2 \cdot \frac{π}{2} - 4 + 2) - 0 = π - 2$ . So the required area is $(π - 2)$ sq. units.
1 mark

Answer: $(π - 2)$ sq. units

Where the marks are

Correct figure showing the segment between the arc and the line (1 mark)
Finds the intersection points $(0, 2)$ and $(2, 0)$ (1 mark)
Sets up the area as $\int_{0}^{2} (4 - x^{2} - (2 - x)) d x$ (1 mark)
Correct antiderivative (1 mark)
Evaluates the limits to get $π - 2$ sq. units (1 mark)

Revise the chapter

Chapter: Application of Integrals (notes, formulas, where marks are lost and every past question)
Topic: Area of Circle and Ellipse by Integration

Practise this chapterBoard-style questions from Application of Integrals, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27: every question, by section.

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CBSE Class 12 Mathematics sample paper 2026-27: question 34(A)

Hints

Step-by-step solution

Revise the chapter

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