15 multiple-choice questions and 2 case-based questions on the current (rationalised) syllabus, in the style of the board paper's Sections A and E. Try each one first; the answer, a one-line reason and a worked solution open on a tap.
Choose one option. Section A of the board paper has 18 MCQs of 1 mark each, spread over all chapters.
The area of the region bounded by the curve \(y = 3x^2\), the \(x\)-axis and the ordinates \(x = 0\) and \(x = 2\) is
Why: ∫₀² 3x² dx = [x³]₀².
\(\int_0^2 3x^2\,dx = \left[x^3\right]_0^2 = 8\) sq units.
Find the area of the triangular region under the line \(y = 2x\), above the \(x\)-axis, from \(x = 0\) to \(x = 5\).
Why: ∫₀⁵ 2x dx = 25 (a triangle with base 5 and height 10).
\(\int_0^5 2x\,dx = \left[x^2\right]_0^5 = 25\) sq units (check: \(\tfrac12 \times 5 \times 10\)).
The area enclosed by the circle \(x^2 + y^2 = 49\) is
Why: 4∫₀⁷ √(49 − x²) dx = 4 × 49π/4.
Area \(= 4\int_0^7\sqrt{49 - x^2}\,dx = 4\left[\dfrac{x}{2}\sqrt{49 - x^2} + \dfrac{49}{2}\sin^{-1}\dfrac{x}{7}\right]_0^7 = 4 \times \dfrac{49\pi}{4} = 49\pi\).
The ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{4} = 1\) (\(a \gt 0\)) encloses an area of \(10\pi\) square units. Then \(a\) is
Why: Area = 4 × (2/a)∫₀ᵃ √(a² − x²) dx = 2πa.
\(y = \dfrac{2}{a}\sqrt{a^2 - x^2}\), so area \(= 4 \cdot \dfrac2a\int_0^a\sqrt{a^2 - x^2}\,dx = \dfrac8a \cdot \dfrac{\pi a^2}{4} = 2\pi a = 10\pi \Rightarrow a = 5\).
The area of the region bounded by the parabola \(y^2 = 9x\) and the line \(x = 4\) is
Why: The region is symmetric about the x-axis: 2∫₀⁴ 3√x dx.
Area \(= 2\int_0^4 3\sqrt x\,dx = 6\left[\dfrac23x^{3/2}\right]_0^4 = 6 \times \dfrac{16}{3} = 32\) sq units.
The area of the region bounded by \(y = \sqrt x\), the \(x\)-axis and the line \(x = 4\) is
Why: ∫₀⁴ x1/2 dx = (2/3) × 8.
\(\int_0^4 x^{1/2}\,dx = \left[\dfrac23x^{3/2}\right]_0^4 = \dfrac23 \times 8 = \dfrac{16}{3}\).
The area of the region in the first quadrant bounded by the circle \(x^2 + y^2 = 4\), the \(x\)-axis and the lines \(x = 0\) and \(x = 1\) is
Why: ∫₀¹ √(4 − x²) dx = [x/2 √(4 − x²) + 2 sin⁻¹(x/2)]₀¹.
\(\left[\dfrac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}\dfrac{x}{2}\right]_0^1 = \dfrac12\sqrt3 + 2 \cdot \dfrac{\pi}{6} = \dfrac{\sqrt3}{2} + \dfrac{\pi}{3}\) sq units.
What is the total area between the curve \(y = x^3\) and the \(x\)-axis from \(x = -1\) to \(x = 1\)?
Why: The part below the axis counts positively: 2 × ∫₀¹ x³ dx.
\(\int_{-1}^1 x^3\,dx = 0\), but area is never negative: area \(= \left|\int_{-1}^0 x^3\,dx\right| + \int_0^1 x^3\,dx = \dfrac14 + \dfrac14 = \dfrac12\).
The area of the region bounded by \(y = \cos x\), the \(x\)-axis and the lines \(x = 0\) and \(x = \dfrac{\pi}{2}\) is
Why: ∫₀π/2 cos x dx = [sin x].
\(\int_0^{\pi/2}\cos x\,dx = \left[\sin x\right]_0^{\pi/2} = 1\) sq unit.
Find the area under the curve \(y = e^x\), above the \(x\)-axis, from \(x = 0\) to \(x = 1\).
Why: ∫₀¹ eˣ dx = e − 1.
\(\int_0^1 e^x\,dx = \left[e^x\right]_0^1 = e - 1\) sq units.
The area of the region between the curve \(y = 4 - x^2\) and the \(x\)-axis is
Why: The curve meets the axis at x = ±2: ∫₋₂² (4 − x²) dx.
\(4 - x^2 = 0 \Rightarrow x = \pm 2\). Area \(= \int_{-2}^{2}(4 - x^2)\,dx = \left[4x - \dfrac{x^3}{3}\right]_{-2}^{2} = \dfrac{16}{3} + \dfrac{16}{3} = \dfrac{32}{3}\).
The area of the region bounded by \(y = \dfrac1x\), the \(x\)-axis and the lines \(x = 1\) and \(x = e^2\) is
Why: ∫₁e² dx/x = log e² = 2.
\(\int_1^{e^2}\dfrac{dx}{x} = \left[\log_e x\right]_1^{e^2} = 2 - 0 = 2\) sq units.
The area of the region bounded by the curve \(x = y^2\), the \(y\)-axis and the lines \(y = 0\) and \(y = 3\) is
Why: Integrate along y: ∫₀³ y² dy.
Horizontal strips: area \(= \int_0^3 x\,dy = \int_0^3 y^2\,dy = \left[\dfrac{y^3}{3}\right]_0^3 = 9\) sq units.
The area of the part of the ellipse \(\dfrac{x^2}{4} + \dfrac{y^2}{25} = 1\) that lies in the first quadrant is
Why: (5/2)∫₀² √(4 − x²) dx = (5/2)(π).
\(y = \dfrac52\sqrt{4 - x^2}\). \(\dfrac52\int_0^2\sqrt{4 - x^2}\,dx = \dfrac52 \times \pi = \dfrac{5\pi}{2}\) (a quarter of \(\pi ab = 10\pi\)).
The line \(x = a\) divides the region bounded by \(y = x^2\), the \(x\)-axis and \(x = 2\) into two parts of equal area. Then \(a\) is
Why: ∫₀ᵃ x² dx = half of 8/3, so a³/3 = 4/3.
Total area \(= \dfrac83\). Need \(\dfrac{a^3}{3} = \dfrac43 \Rightarrow a^3 = 4 \Rightarrow a = \sqrt[3]{4}\).
Section E of the board paper has three case-based questions of 4 marks: a real-life passage, then parts of 1, 1 and 2 marks, with a choice (OR) on the 2-mark part.
A park has an ornamental pond whose edge, in a coordinate system with the centre of the pond at the origin and units in metres, is the ellipse \(\dfrac{x^2}{25} + \dfrac{y^2}{9} = 1\). The park engineer uses integration to find areas for tiling and lighting.
A landscape designer plans a flower bed bounded by a straight path along the \(x\)-axis, a straight hedge along the line \(x = 9\), and a curved edging along \(y = \sqrt x\) (all lengths in metres, starting from the corner at the origin).
This free set is separate from the chapter's question bank. On the Application of Integrals chapter page the revision notes and the first step of the Route to 95 are free for everyone. CBSE Essentials adds all 39 questions in the chapter bank (short and long answers, assertion–reason and more case studies) with their full step-marking schemes, the Route to 95 checkpoints with your progress saved, Skill Builders and the full common-mistakes library. There is no AI marking on CBSE Math Revision: you check your work against the marking scheme.
Original questions written by CBSE Math Revision for the CBSE 2026-27 syllabus; not taken from NCERT, NCERT Exemplar or CBSE papers. Every answer was re-checked by computer algebra and by an independent reviewer. CBSE Math Revision is independent and not affiliated with CBSE or NCERT. Spotted a slip? Tell us and it goes in the corrections log.