Board Exam-Pattern Questions

How This Bank Works

Note: These are original practice questions written in the CBSE/State Board pattern — matched to the style, difficulty and marking scheme of board questions on Application of Integrals. They are not reproductions of any specific year's paper.

This chapter's board slot is dependable: usually one sketch-and-find-the-area question. The recurring shapes:

1. 2-mark: direct area under a named curve between given ordinates, or a circle/ellipse formula application with the setup shown.
2. 3-mark: a region needing one idea — a symmetry doubling, a sign split, a horizontal strip, or a latus-rectum region.
3. 5-mark: the full production — sketch, intercepts, split or segment setup, root-formula evaluation, exact answer.

Marking-scheme habits that earn full credit:

1. Draw the figure and shade the region — the sketch carries a mark of its own.
2. State the strip and limits ("taking vertical strips, $A = \int_{-1}^{1} y\,dx$") before integrating.
3. Show the split at every axis crossing, with the absolute value written explicitly.
4. End with units: "square units" belongs in the final line.

Attempt each question before opening its solution.

2-Mark Questions

Q1. Find the area bounded by $y = 3x^2$, the $x$-axis and the ordinates $x = 1$, $x = 2$.

Solution:

1. Set up: the curve is above the axis: $A = \int_1^2 3x^2\,dx$.
2. Evaluate: $\left[x^3\right]_1^2 = 8 - 1 = 7$.

Answer: $7$ square units.

Q2. Using integration, find the area enclosed by the circle $x^2 + y^2 = 25$.

Solution:

1. Symmetry: $A = 4\int_0^5\sqrt{25 - x^2}\,dx$.
2. Quarter-circle value: $\int_0^5\sqrt{25 - x^2}\,dx = \frac{25\pi}{4}$, so $A = 25\pi$.

Answer: $25\pi$ square units.

Q3. Find the area bounded by $y = \cos x$, the $x$-axis, $x = 0$ and $x = \frac{\pi}{2}$.

Solution:

1. Sign check: $\cos x \geq 0$ on $\left[0, \frac{\pi}{2}\right]$ — no split.
2. Evaluate: $[\sin x]_0^{\pi/2} = 1$.

Answer: $1$ square unit.

Q4. Find the area of the region bounded by the ellipse $\dfrac{x^2}{36} + \dfrac{y^2}{4} = 1$.

Solution:

1. Semi-axes: $a = 6$, $b = 2$.
2. Apply $\pi ab$: $A = \pi\cdot6\cdot2$.

Answer: $12\pi$ square units.

3-Mark Questions

Q5. Find the area of the region bounded by the curve $y = x^2 - x$ and the $x$-axis.

Solution:

1. Roots: $x(x - 1) = 0$ gives $x = 0, 1$; between them the parabola dips below the axis.
2. Integrate: $\int_0^1\left(x^2 - x\right)dx = \frac13 - \frac12 = -\frac16$.
3. Absolute value: $A = \frac16$.

Answer: $\dfrac{1}{6}$ square units.

Q6. Find the area of the region bounded by the parabola $y^2 = 16x$ and its latus rectum.

Solution:

1. Identify: $y^2 = 16x = 4(4)x$, so $a = 4$; the latus rectum is $x = 4$.
2. Symmetry about the $x$-axis: $A = 2\int_0^4 4\sqrt x\,dx$ (upper branch $y = 4\sqrt x$).
3. Evaluate: $8\cdot\frac23\left[x^{3/2}\right]_0^4 = \frac{16}{3}\cdot8 = \frac{128}{3}$.

Answer: $\dfrac{128}{3}$ square units.

Q7. Sketch $y = \sin x$ on $[0, 2\pi]$ and find the area bounded by the curve and the $x$-axis over this interval.

Solution:

1. Sketch: one arch above the axis on $[0, \pi]$, one below on $[\pi, 2\pi]$.
2. Pieces: $\int_0^{\pi}\sin x\,dx = 2$; $\left|\int_{\pi}^{2\pi}\sin x\,dx\right| = 2$.
3. Add: $2 + 2 = 4$.

Answer: $4$ square units.

Q8. Find the area of the region bounded by the curve $x = 4 - y^2$ and the $y$-axis.

Solution:

1. Intersections with the $y$-axis: $4 - y^2 = 0$ at $y = \pm2$; between them $x > 0$.
2. Horizontal strips: $A = \int_{-2}^{2}\left(4 - y^2\right)dy = 2\int_0^2\left(4 - y^2\right)dy$ (even in $y$).
3. Evaluate: $2\left[4y - \frac{y^3}{3}\right]_0^2 = 2\left(8 - \frac83\right) = \frac{32}{3}$.

Answer: $\dfrac{32}{3}$ square units.

5-Mark Questions

Q9. Sketch the region and find the area bounded by the line $y = 2x + 3$, the $x$-axis and the ordinates $x = -3$ and $x = 0$.

Solution:

1. Intercept: $2x + 3 = 0$ at $x = -\frac32$, inside $[-3, 0]$ — the line is below the axis to its left, above to its right.
2. Left piece: $\int_{-3}^{-3/2}(2x+3)dx = \left[x^2 + 3x\right]_{-3}^{-3/2} = -\frac94 - 0 = -\frac94$; absolute value $\frac94$.
3. Right piece: $\int_{-3/2}^{0}(2x+3)dx = 0 - \left(-\frac94\right) = \frac94$.
4. Add: $\frac94 + \frac94 = \frac92$. (Check: two congruent triangles, base $\frac32$, height $3$: $2\cdot\frac12\cdot\frac32\cdot3 = \frac92$. ✓)

Answer: $\dfrac{9}{2}$ square units.

Q10. Find the area of the smaller region cut off from the circle $x^2 + y^2 = 32$ by the line $x = 4$.

Solution:

1. Geometry: the circle has radius $\sqrt{32} = 4\sqrt2$; the line $x = 4$ cuts a minor segment to its right, spanning $4 \leq x \leq 4\sqrt2$.
2. Strip: height $2\sqrt{32 - x^2}$: $A = 2\int_4^{4\sqrt2}\sqrt{32 - x^2}\,dx$.
3. Root formula: $\left[\frac x2\sqrt{32 - x^2} + 16\sin^{-1}\frac{x}{4\sqrt2}\right]_4^{4\sqrt2} = \left(0 + 16\cdot\frac{\pi}{2}\right) - \left(8 + 16\cdot\frac{\pi}{4}\right) = 4\pi - 8$.
4. Double: $A = 8\pi - 16$.

Answer: $8\pi - 16$ square units (about $9.13$ — positive, as required).

Q11. Using integration, find the area of the region in the first quadrant enclosed by the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ and the coordinate axes. Show the complete working.

Solution:

1. Strip: in the first quadrant $y = \frac34\sqrt{16 - x^2}$: $A = \frac34\int_0^4\sqrt{16 - x^2}\,dx$.
2. Root formula: $\int_0^4\sqrt{16 - x^2}\,dx = \left[\frac x2\sqrt{16 - x^2} + 8\sin^{-1}\frac x4\right]_0^4 = 8\cdot\frac{\pi}{2} = 4\pi$.
3. Multiply: $A = \frac34\cdot4\pi = 3\pi$. (Check: quarter of the full ellipse $\pi\cdot4\cdot3 = 12\pi$. ✓)

Answer: $3\pi$ square units.

Q12. Sketch $y = x^3$ and find the area bounded by the curve, the $x$-axis and the ordinates $x = -1$, $x = 1$.

Solution:

1. Sketch: the cubic passes through the origin, below the axis for $x < 0$, above for $x > 0$ — split at $0$.
2. Left piece: $\left|\int_{-1}^0 x^3dx\right| = \left|-\frac14\right| = \frac14$.
3. Right piece: $\int_0^1 x^3dx = \frac14$.
4. Add: $\frac14 + \frac14 = \frac12$. (The signed integral is $0$ — the odd function's trap, dodged by the split.)

Answer: $\dfrac{1}{2}$ square units.