Questions and Answers (Board Exam)

Introduction to Board Exam PYQs

This section contains 30 important board-style questions on Area under Curves. In board exams, drawing a rough sketch of the curve, identifying the limits of integration, and clearly stating the upper curve/lower curve or right curve/left curve is essential. Whenever the curve crosses the x-axis or when modulus is involved, split the interval carefully so that the geometrical area is always taken as positive.

Question 1 [CBSE 2026]

Find the area of the region bounded by the curve $y = x^2$ and the lines $x = 1$, $x = 2$, and the x-axis.

Solution: Step 1: Sketch the region. The curve $y=x^2$ is an upward-opening parabola. Between $x=1$ and $x=2$, the curve lies above the x-axis.

Step 2: Set up the integral. Since the curve is above the x-axis throughout $[1,2]$, the required area is $$\text{Area} = \int_1^2 x^2\,dx.$$

Step 3: Integrate. Using the power rule, $$\int x^2\,dx = \frac{x^3}{3}.$$ Hence, $$\text{Area} = \left[\frac{x^3}{3}\right]_1^2.$$

Step 4: Evaluate the limits. $$= \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.$$

Answer: $\frac{7}{3}$ square units.

Question 2 [CBSE 2024]

Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Solution: Step 1: Use symmetry. The ellipse is symmetric about both coordinate axes, so its total area is four times the area in the first quadrant.

Step 2: Express $y$ in terms of $x$. From $$\frac{x^2}{16} + \frac{y^2}{9} = 1,$$ we get $$\frac{y^2}{9} = 1 - \frac{x^2}{16} = \frac{16-x^2}{16},$$ so in the first quadrant, $$y = \frac{3}{4}\sqrt{16-x^2}.$$

Step 3: Set up the integral. In the first quadrant, $x$ varies from $0$ to $4$. Therefore, $$\text{Total Area} = 4\int_0^4 \frac{3}{4}\sqrt{16-x^2}\,dx = 3\int_0^4 \sqrt{16-x^2}\,dx.$$

Step 4: Use the standard integral formula. $$\int \sqrt{a^2-x^2}\,dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right).$$ Here $a=4$, so $$\text{Area} = 3\left[ \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right) \right]_0^4.$$

Step 5: Evaluate. At $x=4$, $$\frac{4}{2}\sqrt{16-16} + 8\sin^{-1}(1) = 0 + 8\cdot\frac{\pi}{2} = 4\pi.$$ At $x=0$, the value is $0$. Thus, $$\text{Area} = 3(4\pi) = 12\pi.$$

Answer: $12\pi$ square units.

Question 3 [CBSE 2025]

Find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$.

Solution: Step 1: Understand the geometry. The parabola $y^2=4x$ opens to the right and is symmetric about the x-axis. The vertical line $x=3$ cuts the parabola and forms a closed region.

Step 2: Find the points where $x=3$ meets the parabola. Substituting $x=3$ in $y^2=4x$ gives $$y^2 = 12 \implies y = \pm 2\sqrt{3}.$$

Step 3: Use symmetry. The total area is twice the area in the first quadrant. In the first quadrant, from $x=0$ to $x=3$, the upper boundary is $$y = \sqrt{4x} = 2\sqrt{x}.$$ So, $$\text{Area} = 2\int_0^3 2\sqrt{x}\,dx = 4\int_0^3 x^{1/2}dx.$$

Step 4: Integrate. $$= 4\left[\frac{x^{3/2}}{3/2}\right]_0^3 = 4\cdot\frac{2}{3}[x^{3/2}]_0^3 = \frac{8}{3}(3^{3/2}).$$

Step 5: Simplify. $$3^{3/2} = 3\sqrt{3}.$$ Hence, $$\text{Area} = \frac{8}{3}(3\sqrt{3}) = 8\sqrt{3}.$$

Answer: $8\sqrt{3}$ square units.

Question 4 [CBSE 2026]

Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Solution: Step 1: Determine where the curve is above and below the x-axis. The curve $y=\sin x$ is positive on $[0,\pi]$ and negative on $[\pi,2\pi]$.

Step 2: Split the interval. For geometrical area, we add the positive area and the absolute value of the negative area: $$\text{Area} = \int_0^\pi \sin x\,dx + \left|\int_\pi^{2\pi} \sin x\,dx\right|.$$

Step 3: Evaluate the first integral. $$\int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = -\cos\pi - (-\cos0) = 1 + 1 = 2.$$

Step 4: Evaluate the second integral. $$\int_\pi^{2\pi} \sin x\,dx = [-\cos x]_\pi^{2\pi} = -\cos2\pi - (-\cos\pi) = -1 - 1 = -2.$$ So its area contribution is $|-2|=2$.

Step 5: Add both parts. $$\text{Total Area} = 2+2 = 4.$$

Answer: $4$ square units.

Question 5 [CBSE 2023]

Find the area of the region bounded by $y = \sqrt{x}$ and $y = x$.

Solution: Step 1: Find the points of intersection. Set $$\sqrt{x} = x.$$ Squaring both sides gives $$x = x^2 \implies x(x-1)=0.$$ Hence, $$x=0,1.$$

Step 2: Identify the upper and lower curves. Take a test point, say $x=\frac14$: $$\sqrt{\frac14}=\frac12, \qquad x=\frac14.$$ Since $\frac12 > \frac14$, the upper curve is $y=\sqrt{x}$ and the lower curve is $y=x$ on $[0,1]$.

Step 3: Set up the area integral. $$\text{Area} = \int_0^1 (\sqrt{x} - x)\,dx = \int_0^1 (x^{1/2} - x)\,dx.$$

Step 4: Integrate. $$= \left[\frac{2}{3}x^{3/2} - \frac{x^2}{2}\right]_0^1 = \frac{2}{3} - \frac{1}{2}.$$

Step 5: Simplify. $$= \frac{4-3}{6} = \frac{1}{6}.$$

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

Question 6 [CBSE 2026]

Find the area bounded by the parabola $y^2 = 8x$ and its latus rectum.

Solution: Step 1: Identify $a$. Comparing $y^2=8x$ with the standard form $y^2=4ax$, we get $$4a = 8 \implies a=2.$$ So the latus rectum is the vertical line $$x=a=2.$$

Step 2: Find the endpoints of the latus rectum. Substitute $x=2$ in the parabola: $$y^2 = 8(2)=16 \implies y=\pm 4.$$

Step 3: Use symmetry. The region is symmetric about the x-axis. Hence, $$\text{Area} = 2\int_0^2 y\,dx.$$ In the first quadrant, $$y = \sqrt{8x} = 2\sqrt{2}\,x^{1/2}.$$ So, $$\text{Area} = 2\int_0^2 2\sqrt{2}\,x^{1/2}dx = 4\sqrt{2}\int_0^2 x^{1/2}dx.$$

Step 4: Integrate. $$= 4\sqrt{2}\left[\frac{2}{3}x^{3/2}\right]_0^2 = \frac{8\sqrt{2}}{3}(2^{3/2}).$$

Step 5: Simplify. Since $$2^{3/2} = 2\sqrt{2},$$ we get $$\text{Area} = \frac{8\sqrt{2}}{3}\cdot 2\sqrt{2} = \frac{32}{3}.$$

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

Question 7 [CBSE 2025]

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

Solution: Step 1: Find where the curve meets the x-axis. Set $y=0$: $$4x - x^2 = 0 \implies x(4-x)=0.$$ Thus, $$x=0,4.$$

Step 2: Check whether the curve is above the x-axis in this interval. At $x=2$, $$y = 4(2)-2^2 = 8-4 = 4 > 0.$$ So the curve lies above the x-axis on $[0,4]$.

Step 3: Set up the integral. $$\text{Area} = \int_0^4 (4x-x^2)\,dx.$$

Step 4: Integrate. $$= \left[2x^2 - \frac{x^3}{3}\right]_0^4 = 2(16) - \frac{64}{3}.$$

Step 5: Simplify. $$= 32 - \frac{64}{3} = \frac{96-64}{3} = \frac{32}{3}.$$

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

Question 8 [CBSE 2022]

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

Solution: Step 1: Determine where the curve changes sign. The curve $y=x^3$ crosses the x-axis at $$x=0.$$ It is below the x-axis on $[-2,0]$ and above the x-axis on $[0,1]$.

Step 2: Split the integral. The geometrical area is $$\text{Area} = \left|\int_{-2}^0 x^3\,dx\right| + \int_0^1 x^3\,dx.$$

Step 3: Evaluate the first integral. $$\int_{-2}^0 x^3\,dx = \left[\frac{x^4}{4}\right]_{-2}^0 = 0 - \frac{16}{4} = -4.$$ So the area contribution is $|-4|=4$.

Step 4: Evaluate the second integral. $$\int_0^1 x^3\,dx = \left[\frac{x^4}{4}\right]_0^1 = \frac{1}{4}.$$

Step 5: Add both parts. $$\text{Area} = 4 + \frac{1}{4} = \frac{17}{4}.$$

Answer: $\frac{17}{4}$ square units.

Question 9 [CBSE 2026]

Find the area bounded by the parabola $y^2 = 4ax$ and the line $y = mx$.

Solution: Step 1: Find the points of intersection. Substitute $y=mx$ into $y^2=4ax$: $$(mx)^2 = 4ax \implies m^2x^2 - 4ax = 0 \implies x(m^2x - 4a)=0.$$ Thus, $$x=0 \quad \text{or} \quad x=\frac{4a}{m^2}.$$

Step 2: Write the parabola in the upper form. From $y^2=4ax$, $$y = 2\sqrt{a}\,x^{1/2}.$$ On the bounded interval, this lies above the line $y=mx$.

Step 3: Set up the area integral. $$\text{Area} = \int_0^{4a/m^2} \left(2\sqrt{a}\,x^{1/2} - mx\right)dx.$$

Step 4: Integrate. $$= \left[ \frac{4\sqrt{a}}{3}x^{3/2} - \frac{m}{2}x^2 \right]_0^{4a/m^2}.$$

Step 5: Evaluate at the upper limit. Using $$\left(\frac{4a}{m^2}\right)^{3/2} = \frac{8a\sqrt{a}}{m^3},$$ we get $$\text{Area} = \frac{4\sqrt{a}}{3}\cdot \frac{8a\sqrt{a}}{m^3} - \frac{m}{2}\cdot \frac{16a^2}{m^4}.$$ This becomes $$\frac{32a^2}{3m^3} - \frac{8a^2}{m^3} = \frac{32a^2 - 24a^2}{3m^3} = \frac{8a^2}{3m^3}.$$

Answer: $\frac{8a^2}{3m^3}$ square units.

Question 10 [CBSE 2024]

Find the area of the region in the first quadrant enclosed by the x-axis, the line $x = \sqrt{3}y$, and the circle $x^2 + y^2 = 4$.

Solution: Step 1: Rewrite the line. From $$x = \sqrt{3}y,$$ we get $$y = \frac{x}{\sqrt{3}}.$$

Step 2: Find the point where the line meets the circle in the first quadrant. Substitute $x=\sqrt{3}y$ into the circle: $$(\sqrt{3}y)^2 + y^2 = 4 \implies 3y^2 + y^2 = 4 \implies 4y^2 = 4 \implies y=1.$$ Hence, $$x = \sqrt{3}.$$ So the intersection point is $(\sqrt{3},1)$.

Step 3: Split the region. From $x=0$ to $x=\sqrt{3}$, the upper boundary is the line $$y = \frac{x}{\sqrt3}.$$ From $x=\sqrt3$ to $x=2$, the upper boundary is the circle $$y = \sqrt{4-x^2}.$$

Step 4: Set up the integral. $$\text{Area} = \int_0^{\sqrt3} \frac{x}{\sqrt3}\,dx + \int_{\sqrt3}^2 \sqrt{4-x^2}\,dx.$$

Step 5: Evaluate the first integral. $$\int_0^{\sqrt3} \frac{x}{\sqrt3}\,dx = \frac{1}{\sqrt3}\left[\frac{x^2}{2}\right]_0^{\sqrt3} = \frac{1}{\sqrt3}\cdot\frac{3}{2} = \frac{\sqrt3}{2}.$$

Step 6: Evaluate the second integral. Using the standard formula, $$\int \sqrt{4-x^2}\,dx = \frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right).$$ So, $$\int_{\sqrt3}^2 \sqrt{4-x^2}\,dx = \left[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\left(\frac{x}{2}\right)\right]_{\sqrt3}^2.$$ At $x=2$, the value is $$0 + 2\cdot\frac{\pi}{2} = \pi.$$ At $x=\sqrt3$, the value is $$\frac{\sqrt3}{2}\cdot 1 + 2\cdot\frac{\pi}{3} = \frac{\sqrt3}{2} + \frac{2\pi}{3}.$$ Thus, $$\int_{\sqrt3}^2 \sqrt{4-x^2}\,dx = \pi - \left(\frac{\sqrt3}{2} + \frac{2\pi}{3}\right) = \frac{\pi}{3} - \frac{\sqrt3}{2}.$$

Step 7: Add the two parts. $$\text{Area} = \frac{\sqrt3}{2} + \left(\frac{\pi}{3} - \frac{\sqrt3}{2}\right) = \frac{\pi}{3}.$$

Answer: $\frac{\pi}{3}$ square units.

Question 11 [CBSE 2025]

Find the area bounded by the parabola $x^2 = 4y$ and the straight line $x = 4y - 2$.

Solution: Step 1: Express both in terms of $y$ or $x$. It is convenient to use $y$ as a function of $x$. From the parabola, $$y = \frac{x^2}{4}.$$ From the line, $$x = 4y-2 \implies y = \frac{x+2}{4}.$$

Step 2: Find the points of intersection. Set the two expressions equal: $$\frac{x^2}{4} = \frac{x+2}{4} \implies x^2 = x+2 \implies x^2-x-2=0.$$ Factorizing, $$(x-2)(x+1)=0,$$ so $$x=-1,2.$$

Step 3: Determine upper and lower curves. At $x=0$, $$y_{\text{line}} = \frac{0+2}{4} = \frac12, \qquad y_{\text{parabola}} = 0.$$ So the line is above the parabola on $[-1,2]$.

Step 4: Set up the integral. $$\text{Area} = \int_{-1}^2 \left(\frac{x+2}{4} - \frac{x^2}{4}\right)dx = \frac14\int_{-1}^2 (x+2-x^2)dx.$$

Step 5: Integrate. $$= \frac14\left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^2.$$ At $x=2$, the value is $$\frac{4}{2}+4-\frac{8}{3} = 6-\frac{8}{3} = \frac{10}{3}.$$ At $x=-1$, the value is $$\frac12 - 2 + \frac13 = -\frac76.$$ So, $$\text{Area} = \frac14\left(\frac{10}{3} - \left(-\frac76\right)\right) = \frac14\cdot\frac{27}{6} = \frac{9}{8}.$$

Answer: $\frac{9}{8}$ square units.

Question 12 [CBSE 2023]

Find the area bounded by the curve $y = \cos x$ between $x = 0$ and $x = \pi$.

Solution: Step 1: Note where the curve changes sign. $\cos x$ is positive on $[0,\pi/2]$ and negative on $[\pi/2,\pi]$.

Step 2: Split the integral. $$\text{Area} = \int_0^{\pi/2} \cos x\,dx + \left|\int_{\pi/2}^{\pi} \cos x\,dx\right|.$$

Step 3: Evaluate. $$\int_0^{\pi/2} \cos x\,dx = [\sin x]_0^{\pi/2} = 1.$$ $$\int_{\pi/2}^{\pi} \cos x\,dx = [\sin x]_{\pi/2}^{\pi} = -1.$$ Taking absolute value of the second part gives $1$.

Step 4: Add. $$\text{Area} = 1+1=2.$$

Answer: $2$ square units.

Question 13 [CBSE 2026]

Find the area bounded by the curve $y = |x + 3|$ and the x-axis, between $x = -6$ and $x = 0$.

Solution: Step 1: Find the point where the modulus changes form. $$x+3=0 \implies x=-3.$$ So the interval must be split at $x=-3$.

Step 2: Write the modulus piecewise. For $x<-3$, $$|x+3| = -(x+3).$$ For $x>-3$, $$|x+3| = x+3.$$

Step 3: Set up the integral. $$\text{Area} = \int_{-6}^{-3} -(x+3)\,dx + \int_{-3}^{0} (x+3)\,dx.$$

Step 4: Evaluate the first part. $$\int_{-6}^{-3} -(x+3)dx = \left[-\frac{x^2}{2} - 3x\right]_{-6}^{-3}.$$ At $x=-3$, value $= -\frac{9}{2} + 9 = \frac{9}{2}$. At $x=-6$, value $= -18 + 18 = 0$. So the first area is $$\frac{9}{2}.$$

Step 5: Evaluate the second part. $$\int_{-3}^{0} (x+3)dx = \left[\frac{x^2}{2} + 3x\right]_{-3}^{0}.$$ At $x=0$, value $=0$. At $x=-3$, value $= \frac{9}{2} - 9 = -\frac{9}{2}$. So the second area is $$0 - \left(-\frac{9}{2}\right) = \frac{9}{2}.$$

Step 6: Add both parts. $$\text{Area} = \frac{9}{2} + \frac{9}{2} = 9.$$

Answer: $9$ square units.

Question 14 [CBSE 2022]

Find the area bounded by $y^2 = x$ and the line $x + y = 2$.

Solution: Step 1: Write both curves in terms of $x$ and $y$. From the parabola, $$x = y^2.$$ From the line, $$x = 2-y.$$ Since both are already in the form $x=f(y)$, integrating with respect to $y$ is easier.

Step 2: Find the intersection points. Set $$y^2 = 2-y \implies y^2 + y - 2 = 0.$$ Factorizing, $$(y+2)(y-1)=0,$$ so $$y=-2,1.$$

Step 3: Determine right and left curves. At $y=0$, $$x_{\text{line}} = 2, \qquad x_{\text{parabola}} = 0.$$ So the line is the right curve and the parabola is the left curve.

Step 4: Set up the integral. $$\text{Area} = \int_{-2}^{1} \left[(2-y) - y^2\right]dy.$$

Step 5: Integrate. $$= \left[2y - \frac{y^2}{2} - \frac{y^3}{3}\right]_{-2}^{1}.$$ At $y=1$, value is $$2 - \frac12 - \frac13 = \frac76.$$ At $y=-2$, value is $$-4 - 2 + \frac83 = -\frac{10}{3}.$$ Thus, $$\text{Area} = \frac76 - \left(-\frac{10}{3}\right) = \frac{7+20}{6} = \frac{27}{6} = \frac92.$$

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

Question 15 [CBSE 2025]

Find the area bounded by the circle $x^2 + y^2 = 16$ and the parabola $y^2 = 6x$.

Solution: Step 1: Find the points of intersection. Substitute $y^2 = 6x$ into the circle: $$x^2 + 6x = 16 \implies x^2 + 6x - 16 = 0.$$ Factorizing, $$(x+8)(x-2)=0.$$ Since the parabola has $x\ge 0$, the relevant value is $$x=2.$$ Then $$y^2 = 6(2)=12 \implies y = \pm 2\sqrt3.$$

Step 2: Identify the enclosed region. The bounded region lies between the circle and the parabola, and is symmetric about the x-axis. For $0 \le x \le 2$, the upper semicircle is $$y = \sqrt{16-x^2},$$ and the upper branch of the parabola is $$y = \sqrt{6x}.$$ Hence total area is $$\text{Area} = 2\int_0^2 \left(\sqrt{16-x^2} - \sqrt{6x}\right)dx.$$

Step 3: Evaluate the circle part. Using the standard formula, $$\int_0^2 \sqrt{16-x^2}\,dx = \left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_0^2.$$ This gives $$\sqrt{12} + 8\sin^{-1}\left(\frac12\right) = 2\sqrt3 + \frac{4\pi}{3}.$$

Step 4: Evaluate the parabola part. $$\int_0^2 \sqrt{6x}\,dx = \sqrt6\int_0^2 x^{1/2}dx = \sqrt6\left[\frac{2}{3}x^{3/2}\right]_0^2.$$ Now, $$2^{3/2} = 2\sqrt2,$$ so $$\int_0^2 \sqrt{6x}\,dx = \frac{2\sqrt6}{3}\cdot 2\sqrt2 = \frac{8\sqrt3}{3}.$$

Step 5: Subtract and multiply by 2. $$\text{Area} = 2\left[\left(2\sqrt3 + \frac{4\pi}{3}\right) - \frac{8\sqrt3}{3}\right] = 2\left(\frac{4\pi}{3} - \frac{2\sqrt3}{3}\right).$$ Hence, $$\text{Area} = \frac{8\pi - 4\sqrt3}{3}.$$

Answer: $\frac{8\pi - 4\sqrt{3}}{3}$ square units.

Question 16 [CBSE 2026]

Find the area of the region bounded by the curve $x^2+y^2=a^2$ and the line $x+y=a$ in the first quadrant.

Solution: Step 1: Find the intersection points. The line meets the axes at $(a,0)$ and $(0,a)$. Both points also lie on the circle $x^2+y^2=a^2$.

Step 2: Identify upper and lower curves. In the first quadrant, the circle is $$y = \sqrt{a^2-x^2},$$ and the line is $$y = a-x.$$ The circle lies above the line between $x=0$ and $x=a$.

Step 3: Set up the integral. $$\text{Area} = \int_0^a \left(\sqrt{a^2-x^2} - (a-x)\right)dx.$$

Step 4: Integrate. $$= \left[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) - ax + \frac{x^2}{2}\right]_0^a.$$

Step 5: Evaluate. At $x=a$, $$0 + \frac{a^2}{2}\cdot\frac{\pi}{2} - a^2 + \frac{a^2}{2} = \frac{\pi a^2}{4} - \frac{a^2}{2}.$$ At $x=0$, value is $0$. Thus, $$\text{Area} = \frac{\pi a^2}{4} - \frac{a^2}{2} = \frac{a^2}{4}(\pi - 2).$$

Answer: $\frac{a^2}{4}(\pi - 2)$ square units.

Question 17 [CBSE 2024]

Find the area bounded by the two parabolas $y^2 = 4x$ and $x^2 = 4y$.

Solution: Step 1: Rewrite the second parabola. From $$x^2 = 4y,$$ we get $$y = \frac{x^2}{4}.$$ From the first parabola, $$y = 2\sqrt{x}.$$

Step 2: Find the points of intersection. Substitute $y=\frac{x^2}{4}$ into $y^2=4x$: $$\left(\frac{x^2}{4}\right)^2 = 4x \implies \frac{x^4}{16} = 4x \implies x^4 - 64x = 0.$$ So, $$x(x^3-64)=0 \implies x=0,4.$$

Step 3: Determine upper and lower curves. On $0<x<4$, the upper curve is $$y = 2\sqrt{x},$$ and the lower curve is $$y = \frac{x^2}{4}.$$

Step 4: Set up the integral. $$\text{Area} = \int_0^4 \left(2\sqrt{x} - \frac{x^2}{4}\right)dx.$$

Step 5: Integrate. $$= \left[\frac{4}{3}x^{3/2} - \frac{x^3}{12}\right]_0^4 = \frac{4}{3}(4^{3/2}) - \frac{64}{12}.$$ Now, $$4^{3/2} = 8.$$ Therefore, $$\text{Area} = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}.$$

Answer: $\frac{16}{3}$ square units.

Question 18 [CBSE 2026]

Find the area of the region $\{(x,y): y^2 \le 4x,\; 4x^2+4y^2 \le 9\}$.

Solution: Step 1: Interpret the inequalities. The region lies to the right of the parabola $$y^2 = 4x$$ and inside the circle $$x^2+y^2 = \frac94.$$ The figure is symmetric about the x-axis.

Step 2: Find the point of intersection in the first quadrant. Substitute $y^2=4x$ into the circle: $$4x^2 + 4(4x) = 9 \implies 4x^2 + 16x - 9 = 0.$$ This gives the relevant positive root $$x = \frac12.$$

Step 3: Split the first-quadrant area. For $0 \le x \le \frac12$, the upper boundary is the parabola $$y = \sqrt{4x} = 2\sqrt{x}.$$ For $\frac12 \le x \le \frac32$, the upper boundary is the circle $$y = \sqrt{\frac94 - x^2}.$$ Thus, $$\text{Area} = 2\left[\int_0^{1/2} 2\sqrt{x}\,dx + \int_{1/2}^{3/2} \sqrt{\frac94 - x^2}\,dx\right].$$

Step 4: Evaluate the first part. $$2\int_0^{1/2} 2\sqrt{x}\,dx = 4\left[\frac{2}{3}x^{3/2}\right]_0^{1/2} = \frac{8}{3}\left(\frac12\right)^{3/2} = \frac{2\sqrt2}{3}.$$

Step 5: Evaluate the second part. Using the standard formula, $$2\int_{1/2}^{3/2} \sqrt{\frac94 - x^2}\,dx = \frac{9\pi}{8} - \frac{\sqrt2}{2} - \frac{9}{4}\sin^{-1}\left(\frac13\right).$$

Step 6: Add the two contributions. $$\text{Area} = \frac{2\sqrt2}{3} - \frac{\sqrt2}{2} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac13\right).$$ Combine the $\sqrt2$ terms: $$\frac{2\sqrt2}{3} - \frac{\sqrt2}{2} = \frac{\sqrt2}{6}.$$ So, $$\text{Area} = \frac{\sqrt2}{6} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac13\right).$$

Answer: $\frac{\sqrt{2}}{6} + \frac{9\pi}{8} - \frac{9}{4}\sin^{-1}\left(\frac{1}{3}\right)$ square units.

Question 19 [CBSE 2023]

Find the area bounded by $y = x^2+2$, $y=x$, $x=0$, and $x=3$.

Solution: Step 1: Check whether the two curves intersect. Set $$x^2+2 = x \implies x^2 - x + 2 = 0.$$ Its discriminant is $$D = (-1)^2 - 4(1)(2) = 1-8 = -7 < 0.$$ So the curves do not intersect.

Step 2: Determine the upper curve. At $x=0$, $$y=x^2+2=2, \qquad y=x=0.$$ Hence $y=x^2+2$ lies above $y=x$ throughout the interval $[0,3]$.

Step 3: Set up the integral. $$\text{Area} = \int_0^3 \left[(x^2+2) - x\right]dx = \int_0^3 (x^2 - x + 2)dx.$$

Step 4: Integrate. $$= \left[\frac{x^3}{3} - \frac{x^2}{2} + 2x\right]_0^3.$$ At $x=3$, $$\frac{27}{3} - \frac{9}{2} + 6 = 9 - \frac92 + 6 = \frac{21}{2}.$$ At $x=0$, value is $0$. Thus, $$\text{Area} = \frac{21}{2}.$$

Answer: $\frac{21}{2}$ square units.

Question 20 [CBSE 2025]

Using integration, find the area of the region bounded by the triangle whose vertices are $(1,0)$, $(2,2)$ and $(3,1)$.

Solution: Step 1: Find the equations of the three sides. For the line through $(1,0)$ and $(2,2)$: $$y = 2(x-1) = 2x-2.$$ For the line through $(2,2)$ and $(3,1)$: $$y-2 = -1(x-2) \implies y = -x+4.$$ For the line through $(1,0)$ and $(3,1)$: $$y = \frac12(x-1).$$

Step 2: Split the triangle into two vertical strips. From $x=1$ to $x=2$, the upper boundary is $y=2x-2$ and the lower boundary is $y=\frac12(x-1)$. From $x=2$ to $x=3$, the upper boundary is $y=-x+4$ and the lower boundary remains $y=\frac12(x-1)$.

Step 3: Set up the area integral. $$\text{Area} = \int_1^2 \left[(2x-2) - \frac12(x-1)\right]dx + \int_2^3 \left[(-x+4) - \frac12(x-1)\right]dx.$$

Step 4: Simplify the first integrand. $$(2x-2) - \frac12(x-1) = \frac32(x-1).$$ So, $$\int_1^2 \frac32(x-1)dx = \frac32\left[\frac{(x-1)^2}{2}\right]_1^2 = \frac34.$$

Step 5: Simplify the second integrand. $$(-x+4) - \frac12(x-1) = -\frac32x + \frac92.$$ Thus, $$\int_2^3 \left(-\frac32x + \frac92\right)dx = \left[-\frac34x^2 + \frac92x\right]_2^3 = \frac34.$$

Step 6: Add both parts. $$\text{Area} = \frac34 + \frac34 = \frac32.$$

Answer: $\frac{3}{2}$ square units.

Question 21 [CBSE 2026]

Find the area of the smaller region bounded by the circle $x^2 + y^2 = 16$ and the line $x = 2$.

Solution: Step 1: Understand the geometry. The circle has radius $4$. The vertical line $x=2$ cuts the circle, producing a smaller segment on the right side. The region is symmetric about the x-axis.

Step 2: Set up the integral. In the first quadrant, the smaller region extends from $x=2$ to $x=4$, with upper boundary $$y = \sqrt{16-x^2}.$$ So, $$\text{Area} = 2\int_2^4 \sqrt{16-x^2}\,dx.$$

Step 3: Use the standard formula. $$\int \sqrt{16-x^2}\,dx = \frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right).$$ Thus, $$\text{Area} = 2\left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}\right)\right]_2^4.$$

Step 4: Evaluate. At $x=4$, $$0 + 8\sin^{-1}(1) = 8\cdot\frac{\pi}{2} = 4\pi.$$ At $x=2$, $$\frac{2}{2}\sqrt{12} + 8\sin^{-1}\left(\frac12\right) = 2\sqrt3 + \frac{4\pi}{3}.$$ Therefore, $$\text{Area} = 2\left(4\pi - 2\sqrt3 - \frac{4\pi}{3}\right) = 2\left(\frac{8\pi}{3} - 2\sqrt3\right).$$ Hence, $$\text{Area} = \frac{16\pi}{3} - 4\sqrt3.$$

Answer: $\frac{16\pi}{3} - 4\sqrt{3}$ square units.

Question 22 [CBSE 2024]

Find the area of the region bounded by $y = \sin x$ and $y = \cos x$ between $x = 0$ and $x = \pi/4$.

Solution: Step 1: Determine the upper curve. On the interval $\left[0,\frac{\pi}{4}\right]$, $$\cos x \ge \sin x.$$ So the upper curve is $y=\cos x$ and the lower curve is $y=\sin x$.

Step 2: Set up the integral. $$\text{Area} = \int_0^{\pi/4} (\cos x - \sin x)dx.$$

Step 3: Integrate. $$= [\sin x + \cos x]_0^{\pi/4}.$$

Step 4: Evaluate. $$= \left(\frac{1}{\sqrt2} + \frac{1}{\sqrt2}\right) - (0+1) = \sqrt2 - 1.$$

Answer: $\sqrt{2} - 1$ square units.

Question 23 [CBSE 2022]

Find the area of the region bounded by the curves $x = y^2$ and $x = 3 - 2y^2$.

Solution: Step 1: Find the points of intersection. Set $$y^2 = 3 - 2y^2 \implies 3y^2 = 3 \implies y^2=1.$$ Hence, $$y = \pm 1.$$

Step 2: Determine right and left curves. At $y=0$, $$x=y^2=0, \qquad x=3-2y^2=3.$$ So $x=3-2y^2$ is the right curve and $x=y^2$ is the left curve.

Step 3: Set up the integral with respect to $y$. $$\text{Area} = \int_{-1}^{1} \left[(3-2y^2) - y^2\right]dy = \int_{-1}^{1} (3-3y^2)dy.$$

Step 4: Use symmetry. The integrand is even, so $$\text{Area} = 2\int_0^1 (3-3y^2)dy = 6\int_0^1 (1-y^2)dy.$$

Step 5: Evaluate. $$= 6\left[y - \frac{y^3}{3}\right]_0^1 = 6\left(1 - \frac13\right) = 6\cdot\frac23 = 4.$$

Answer: $4$ square units.

Question 24 [CBSE 2026]

Find the area bounded by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ and the line $\frac{x}{3} + \frac{y}{2} = 1$.

Solution: Step 1: Find the intercept points. The line meets the axes at $(3,0)$ and $(0,2)$, and these points also lie on the ellipse. So the smaller bounded region lies in the first quadrant.

Step 2: Write both curves as functions of $x$. From the ellipse, $$y = \frac{2}{3}\sqrt{9-x^2}.$$ From the line, $$y = \frac{2}{3}(3-x).$$

Step 3: Set up the integral. The ellipse lies above the line between $x=0$ and $x=3$, so $$\text{Area} = \int_0^3 \frac{2}{3}\left(\sqrt{9-x^2} - (3-x)\right)dx.$$

Step 4: Integrate. $$\text{Area} = \frac{2}{3}\left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\sin^{-1}\left(\frac{x}{3}\right) - 3x + \frac{x^2}{2} \right]_0^3.$$

Step 5: Evaluate. At $x=3$, $$0 + \frac{9}{2}\cdot\frac{\pi}{2} - 9 + \frac{9}{2} = \frac{9\pi}{4} - \frac{9}{2}.$$ At $x=0$, value is $0$. Hence, $$\text{Area} = \frac{2}{3}\left(\frac{9\pi}{4} - \frac{9}{2}\right) = \frac{3\pi}{2} - 3 = \frac{3}{2}(\pi - 2).$$

Answer: $\frac{3}{2}(\pi - 2)$ square units.