NCERT Solutions For Class 12 Maths Chapter 8 Application Of Integrals Exercise 8.1 - 2025-26

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

Summary: To find the area enclosed by the ellipse, use the property of symmetry and integrate over one quadrant, then multiply by 4.

• Rewrite the given equation as \(\frac{x^2}{4^2}+\frac{y^2}{3^2}=1\).
• Ellipse meets x-axis at \(x = \pm4\), and y-axis at \(y = \pm3\).
• The ellipse is symmetric about both axes.
• Area of ellipse = \(4\) × (area in first quadrant).
• Area in first quadrant = \(\int_{0}^{4} y dx\).
• From the ellipse, \(y = 3\sqrt{1 - \frac{x^2}{16}}\).
• So, Area = \(4 \int_{0}^{4} 3\sqrt{1 - \frac{x^2}{16}} dx\)
• Let’s simplify: \(3\sqrt{1 - \frac{x^2}{16}} = \frac{3}{4}\sqrt{16 - x^2}\).
• So, area in first quadrant = \(\frac{3}{4}\int_{0}^{4} \sqrt{16 - x^2} dx\)
• Apply standard integral: \(\int \sqrt{a^2-x^2}dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\).
• Area in first quadrant = \(\frac{3}{4}\left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}(\frac{x}{4})\right]_{0}^{4}\).
• Plug in limits: \(x=4\) and \(x=0\):
• \(\sin^{-1}(1) = \frac{\pi}{2}\), \(\sin^{-1}(0) = 0\)
• Area in first quadrant = \(\frac{3}{4} \times 8 \times \frac{\pi}{2} = 3\pi\)
• Total area = \(4 \times 3\pi = 12\pi\) sq. units.

2. Find the area of the region bounded by the ellipse \[\frac{{{x}^{2}}}{4}+\frac{{{y}^{2}}}{9}=1\].

Summary: To find the area inside this ellipse, use symmetry, integrate for one quadrant, and multiply by 4.

• Rewrite: \(\frac{x^2}{2^2} + \frac{y^2}{3^2} = 1\).
• Ellipse crosses x-axis at \(x = \pm2\), y-axis at \(y = \pm3\).
• Equation is symmetric in both axes.
• Area of ellipse = \(4\) × (area in first quadrant).
• Area in first quadrant = \(\int_{0}^{2} y dx\).
• From the ellipse: \(y = 3 \sqrt{1 - \frac{x^2}{4}}\).
• Or, \(y = \frac{3}{2}\sqrt{4 - x^2}\).
• Area = \(\int_{0}^{2} \frac{3}{2}\sqrt{4 - x^2} dx\).
• Factor out: \(\frac{3}{2}\int_{0}^{2}\sqrt{4 - x^2}dx\).
• Use the standard formula as before: \(\int \sqrt{a^2-x^2}dx\).
• Apply limits: get area in first quadrant = \(\frac{3}{2}\left[\frac{x}{2}\sqrt{4 - x^2} + 2\sin^{-1}(\frac{x}{2})\right]_{0}^{2}\).
• With limits: area in first quadrant = \(\frac{3}{2} \cdot \frac{2\pi}{2} = \frac{3\pi}{2}\).
• Total area = \(4 \cdot \frac{3\pi}{2} = 6\pi\) sq. units.

3. Area lying in the first quadrant and bounded by the circle \[{{\text{x}}^2} + {y^2} = 4\] and the lines \[x = 0\] and \[x = 2\] is

\[\pi \] \[\frac{\pi }{2}\] \[\frac{\pi }{3}\] \[\frac{\pi }{4}\]
Ans: Area of OAB = \(\int_{0}^{2} y dx\) Given, \(x^2 + y^2 = 4\) (Equation of circle) So, \(y^2 = 4 - x^2\) ⇒ \(y = \sqrt{4 - x^2}\) Calculate: Area = \(\int_{0}^{2} \sqrt{4 - x^2} dx\) Use the standard integral: \(\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\) With \(a=2\): Area = \([\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2}]_0^2\) Plug in values: At \(x=2\): \(\sin^{-1}(1)=\frac{\pi}{2}\), so expression is \(2 \times \frac{\pi}{2} = \pi\). Final answer: \(\pi\) square units. Hence, option (A) is correct.

4. Area of the region bounded by the curve \({{y}^{2}}=4x\), y-axis and the line \(y=3\) is

2
\(\frac{9}{4}\)
\(\frac{9}{3}\)
\(\frac{9}{2}\)

Ans: The area bounded by curve \(y^2=4x\), y-axis and \(y=3\) is the region OAB. Area = \(\int_0^3 x\,dy\) From the curve: \(x = \frac{y^2}{4}\) So, Area = \(\int_0^3 \frac{y^2}{4} dy\) = \(\frac{1}{4}\int_0^3 y^2 dy\) = \(\frac{1}{4}\left[\frac{y^3}{3}\right]_0^3\) = \(\frac{1}{4} \left[\frac{27}{3} - 0\right]\) = \(\frac{1}{4} [9]\) = \(\frac{9}{4}\) square units So, option (B) is the correct answer.

What You Will Learn from Class 12 Maths Exercise 8.1 Solutions

• You will practice finding area under curves and between curves using integrals.
• ex 8.1 class 12 maths ncert solutions teach you how to set correct limits for area questions.
• Questions cover calculations for ellipses, circles, and parabolas.
• It’s important to identify upper and lower curves before setting up an integral.
• Drawing a sketch helps to visualize all bounded regions for each problem.