Mathematics | Class 12 Mathematics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 min read
Mathematics – this guide gives you a concise, exam-ready overview of Mathematics from Class 12 Mathematics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
8.1 Introduction
This section introduces the fundamental motivation behind studying the application of integrals in mathematics, particularly in geometry and real-life problems. It begins by recalling that in earlier studies of geometry, students learned formulae to calculate areas of simple geometric figures such as triangles, rectangles, trapeziums, and circles. These formulae are essential for many practical applications but are limited to simple shapes with straight edges or standard curves. However, when dealing with areas enclosed by more complex curves, these elementary formulae become inadequate. To overcome this limitation, concepts from Integral Calculus are necessary. The chapter builds on the previous chapter where the definite integral was introduced as the limit of a sum, specifically for finding the area bounded by a curve y = f(x), the ordinates x = a and x = b, and the x-axis. The current chapter focuses on applying integrals to find areas under simple curves, areas between lines and arcs of circles, parabolas, and ellipses in their standard forms. It also covers finding areas bounded by these curves. The section includes a historical reference to A.L. Cauchy, who contributed significantly to the rigorous foundation of integral calculus. The introduction sets the stage for understanding how integration serves as a powerful tool for calculating areas that are otherwise difficult or impossible to determine using elementary geometry.
📊 Diagram: The chapter begins with a portrait of A.L. Cauchy (1789-1857), highlighting his contribution to integral calculus.
🧪 Activity: No specific activity in this introductory section.
🔗 Connection: Leads to Section 8.2 where the method of finding area under simple curves using integrals is explained in detail.
Frequently asked questions
1. Find the area of the region bounded by the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\). 2. Find the area of the region bounded by the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\).
1. The given ellipse is \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).
This ellipse has semi-major axis \(a = 4\) (since \(16 = 4^2\)) and semi-minor axis \(b = 3\) (since \(9 = 3^2\)).
Area of ellipse = \(\pi a b = \pi \times 4 \times 3 = 12\pi\).
2. The given ellipse is \(\frac{x^2}{4} + \frac{y^2}{9} = 1\).
Here, semi-major axis \(a = 3\) (since \(9 = 3^2\)) and semi-minor axis \(b = 2\) (since \(4 = 2^2\)).
Area of ellipse = \(\pi a b = \pi \times 3 \times 2 = 6\pi\).
3. Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is
The circle is $x^2 + y^2 = 4$, radius $r=2$. The area in the first quadrant bounded by $x=0$ and $x=2$ is the quarter circle sector from $x=0$ to $x=2$. Since $x=2$ is the rightmost point of the circle, the area is one quarter of the circle area.
Area of full circle = $\pi r^2 = \pi \times 2^2 = 4\pi$
Area in first quadrant bounded by $x=0$ and $x=2$ = $\frac{1}{4} \times 4\pi = \pi$
Hence, the answer is (A) $\pi$.
4. Area of the region bounded by the curve $y^2 = 4x$, $y$-axis and the line $y = 3$ is
Given curve: $y^2 = 4x$ or $x = \frac{y^2}{4}$.
Bounded by $y$-axis ($x=0$), curve, and line $y=3$.
Area = $\int_{y=0}^{3} x \, dy = \int_0^3 \frac{y^2}{4} dy = \frac{1}{4} \int_0^3 y^2 dy = \frac{1}{4} \left[ \frac{y^3}{3} \right]_0^3 = \frac{1}{4} \times \frac{27}{3} = \frac{27}{12} = \frac{9}{4}$.
Hence, answer is (B) $\frac{9}{4}$.
Find the area under the given curves and given lines: (i) y = x^2, x = 1, x = 2 and x-axis (ii) y = x^4, x = 1, x = 5 and x-axis
Solution: (i) Area under y = x^2 from x = 1 to x = 2 and x-axis: Area = ∫ from 1 to 2 of x^2 dx = [x^3/3] from 1 to 2 = (8/3) - (1/3) = 7/3.
(ii) Area under y = x^4 from x = 1 to x = 5 and x-axis: Area = ∫ from 1 to 5 of x^4 dx = [x^5/5] from 1 to 5 = (3125/5) - (1/5) = (3124/5) = 624.8.
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