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.
Historical Note
This section traces the historical development of integral calculus, emphasizing its origins and evolution. The concept of integration dates back to ancient Greece with the method of exhaustion developed by mathematicians such as Eudoxus (440 B.C.) and Archimedes (300 B.C.). This method involved approximating areas and volumes by inscribing and circumscribing polygons, regarded as an early form of integration. The systematic development of calculus began in the 17th century with Isaac Newton and Gottfried Wilhelm Leibniz. Newton introduced the theory of fluxions, focusing on rates of change and tangents, and developed the concept of the anti-derivative (indefinite integral) as the inverse operation of differentiation. Leibniz independently developed calculus, emphasizing the summation of infinitely small quantities and introducing the integral symbol ∫. He recognized the connection between differentiation and integration, now known as the Fundamental Theorem of Calculus. The section notes that although Newton and Leibniz had different approaches, their theories were practically equivalent. The rigorous justification of limits and the foundation of calculus were later provided by Augustin-Louis Cauchy in the early 19th century. The section concludes with a quotation from Lie Sophie, acknowledging the contributions of earlier mathematicians like Kepler, Descartes, Cavalieri, Fermat, and Wallis, and crediting Newton and Leibniz with discovering the inverse relationship between differentiation and integration.
📊 Diagram: No diagrams; historical portraits and timelines are typical but not included here.
🧪 Activity: No activity; serves to provide historical context.
🔗 Connection: Concludes the chapter; foundation for further study in calculus.
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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