Concept Scroll LogoConceptScroll
Join or Sign In
MathematicsClass 12Mathematics

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.

8.2 Area under Simple Curves

This section elaborates on the intuitive and formal method of finding the area bounded by a curve y = f(x), the x-axis, and vertical lines (ordinates) x = a and x = b using definite integrals. It recalls the concept of definite integral as the limit of sums and the Fundamental Theorem of Calculus for evaluating these integrals. The area under the curve can be visualized as composed of a large number of very thin vertical strips, each of infinitesimal width dx and height y = f(x). The area of each thin strip is dA = y dx. Summing these elementary areas from x = a to x = b gives the total area A = ∫ from a to b of f(x) dx. Similarly, if the curve is expressed as x = g(y), the area bounded by this curve, the y-axis, and horizontal lines y = c and y = d can be found by considering horizontal strips of width dy and length x = g(y), so that dA = x dy. The total area is then A = ∫ from c to d of g(y) dy. The section also discusses the sign of the area when the curve lies below the x-axis. Since f(x) < 0 in such cases, the definite integral yields a negative value, but the area is always taken as the absolute value of this integral. When the curve crosses the x-axis, the total area is the sum of the absolute values of the integrals over the intervals where the curve lies above or below the axis. This section includes diagrams illustrating vertical and horizontal strips and the effect of the curve lying below the x-axis.

📊 Diagram: Fig 8.1 shows vertical strips under a curve y = f(x) between x = a and x = b. Fig 8.2 illustrates horizontal strips under a curve x = g(y) between y = c and y = d. Fig 8.3 depicts a curve lying below the x-axis, showing the negative integral value. Fig 8.4 shows a curve crossing the x-axis with areas above and below the axis.

🧪 Activity: No specific activity, but students are encouraged to visualize areas as sums of strips.

🔗 Connection: Prepares for examples calculating areas bounded by circles and ellipses in the next sections.

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.

Ready to ace this chapter?

Get the full Mathematics chapter — interactive notes, diagrams, worked solutions, polls and a free practice quiz — in the ConceptScroll app.

Open in ConceptScroll →

Study smarter with ConceptScroll

Daily NCERT-aligned reels, AI doubt solving and chapter quizzes — all free.

Start learning free
#cbse notes#class 12#mathematics#ncert

Continue reading

← All articles