This MCQ module is based on: Area Under Curves – Lines, Parabolas, Circles, Ellipses
TOPIC 8 OF 25
Area Under Curves – Lines, Parabolas, Circles, Ellipses
🎓 Class 12
Mathematics
CBSE
Theory
Ch 8 — Application of Integrals
⏱ ~25 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Area Under Curves – Lines, Parabolas, Circles, Ellipses
Targeting Class 12 level in Calculus, with Advanced difficulty.
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8.1 Introduction
In geometry, we have learnt formulae to calculate areas of various geometrical figures including triangles, rectangles, trapezias and circles. Such formulae are fundamental in many real-life applications. However, these formulae are inadequate for calculating the areas enclosed by curves. For that, we shall need some concepts of Integral Calculus?.
In the previous chapter, we have studied how to find the area bounded by the curve \(y = f(x)\), the ordinates \(x = a\), \(x = b\), and the \(x\)-axis, while calculating the definite integral as the limit of a sum. Here, in this chapter, we shall study a specific application of integrals to find the area under simple curves, area between lines and arcs of circles, parabolas, and ellipses (standard forms only).
Historical Note
A. L. Cauchy (1789–1857) formalised the modern notion of the definite integral. The origin of Integral Calculus goes back to the early period of development of Mathematics and is related to the method of exhaustion developed by the mathematicians of ancient Greece (Eudoxus, 440 B.C., and Archimedes, 300 B.C.).
8.2 Area under Simple Curves
We have studied in the previous chapter that the definite integral is the limit of a sum, and we now consider the easy and intuitive way of finding the area bounded by the curve \(y = f(x)\), the \(x\)-axis, and the ordinates \(x = a\) and \(x = b\).
Fig 8.1 — Area under the curve \(y = f(x)\) from \(x = a\) to \(x = b\) computed as thin vertical strips of width \(dx\)
Area Formula (Vertical Strips)
The area A of the region bounded by the curve \(y = f(x)\), the \(x\)-axis, and the lines \(x = a\) and \(x = b\) (where \(b > a\)) is:
\[\text{Area} = \int_a^b y\,dx = \int_a^b f(x)\,dx\]
Area Formula (Horizontal Strips)
The area A of the region bounded by the curve \(x = g(y)\), the \(y\)-axis, and the lines \(y = c\) and \(y = d\) is:
\[\text{Area} = \int_c^d x\,dy = \int_c^d g(y)\,dy\]
Fig 8.2 — Area using horizontal strips when \(x = g(y)\)
Important — Curve Below x-axis
If the position of the curve is below the \(x\)-axis, then since \(f(x) < 0\) from \(x = a\) to \(x = b\), the area bounded by the curve, the \(x\)-axis, and the ordinates comes out to be negative. But we take the absolute value:
\[\text{Area} = \left|\int_a^b f(x)\,dx\right|\]
If part of the curve is above and part below the \(x\)-axis, split at the zero-crossing points and add the absolute values of each portion.
Worked Examples
Example 1: Area Enclosed by a Circle
Find the area enclosed by the circle \(x^2 + y^2 = a^2\).
Solution
Fig 8.5 — Circle \(x^2 + y^2 = a^2\), shading one quadrant
From the equation: \(y = \pm\sqrt{a^2 - x^2}\). In the first quadrant, \(y = \sqrt{a^2 - x^2} \geq 0\).
Since the circle is symmetrical about both axes, total area = 4 × area in the first quadrant:
\[\text{Area} = 4\int_0^a \sqrt{a^2 - x^2}\,dx\]Using the standard formula \(\int \sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\):
\[\text{Area} = 4\left[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a = 4\left[0 + \frac{a^2}{2}\cdot\frac{\pi}{2} - 0\right] = 4\cdot\frac{\pi a^2}{4} = \pi a^2\]Example 2: Area Enclosed by an Ellipse
Find the area enclosed by the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).
Solution
Fig 8.7 — Ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
From the equation: \(y = \frac{b}{a}\sqrt{a^2 - x^2}\) (first quadrant).
Area = 4 × (area in first quadrant):
\[\text{Area} = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2}\,dx = \frac{4b}{a}\left[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a\] \[= \frac{4b}{a}\cdot\frac{a^2}{2}\cdot\frac{\pi}{2} = \pi ab\]8.3 Area Between Two Curves
Formula
The area of the region enclosed between the curves \(y = f(x)\) and \(y = g(x)\), and the lines \(x = a\) and \(x = b\) (where \(f(x) \geq g(x)\) in \([a, b]\)) is:
\[\text{Area} = \int_a^b [f(x) - g(x)]\,dx\]
If the curves interchange positions (i.e., sometimes \(f > g\) and sometimes \(g > f\)), split the integral at the intersection points and take absolute values.
Example 3: Area Between a Line and a Parabola
Find the area of the region bounded by the line \(y = 3x + 2\), the \(x\)-axis, and the ordinates \(x = -1\) and \(x = 1\).
Solution
The line \(y = 3x + 2\) crosses the \(x\)-axis when \(3x + 2 = 0\), i.e., \(x = -\frac{2}{3}\).
For \(x \in [-1, -\frac{2}{3}]\): the line is below the \(x\)-axis (\(y < 0\)).
For \(x \in [-\frac{2}{3}, 1]\): the line is above the \(x\)-axis (\(y > 0\)).
Area of the region ACBA (below \(x\)-axis): \[\left|\int_{-1}^{-2/3}(3x+2)\,dx\right| = \left|\left[\frac{3x^2}{2}+2x\right]_{-1}^{-2/3}\right| = \left|\left(\frac{2}{3} - \frac{4}{3}\right) - \left(\frac{3}{2} - 2\right)\right| = \left|-\frac{2}{3} + \frac{1}{2}\right| = \frac{1}{6}\] Area of the region ADEA (above \(x\)-axis): \[\int_{-2/3}^1(3x+2)\,dx = \left[\frac{3x^2}{2}+2x\right]_{-2/3}^1 = \left(\frac{3}{2}+2\right) - \left(\frac{2}{3}-\frac{4}{3}\right) = \frac{7}{2} + \frac{2}{3} = \frac{25}{6}\] Total area = \(\frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}\) sq. units.
For \(x \in [-1, -\frac{2}{3}]\): the line is below the \(x\)-axis (\(y < 0\)).
For \(x \in [-\frac{2}{3}, 1]\): the line is above the \(x\)-axis (\(y > 0\)).
Area of the region ACBA (below \(x\)-axis): \[\left|\int_{-1}^{-2/3}(3x+2)\,dx\right| = \left|\left[\frac{3x^2}{2}+2x\right]_{-1}^{-2/3}\right| = \left|\left(\frac{2}{3} - \frac{4}{3}\right) - \left(\frac{3}{2} - 2\right)\right| = \left|-\frac{2}{3} + \frac{1}{2}\right| = \frac{1}{6}\] Area of the region ADEA (above \(x\)-axis): \[\int_{-2/3}^1(3x+2)\,dx = \left[\frac{3x^2}{2}+2x\right]_{-2/3}^1 = \left(\frac{3}{2}+2\right) - \left(\frac{2}{3}-\frac{4}{3}\right) = \frac{7}{2} + \frac{2}{3} = \frac{25}{6}\] Total area = \(\frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}\) sq. units.
Example 4: Area Between a Parabola and a Line
Find the area bounded by the curve \(y = \cos x\) between \(x = 0\) and \(x = 2\pi\).
Solution
\(\cos x \geq 0\) on \([0, \frac{\pi}{2}]\) and \([ \frac{3\pi}{2}, 2\pi]\); \(\cos x \leq 0\) on \([\frac{\pi}{2}, \frac{3\pi}{2}]\).
\[\text{Area} = \int_0^{\pi/2}\cos x\,dx + \left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right| + \int_{3\pi/2}^{2\pi}\cos x\,dx\]
\[= [\sin x]_0^{\pi/2} + |[\sin x]_{\pi/2}^{3\pi/2}| + [\sin x]_{3\pi/2}^{2\pi} = 1 + |-1 - 1| + |0 - (-1)| = 1 + 2 + 1 = 4\]
Interactive: Area Under \(y = x^n\) from 0 to \(a\)
Compute the area under a power curve and see how it changes with \(n\)
Enter values and click Compute.
Activity: Visualising Area as Vertical Strips
L2 Understand
Predict: The area under \(y = x^2\) from 0 to 1 is less than the area of the unit square. Estimate what fraction of the square is shaded.
- Draw a 1 × 1 square on graph paper. Plot \(y = x^2\) from \(x = 0\) to \(x = 1\).
- Count the number of small grid squares below the curve and compare to the total (100 if using a 10 × 10 grid).
- Calculate exactly: \(\int_0^1 x^2\,dx = \frac{1}{3}\). So the area is \(\frac{1}{3}\) of the unit square.
- Repeat for \(y = x^3\). The area is \(\frac{1}{4}\). Notice the pattern: for \(y = x^n\), area = \(\frac{1}{n+1}\).
Observation: As \(n\) increases, the curve \(y = x^n\) becomes flatter near \(x = 0\) and steeper near \(x = 1\). The area under the curve from 0 to 1 is \(\frac{1}{n+1}\), which approaches 0 as \(n \to \infty\). This makes geometric sense: for very large \(n\), \(x^n \approx 0\) for most of \([0, 1)\).
Competency-Based Questions
Scenario: A landscape architect designs a pond whose boundary follows the curve \(y = 4 - x^2\) (a downward parabola) above the \(x\)-axis. The pond's surface is at ground level (the \(x\)-axis). One unit on the plan equals 1 metre.
Q1. The pond extends from \(x = -2\) to \(x = 2\) because \(y = 0\) at those points. The surface area of the pond is:
L3 ApplyAnswer: (b). \(\int_{-2}^2 (4-x^2)\,dx = 2\int_0^2(4-x^2)\,dx = 2\left[4x-\frac{x^3}{3}\right]_0^2 = 2\left(8-\frac{8}{3}\right) = 2\cdot\frac{16}{3} = \frac{32}{3} \approx 10.67\) m².
Q2. The architect wants to build a rectangular walkway of maximum area entirely inside the pond (inscribed under the parabola). If the rectangle has width \(2x\) and height \(4 - x^2\), find the maximum area.
L4 AnalyseAnswer: Area of rectangle = \(A(x) = 2x(4-x^2) = 8x - 2x^3\). \(A'(x) = 8 - 6x^2 = 0 \Rightarrow x = \frac{2}{\sqrt{3}}\). \(A'' = -12x < 0\), confirming maximum.
\(A_{\max} = 2\cdot\frac{2}{\sqrt{3}}\left(4 - \frac{4}{3}\right) = \frac{4}{\sqrt{3}}\cdot\frac{8}{3} = \frac{32}{3\sqrt{3}} = \frac{32\sqrt{3}}{9} \approx 6.16\) m².
\(A_{\max} = 2\cdot\frac{2}{\sqrt{3}}\left(4 - \frac{4}{3}\right) = \frac{4}{\sqrt{3}}\cdot\frac{8}{3} = \frac{32}{3\sqrt{3}} = \frac{32\sqrt{3}}{9} \approx 6.16\) m².
Q3. What fraction of the total pond area does the maximum inscribed rectangle occupy? Evaluate whether this is a good use of space.
L5 EvaluateAnswer: Fraction = \(\frac{32\sqrt{3}/9}{32/3} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \approx 0.577\), or about 57.7%. This means the walkway covers just over half the pond area. For practical purposes, this is a reasonable use of space — the curved edges naturally reduce the usable rectangular area.
Q4. If the architect adds a fountain island shaped as the region between \(y = x^2\) and \(y = 4 - x^2\), compute the area of this lens-shaped island.
L6 CreateAnswer: The curves intersect where \(x^2 = 4 - x^2 \Rightarrow x^2 = 2 \Rightarrow x = \pm\sqrt{2}\).
Area = \(\int_{-\sqrt{2}}^{\sqrt{2}}[(4-x^2) - x^2]\,dx = \int_{-\sqrt{2}}^{\sqrt{2}}(4 - 2x^2)\,dx = 2\int_0^{\sqrt{2}}(4 - 2x^2)\,dx\) \[= 2\left[4x - \frac{2x^3}{3}\right]_0^{\sqrt{2}} = 2\left(4\sqrt{2} - \frac{4\sqrt{2}}{3}\right) = 2\cdot\frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3} \approx 7.54\] m².
Area = \(\int_{-\sqrt{2}}^{\sqrt{2}}[(4-x^2) - x^2]\,dx = \int_{-\sqrt{2}}^{\sqrt{2}}(4 - 2x^2)\,dx = 2\int_0^{\sqrt{2}}(4 - 2x^2)\,dx\) \[= 2\left[4x - \frac{2x^3}{3}\right]_0^{\sqrt{2}} = 2\left(4\sqrt{2} - \frac{4\sqrt{2}}{3}\right) = 2\cdot\frac{8\sqrt{2}}{3} = \frac{16\sqrt{2}}{3} \approx 7.54\] m².
Assertion–Reason Questions
Assertion (A): The area enclosed by the circle \(x^2 + y^2 = 9\) is \(9\pi\).
Reason (R): The area enclosed by the circle \(x^2 + y^2 = r^2\) is \(\pi r^2\).
Reason (R): The area enclosed by the circle \(x^2 + y^2 = r^2\) is \(\pi r^2\).
Answer: (a) — Both true. Here \(r = 3\), so area \(= \pi(3)^2 = 9\pi\). R gives the general formula that directly yields A.
Assertion (A): \(\int_{-1}^1 x^3\,dx = 0\), so the area bounded by \(y = x^3\), the \(x\)-axis, and \(x = -1, x = 1\) is 0.
Reason (R): When a curve lies partly above and partly below the \(x\)-axis, the definite integral gives the net signed area, not the total area.
Reason (R): When a curve lies partly above and partly below the \(x\)-axis, the definite integral gives the net signed area, not the total area.
Answer: (d) — The first part of A is true (\(\int_{-1}^1 x^3\,dx = 0\) since \(x^3\) is odd). But the conclusion that the area is 0 is false. The actual area = \(2\int_0^1 x^3\,dx = \frac{1}{2}\). R is true and explains why the integral is zero while the area is not: the negative and positive portions cancel in the signed integral but not in the total area.
Assertion (A): The area of the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\) is \(6\pi\).
Reason (R): The area of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(\pi ab\).
Reason (R): The area of the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) is \(\pi ab\).
Answer: (a) — Both true. Here \(a = 2, b = 3\), so area \(= \pi(2)(3) = 6\pi\). R provides the general formula.
Frequently Asked Questions
How to find area under a curve using integration?
The area under y = f(x) from a to b is the definite integral. If the curve is below the x-axis, take absolute value. Split at x-intercepts if the curve crosses the axis.
How to find area between two curves?
Area = integral of |f(x) - g(x)| from a to b. Find intersection points, determine which curve is above, then integrate the difference.
What is the area bounded by a parabola?
For y^2 = 4ax bounded by latus rectum x = a, the area is 8a^2/3. This is a standard NCERT application solved using integration.
How to calculate area of a circle using integration?
For x^2 + y^2 = r^2, area = 4 * integral from 0 to r of sqrt(r^2-x^2)dx = pi*r^2. Use the substitution x = r*sin(theta).
How to find area of an ellipse using integration?
For x^2/a^2 + y^2/b^2 = 1, area = 4 * integral from 0 to a of (b/a)*sqrt(a^2-x^2)dx = pi*a*b. Same substitution as circle.
Frequently Asked Questions — Application of Integrals
What is Area Under Curves - Lines, Parabolas, Circles, Ellipses in NCERT Class 12 Mathematics?
Area Under Curves - Lines, Parabolas, Circles, Ellipses is a key concept covered in NCERT Class 12 Mathematics, Chapter 8: Application of Integrals. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.
How do I solve problems on Area Under Curves - Lines, Parabolas, Circles, Ellipses step by step?
To solve problems on Area Under Curves - Lines, Parabolas, Circles, Ellipses, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 12 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.
What are the most important formulas for Chapter 8: Application of Integrals?
The essential formulas of Chapter 8 (Application of Integrals) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.
Is Area Under Curves - Lines, Parabolas, Circles, Ellipses important for the Class 12 board exam?
Area Under Curves - Lines, Parabolas, Circles, Ellipses is part of the NCERT Class 12 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.
What mistakes should students avoid in Area Under Curves - Lines, Parabolas, Circles, Ellipses?
Common mistakes in Area Under Curves - Lines, Parabolas, Circles, Ellipses include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.
Where can I find more NCERT practice questions on Area Under Curves - Lines, Parabolas, Circles, Ellipses?
End-of-chapter NCERT exercises for Area Under Curves - Lines, Parabolas, Circles, Ellipses cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 8, and solve at least one previous-year board paper to consolidate your understanding.
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