This MCQ module is based on: Exercises and Summary – Application of Integrals
TOPIC 9 OF 25
Exercises and Summary – Application of Integrals
🎓 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: Exercises and Summary – Application of Integrals
Targeting Class 12 level in Calculus, with Advanced difficulty.
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Exercise 8.1
1. Find the area of the region bounded by the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).
Here \(a = 4, b = 3\). Area of ellipse = \(\pi ab = \pi(4)(3) = 12\pi\) sq. units.
2. Find the area of the region bounded by the ellipse \(\frac{x^2}{4} + \frac{y^2}{9} = 1\).
Here \(a = 2, b = 3\). Area = \(\pi(2)(3) = 6\pi\) sq. units.
Choose the correct answer in the following Exercises 3 and 4:
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:
(A) \(\pi\) (B) \(\frac{\pi}{2}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{\pi}{4}\)
(A) \(\pi\) (B) \(\frac{\pi}{2}\) (C) \(\frac{\pi}{3}\) (D) \(\frac{\pi}{4}\)
\(\int_0^2 \sqrt{4-x^2}\,dx = \left[\frac{x}{2}\sqrt{4-x^2} + 2\sin^{-1}\frac{x}{2}\right]_0^2 = 0 + 2\cdot\frac{\pi}{2} = \pi\). Answer: (A) \(\pi\)
4. Area of the region bounded by the curve \(y^2 = 4x\), \(y\)-axis, and the line \(y = 3\) is:
(A) 2 (B) \(\frac{9}{4}\) (C) \(\frac{9}{3}\) (D) \(\frac{9}{2}\)
(A) 2 (B) \(\frac{9}{4}\) (C) \(\frac{9}{3}\) (D) \(\frac{9}{2}\)
From \(y^2 = 4x\): \(x = \frac{y^2}{4}\). Area = \(\int_0^3 \frac{y^2}{4}\,dy = \frac{1}{4}\left[\frac{y^3}{3}\right]_0^3 = \frac{1}{4}\cdot 9 = \frac{9}{4}\). Answer: (B) \(\frac{9}{4}\)
Miscellaneous Examples
Example 3 (Miscellaneous)
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 at \(x = -\frac{2}{3}\).
For \(x \in [-1, -\frac{2}{3}]\): line is below the \(x\)-axis. For \(x \in [-\frac{2}{3}, 1]\): line is above.
Area below = \(\left|\int_{-1}^{-2/3}(3x+2)\,dx\right| = \left|\left[\frac{3x^2}{2}+2x\right]_{-1}^{-2/3}\right| = \left|(-\frac{2}{3}+\frac{4}{3}) - (\frac{3}{2}-2)\right|\)... After computation = \(\frac{1}{6}\).
Area above = \(\int_{-2/3}^1(3x+2)\,dx = \frac{25}{6}\).
Total area = \(\frac{1}{6} + \frac{25}{6} = \frac{13}{3}\) sq. units.
For \(x \in [-1, -\frac{2}{3}]\): line is below the \(x\)-axis. For \(x \in [-\frac{2}{3}, 1]\): line is above.
Area below = \(\left|\int_{-1}^{-2/3}(3x+2)\,dx\right| = \left|\left[\frac{3x^2}{2}+2x\right]_{-1}^{-2/3}\right| = \left|(-\frac{2}{3}+\frac{4}{3}) - (\frac{3}{2}-2)\right|\)... After computation = \(\frac{1}{6}\).
Area above = \(\int_{-2/3}^1(3x+2)\,dx = \frac{25}{6}\).
Total area = \(\frac{1}{6} + \frac{25}{6} = \frac{13}{3}\) sq. units.
Example 4 (Miscellaneous)
Find the area bounded by the curve \(y = \cos x\) between \(x = 0\) and \(x = 2\pi\).
Solution
The curve \(y = \cos x\) is positive on \([0, \frac{\pi}{2}]\) and \([\frac{3\pi}{2}, 2\pi]\), and negative on \([\frac{\pi}{2}, \frac{3\pi}{2}]\).
Area = \(\int_0^{\pi/2}\cos x\,dx + \left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right| + \int_{3\pi/2}^{2\pi}\cos x\,dx = 1 + 2 + 1 = 4\) sq. units.
Area = \(\int_0^{\pi/2}\cos x\,dx + \left|\int_{\pi/2}^{3\pi/2}\cos x\,dx\right| + \int_{3\pi/2}^{2\pi}\cos x\,dx = 1 + 2 + 1 = 4\) sq. units.
Miscellaneous Exercise on Chapter 8
1. Find the area under the given curves and given lines:
(i) \(y = x^2\), \(x = 1\), \(x = 2\), and the \(x\)-axis
(ii) \(y = x^4\), \(x = 1\), \(x = 5\), and the \(x\)-axis
(i) \(y = x^2\), \(x = 1\), \(x = 2\), and the \(x\)-axis
(ii) \(y = x^4\), \(x = 1\), \(x = 5\), and the \(x\)-axis
(i) \(\int_1^2 x^2\,dx = \left[\frac{x^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}\) sq. units.
(ii) \(\int_1^5 x^4\,dx = \left[\frac{x^5}{5}\right]_1^5 = \frac{3125}{5} - \frac{1}{5} = \frac{3124}{5} = 624.8\) sq. units.
(ii) \(\int_1^5 x^4\,dx = \left[\frac{x^5}{5}\right]_1^5 = \frac{3125}{5} - \frac{1}{5} = \frac{3124}{5} = 624.8\) sq. units.
2. Sketch the graph of \(y = |x + 3|\) and evaluate \(\int_{-6}^0 |x+3|\,dx\).
\(|x+3| = -(x+3)\) when \(x < -3\) and \(= (x+3)\) when \(x \geq -3\).
\[\int_{-6}^{-3}[-(x+3)]\,dx + \int_{-3}^0(x+3)\,dx = \left[-\frac{x^2}{2}-3x\right]_{-6}^{-3} + \left[\frac{x^2}{2}+3x\right]_{-3}^0\] \[= \left[(-\frac{9}{2}+9) - (-18+18)\right] + \left[0 - (\frac{9}{2}-9)\right] = \frac{9}{2} + \frac{9}{2} = 9\] sq. units.
\[\int_{-6}^{-3}[-(x+3)]\,dx + \int_{-3}^0(x+3)\,dx = \left[-\frac{x^2}{2}-3x\right]_{-6}^{-3} + \left[\frac{x^2}{2}+3x\right]_{-3}^0\] \[= \left[(-\frac{9}{2}+9) - (-18+18)\right] + \left[0 - (\frac{9}{2}-9)\right] = \frac{9}{2} + \frac{9}{2} = 9\] sq. units.
3. Find the area bounded by the curve \(y = \sin x\) between \(x = 0\) and \(x = 2\pi\).
\(\sin x \geq 0\) on \([0, \pi]\) and \(\sin x \leq 0\) on \([\pi, 2\pi]\).
Area = \(\int_0^\pi \sin x\,dx + \left|\int_\pi^{2\pi}\sin x\,dx\right| = [-\cos x]_0^\pi + |[-\cos x]_\pi^{2\pi}| = 2 + |-(-1-1)| = 2 + 2 = 4\) sq. units.
Area = \(\int_0^\pi \sin x\,dx + \left|\int_\pi^{2\pi}\sin x\,dx\right| = [-\cos x]_0^\pi + |[-\cos x]_\pi^{2\pi}| = 2 + |-(-1-1)| = 2 + 2 = 4\) sq. units.
4. Area bounded by the curve \(y = x^3\), the \(x\)-axis, and the ordinates \(x = -2\) and \(x = 1\) is:
(A) \(-9\) (B) \(\frac{-15}{4}\) (C) \(\frac{15}{4}\) (D) \(\frac{17}{4}\)
(A) \(-9\) (B) \(\frac{-15}{4}\) (C) \(\frac{15}{4}\) (D) \(\frac{17}{4}\)
\(x^3\) changes sign at \(x = 0\). Area = \(\left|\int_{-2}^0 x^3\,dx\right| + \int_0^1 x^3\,dx = \left|\left[\frac{x^4}{4}\right]_{-2}^0\right| + \left[\frac{x^4}{4}\right]_0^1 = |0 - 4| + \frac{1}{4} = 4 + \frac{1}{4} = \frac{17}{4}\). Answer: (D)
5. The area bounded by the curve \(y = |x|\), \(x\)-axis, and the ordinates \(x = -1\) and \(x = 1\) is given by:
(A) 0 (B) \(\frac{1}{3}\) (C) \(\frac{2}{3}\) (D) 1
(A) 0 (B) \(\frac{1}{3}\) (C) \(\frac{2}{3}\) (D) 1
\(|x|\) is even, so area = \(2\int_0^1 x\,dx = 2\cdot\frac{1}{2} = 1\). Answer: (D) 1
Summary of Chapter 8
Key Formulae
1. Area bounded by curve, x-axis, and vertical lines:
\[\text{Area} = \int_a^b y\,dx = \int_a^b f(x)\,dx\]2. Area bounded by curve, y-axis, and horizontal lines:
\[\text{Area} = \int_c^d x\,dy = \int_c^d g(y)\,dy\]3. Area between two curves:
\[\text{Area} = \int_a^b [f(x) - g(x)]\,dx \quad \text{where } f(x) \geq g(x)\]4. Standard areas:
- Circle \(x^2 + y^2 = r^2\): Area = \(\pi r^2\)
- Ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\): Area = \(\pi ab\)
5. When curve crosses the x-axis: Split at zero-crossing points and add absolute values of each portion.
Historical Note
The origin of the 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.) pioneered early methods for computing areas. The systematic approach to the theory of Calculus began in the 17th century. In 1665, Newton began his work on the Calculus (the theory of fluxions), and in 1684–86 Leibnitz published his work on Calculus summatorius. Both Newton and Leibnitz adopted quite independent lines of approach, though their respective theories accomplished results that were practically identical. Leibnitz used the notion of the definite integral, while Cauchy in the early 19th century formalised the concept of limit that underpins the modern definition.
Activity: Area Estimation by Counting Squares
L2 Understand
Predict: Before integrating, can you estimate the area of a quarter-circle of radius 4 by counting grid squares?
- Draw the quarter-circle \(x^2 + y^2 = 16\) in the first quadrant on a 4 × 4 grid.
- Count grid squares fully inside the curve (these are underestimates).
- Count grid squares that the curve passes through (boundary squares).
- Estimate: area is between (full squares) and (full + boundary squares).
- Now compute exactly: \(\frac{1}{4}\pi(4)^2 = 4\pi \approx 12.57\). How close was your estimate?
Observation: On a unit grid, roughly 9–10 squares are fully inside, and about 6–7 are boundary squares. So the area is between 9 and 16, with the exact value \(4\pi \approx 12.57\) falling in between. This is essentially what integration does — it takes infinitely thin strips instead of grid squares, giving the exact answer.
Competency-Based Questions
Scenario: A farmer has a plot of land bounded by a straight canal (\(x\)-axis), a road (\(y\)-axis), and a curved fence following \(y = 6 - x^2\) (in units of 10 metres). The farmer wants to divide the plot into two equal halves using a vertical fence at \(x = k\).
Q1. The total area of the plot (from \(x = 0\) to where the curve meets the canal) is:
L3 ApplyAnswer: (d). The curve meets the \(x\)-axis when \(6 - x^2 = 0 \Rightarrow x = \sqrt{6}\). Area = \(\int_0^{\sqrt{6}}(6-x^2)\,dx = \left[6x - \frac{x^3}{3}\right]_0^{\sqrt{6}} = 6\sqrt{6} - \frac{6\sqrt{6}}{3} = 6\sqrt{6} - 2\sqrt{6} = 4\sqrt{6}\) sq. units.
Q2. To divide the plot into two equal halves, the farmer needs \(\int_0^k (6-x^2)\,dx = 2\sqrt{6}\). Set up and explain the equation in \(k\) that needs to be solved.
L4 AnalyseAnswer: \(\int_0^k(6-x^2)\,dx = 6k - \frac{k^3}{3} = 2\sqrt{6}\). This gives the cubic equation \(k^3 - 18k + 6\sqrt{6} = 0\). This must be solved numerically. By trial: \(k \approx 1.1\) gives approximately \(6(1.1) - \frac{1.331}{3} \approx 6.6 - 0.44 = 6.16\), but \(2\sqrt{6} \approx 4.90\). So \(k\) is closer to 0.87. The equation requires numerical methods for an exact answer.
Q3. Convert the area to actual land area in square metres (recall: 1 unit = 10 m on the plan).
L3 ApplyAnswer: Each unit on the plan = 10 m. So 1 sq. unit on the plan = \(10 \times 10 = 100\) m². Total area = \(4\sqrt{6} \times 100 = 400\sqrt{6} \approx 979.8\) m², or about 0.098 hectares.
Q4. If the farmer instead wants to enclose a rectangular portion of maximum area within the curved plot (with one side on the x-axis), what are the optimal dimensions?
L6 CreateAnswer: Rectangle with width \(x\) (from 0 to \(x\)) and height \(6 - x^2\). Area \(= x(6 - x^2) = 6x - x^3\). \(A'(x) = 6 - 3x^2 = 0 \Rightarrow x = \sqrt{2}\). Height = \(6 - 2 = 4\). Dimensions: width = \(\sqrt{2} \approx 1.41\) units (14.1 m), height = 4 units (40 m). Maximum area = \(\sqrt{2} \times 4 = 4\sqrt{2} \approx 5.66\) sq. units (566 m²).
Assertion–Reason Questions
Assertion (A): The area bounded by \(y = \sin x\) and the \(x\)-axis from \(x = 0\) to \(x = 2\pi\) is \(4\).
Reason (R): \(\int_0^{2\pi}\sin x\,dx = 0\).
Reason (R): \(\int_0^{2\pi}\sin x\,dx = 0\).
Answer: (b) — Both true. A is true (area = 4 as shown in the exercises). R is also true (\(\int_0^{2\pi}\sin x\,dx = [-\cos x]_0^{2\pi} = 0\)). However, R does NOT explain A; in fact, it shows why the integral value differs from the area — the signed integral is 0 but the actual area is 4 (positive and negative parts are added separately).
Assertion (A): To find the area between \(y = x^2\) and \(y = x\), we compute \(\int_0^1 (x - x^2)\,dx\).
Reason (R): For \(x \in [0, 1]\), the line \(y = x\) lies above the parabola \(y = x^2\).
Reason (R): For \(x \in [0, 1]\), the line \(y = x\) lies above the parabola \(y = x^2\).
Answer: (a) — Both true. The curves intersect at \(x = 0\) and \(x = 1\). For \(x \in (0, 1)\): \(x > x^2\), so \(y = x\) is above \(y = x^2\) (R is true). Therefore, area = \(\int_0^1(x - x^2)\,dx\) (A is true). R explains A because the formula uses \(f(x) - g(x)\) where \(f\) is the upper curve. Area = \(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\).
Assertion (A): The area of the region bounded by \(y^2 = 4x\) and \(x = 3\) is \(8\sqrt{3}\).
Reason (R): From \(y^2 = 4x\), we get \(y = 2\sqrt{x}\) (upper half). The area is \(2\int_0^3 2\sqrt{x}\,dx\).
Reason (R): From \(y^2 = 4x\), we get \(y = 2\sqrt{x}\) (upper half). The area is \(2\int_0^3 2\sqrt{x}\,dx\).
Answer: (a) — Both true. By symmetry about the \(x\)-axis: Area = \(2\int_0^3 2\sqrt{x}\,dx = 4\left[\frac{2}{3}x^{3/2}\right]_0^3 = \frac{8}{3}(3\sqrt{3}) = 8\sqrt{3}\). R correctly describes the method, and A gives the correct value.
Frequently Asked Questions
What exercises are in NCERT Class 12 Chapter 8?
Chapter 8 contains Exercise 8.1 (area under simple curves, between line and parabola/circle/ellipse) and a Miscellaneous Exercise with complex bounded region problems.
How to approach area between line and parabola?
Find intersection points, sketch the region, determine which curve is on top, set up the integral of (upper - lower) between intersections, evaluate and simplify.
How to find area using horizontal strips?
When curves are x = f(y), integrate with respect to y: area = integral of |x_right - x_left| dy. Use when boundaries are clearer in terms of y.
What are common exam questions from Chapter 8?
Area between parabola and line, circle/ellipse area by integration, area between two parabolas, triangle area using integration, and enclosed regions between standard curves.
What are tips for solving area problems?
Draw a rough sketch first, find all intersection points, identify upper and lower curves, show the integral setup clearly, and double-check by visual estimation.
Frequently Asked Questions — Application of Integrals
What is Exercises and Summary - Application of Integrals in NCERT Class 12 Mathematics?
Exercises and Summary - Application of Integrals 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 Exercises and Summary - Application of Integrals step by step?
To solve problems on Exercises and Summary - Application of Integrals, 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 Exercises and Summary - Application of Integrals important for the Class 12 board exam?
Exercises and Summary - Application of Integrals 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 Exercises and Summary - Application of Integrals?
Common mistakes in Exercises and Summary - Application of Integrals 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 Exercises and Summary - Application of Integrals?
End-of-chapter NCERT exercises for Exercises and Summary - Application of Integrals 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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