NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals (NCERT 2026–27)

These Class 12 Maths Chapter 8 solutions cover Application of Integrals from the latest NCERT textbook (Reprint 2026–27). Every question of Exercise 8.1 and the Miscellaneous Exercise on Chapter 8 is reproduced verbatim and solved step by step — finding the area under simple curves, the area bounded by a circle and an ellipse, and the area between a curve, the axes and given ordinates — with each answer cross-checked against the book’s answer key.

Class: 12 Subject: Mathematics Chapter: 8 – Application of Integrals Exercises: Exercise 8.1, Miscellaneous Exercise Topics: Area under curves, circle, ellipse Session: 2026–27
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Chapter 8 Overview

Chapter 8, Application of Integrals, uses the definite integral to compute the area enclosed by a curve. The area bounded by the curve y = f(x), the x-axis and the ordinates x = a, x = b is found by adding up thin vertical strips of area y dx; horizontal strips of area x dy are used when measuring along the y-axis. The chapter applies these ideas to standard figures — the circle and the ellipse — and shows how to handle a curve that dips below the axis by taking the absolute value of the integral. The Class 12 Maths Chapter 8 solutions below work through every Exercise 8.1 and Miscellaneous Exercise question in order.

Key Concepts

Area as a limit of a sum: the region between y = f(x), the x-axis and x = a, x = b is built from elementary vertical strips of area dA = y dx; summing them gives a definite integral.

Vertical vs. horizontal strips: integrate y dx between two x-ordinates, or x dy between two y-ordinates — choose whichever the boundary makes simpler.

Curve below the axis: if f(x) < 0 on [a, b], the integral is negative; area is its absolute value. If a curve crosses the axis, split the integral and add the magnitudes: Area = |A1| + A2.

Symmetry shortcut: a circle or ellipse is symmetric about both axes, so its total area is 4 × the first-quadrant area — this simplifies the integration.

Important Formulas (Chapter 8)

Area under a curve (vertical strips): Area = ∫ab y dx = ∫ab f(x) dx.

Area to the side of the y-axis (horizontal strips): Area = ∫cd x dy = ∫cd g(y) dy.

Standard integral: ∫ √(a2 − x2) dx = (x/2)√(a2 − x2) + (a2/2) sin−1(x/a) + C.

Area of a circle x2 + y2 = a2:  A = πa2.

Area of an ellipse x2/a2 + y2/b2 = 1:  A = πab.

Exercise 8.1 Solutions

Questions are reproduced verbatim from the NCERT textbook; the worked solutions are original and verified against the answers given at the back of the book.

1. Find the area of the region bounded by the ellipse x2/16 + y2/9 = 1.

SOLUTION Here a2 = 16 and b2 = 9, so a = 4, b = 3. The ellipse is symmetric about both axes, so Area = 4 × (area in the first quadrant). From the equation, in the first quadrant y = (3/4)√(16 − x2), with x from 0 to 4. Area = 4 ∫04 (3/4)√(16 − x2) dx = 3 ∫04 √(16 − x2) dx. Using ∫√(a2−x2) dx = (x/2)√(a2−x2) + (a2/2)sin−1(x/a) with a = 4: = 3 [ (x/2)√(16−x2) + 8 sin−1(x/4) ]04 = 3 [ (0 + 8 · π/2) − (0 + 0) ] = 3 × 4π. ∴ Area = 12π square units.

2. Find the area of the region bounded by the ellipse x2/4 + y2/9 = 1.

SOLUTION Here a2 = 4 and b2 = 9, so the semi-axes are 2 and 3. Using the standard result Area of ellipse = π × (semi-axis) × (semi-axis): By symmetry, Area = 4 ∫02 y dx, where y = (3/2)√(4 − x2) in the first quadrant. = 4 × (3/2) ∫02 √(4 − x2) dx = 6 [ (x/2)√(4−x2) + 2 sin−1(x/2) ]02. = 6 [ (0 + 2 · π/2) − 0 ] = 6 × π. ∴ Area = 6π square units.

Choose the correct answer in the following Exercises 3 and 4.

3. Area lying in the first quadrant and bounded by the circle x2 + y2 = 4 and the lines x = 0 and x = 2 is (A) π   (B) π/2   (C) π/3   (D) π/4

SOLUTION The circle has radius 2. The region bounded by x = 0, x = 2 and the circle, in the first quadrant, is exactly one quarter of the disc. Area = ∫02 √(4 − x2) dx = [ (x/2)√(4−x2) + 2 sin−1(x/2) ]02 = (0 + 2 · π/2) − 0 = π. ∴ the correct option is (A) π.

4. Area of the region bounded by the curve y2 = 4x, y-axis and the line y = 3 is (A) 2   (B) 9/4   (C) 9/3   (D) 9/2

SOLUTION The boundary uses the y-axis and the line y = 3, so use horizontal strips and integrate with respect to y. From y2 = 4x, x = y2/4. Area = ∫03 x dy = ∫03 (y2/4) dy. = (1/4)[ y3/3 ]03 = (1/4)(27/3) = (1/4)(9) = 9/4. ∴ the correct option is (B) 9/4.

Miscellaneous Exercise on Chapter 8 Solutions

1. Find the area under the given curves and given lines: (i) y = x2, x = 1, x = 2 and x-axis (ii) y = x4, x = 1, x = 5 and x-axis

SOLUTION (i) Area = ∫12 x2 dx = [ x3/3 ]12 = 8/3 − 1/3 = 7/3 square units. (ii) Area = ∫15 x4 dx = [ x5/5 ]15 = 3125/5 − 1/5 = 3124/5 = 624.8 square units.

2. Sketch the graph of y = |x + 3| and evaluate ∫−60 |x + 3| dx.

SOLUTION Graph. y = |x + 3| is a V-shaped graph with its vertex (lowest point) at (−3, 0). For x ≥ −3 it is the line y = x + 3 (rising); for x < −3 it is the line y = −(x + 3) (falling). Both arms lie on or above the x-axis. Split at the vertex x = −3, since the formula for |x + 3| changes there: ∫−60 |x+3| dx = ∫−6−3 −(x+3) dx + ∫−30 (x+3) dx. First integral: −[ x2/2 + 3x ]−6−3 = −[ (9/2 − 9) − (18 − 18) ] = −(−9/2) = 9/2. Second integral: [ x2/2 + 3x ]−30 = 0 − (9/2 − 9) = 0 − (−9/2) = 9/2. Total = 9/2 + 9/2 = 9 square units.

3. Find the area bounded by the curve y = sin x between x = 0 and x = 2π.

SOLUTION Between 0 and π, sin x ≥ 0; between π and 2π, sin x ≤ 0. So take the absolute value over the second part and add the magnitudes. Area = ∫0π sin x dx + |∫π2π sin x dx|. ∫0π sin x dx = [−cos x]0π = (−cosπ) − (−cos 0) = 1 + 1 = 2. ∫π2π sin x dx = [−cos x]π2π = (−cos 2π) − (−cosπ) = −1 − 1 = −2; its magnitude is 2. Total area = 2 + 2 = 4 square units.

Choose the correct answer in the following Exercises from 4 to 5.

4. Area bounded by the curve y = x3, the x-axis and the ordinates x = −2 and x = 1 is (A) −9   (B) −15/4   (C) 15/4   (D) 17/4

SOLUTION The curve y = x3 is below the x-axis on [−2, 0] and above it on [0, 1]; area must add the magnitudes of the two parts. ∫−20 x3 dx = [ x4/4 ]−20 = 0 − 16/4 = −4; magnitude = 4. ∫01 x3 dx = [ x4/4 ]01 = 1/4. Total area = 4 + 1/4 = 16/4 + 1/4 = 17/4. ∴ the correct option is (D) 17/4.

5. The area bounded by the curve y = x |x|, x-axis and the ordinates x = −1 and x = 1 is given by (A) 0   (B) 1/3   (C) 2/3   (D) 4/3 [Hint: y = x2 if x > 0 and y = −x2 if x < 0].

SOLUTION Using the hint: on [−1, 0], y = −x2 (below the axis); on [0, 1], y = x2 (above the axis). Add the magnitudes. ∫−10 (−x2) dx = −[ x3/3 ]−10 = −(0 − (−1/3)) = −1/3; magnitude = 1/3. ∫01 x2 dx = [ x3/3 ]01 = 1/3. Total area = 1/3 + 1/3 = 2/3. ∴ the correct option is (C) 2/3.

Common Mistakes to Avoid

Watch out for these

  • Forgetting to take the absolute value when the curve dips below the x-axis — a negative integral is not a negative area.
  • Not splitting the integral where the curve crosses the axis (e.g. sin x at x = π, or x3 at x = 0); a single integral can cancel out and give the wrong answer.
  • For boundaries on the y-axis (like Exercise 8.1 Q4), integrating with respect to x instead of using horizontal strips x dy.
  • Misreading the semi-axes of an ellipse — a2 and b2 sit under x2 and y2; take square roots before using A = πab.
  • Slips in the standard integral ∫√(a2−x2) dx; the sin−1 term carries the coefficient a2/2, not a/2.
  • Not identifying the vertex of a modulus graph (e.g. |x + 3| turns at x = −3) before splitting the limits.

Practice MCQs & Assertion–Reason

1. The area enclosed by the circle x2 + y2 = 9 is:

(a) 3π    (b) 6π    (c) 9π    (d) 18π

2. The area of the ellipse x2/25 + y2/16 = 1 is:

(a) 9π    (b) 20π    (c) 40π    (d) 400π

3. The area under y = x2 between x = 0 and x = 3, above the x-axis, is:

(a) 3    (b) 6    (c) 9    (d) 27

4. The area bounded by y = cos x, the x-axis between x = 0 and x = π is:

(a) 1    (b) 2    (c) 0    (d) 4

5. To find the area bounded by y2 = 4x, the y-axis and y = 3, the correct integral is:

(a) ∫03 (y2/4) dy    (b) ∫03 4x dx    (c) ∫03 2√x dx    (d) ∫03 (4/y2) dy

6. The area bounded by y = x3, the x-axis, x = −2 and x = 1 is:

(a) −15/4    (b) 15/4    (c) 17/4    (d) −9

7. The standard integral ∫ √(a2 − x2) dx equals:

(a) (x/2)√(a2−x2) + (a2/2)sin−1(x/a) + C    (b) (a/2)sin−1(x/a) + C    (c) x√(a2−x2) + C    (d) sin−1(x/a) + C

8. The area bounded by y = x|x|, the x-axis, x = −1 and x = 1 is:

(a) 0    (b) 1/3    (c) 2/3    (d) 4/3

9. The area in the first quadrant bounded by x2 + y2 = 4 and the lines x = 0, x = 2 is:

(a) π/4    (b) π/2    (c) π    (d) 2π

10. When computing the area of a region symmetric about both axes (e.g. a circle or ellipse), a useful shortcut is to find:

(a) twice the upper-half area    (b) 4 × the first-quadrant area    (c) the integral over one period    (d) half the first-quadrant area

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(a), 6-(c), 7-(a), 8-(c), 9-(c), 10-(b).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: The area of the ellipse x2/16 + y2/9 = 1 is 12π square units.

Reason: The area of an ellipse with semi-axes a and b is πab.

A-R 2. Assertion: The area bounded by y = sin x between x = 0 and x = 2π is 4 square units.

Reason: ∫02π sin x dx = 0.

A-R 3. Assertion: The area bounded by y = x3, the x-axis, x = −2 and x = 1 is 17/4 square units.

Reason: When a curve lies below the x-axis, the area is the absolute value of the integral over that part.

A-R 4. Assertion: The area bounded by y2 = 4x, the y-axis and y = 3 is found by integrating with respect to y.

Reason: When the bounding lines are horizontal (or along the y-axis), horizontal strips of area x dy are convenient.

A-R 5. Assertion: ∫−60 |x + 3| dx = 9.

Reason: |x + 3| is always non-negative, so the integral equals the area between the graph and the x-axis.

Answer key: 1-(A), 2-(B), 3-(A), 4-(A), 5-(A).

Quick Revision Summary

  • Area between y = f(x), the x-axis and x = a, x = b is ∫ab y dx; between a curve, the y-axis and y = c, y = d it is ∫cd x dy.
  • For a curve below the axis, area = |integral|; if the curve crosses the axis, split and add magnitudes.
  • Use the standard integral ∫√(a2−x2) dx = (x/2)√(a2−x2) + (a2/2)sin−1(x/a) for circles and ellipses.
  • Area of circle x2 + y2 = a2 is πa2; area of ellipse x2/a2 + y2/b2 = 1 is πab.
  • Exploit symmetry: for a figure symmetric about both axes, total area = 4 × first-quadrant area.

How to score full marks in this chapter

Always draw a rough sketch first — it shows where the curve is below the axis and where it crosses, so you know when to split the integral and take absolute values. Decide between vertical strips (y dx) and horizontal strips (x dy) from the bounding lines, and use symmetry to shorten the work for circles and ellipses. Quote the standard integral carefully and evaluate the limits one term at a time to avoid sign slips.

Frequently Asked Questions

What does Class 12 Maths Chapter 8 Application of Integrals cover?

Chapter 8 uses the definite integral to find the area enclosed by a curve — the area under simple curves, the area bounded by a circle and an ellipse, and the area between a curve, the axes and given ordinates — including curves that lie below the axis.

How many exercises are there in Class 12 Maths Chapter 8?

There is one main exercise, Exercise 8.1 (with two area questions and two multiple-choice questions), followed by the Miscellaneous Exercise on Chapter 8. Every question of both is solved step by step on this page.

How do I find the area when a curve goes below the x-axis?

Split the integral wherever the curve crosses the x-axis, evaluate each piece, and add the absolute values. A part below the axis gives a negative integral, so its magnitude is taken as the area.

Are these Class 12 Maths Chapter 8 solutions free?

Yes. All solutions are free and follow the official NCERT Mathematics textbook for the 2026–27 session, with every answer verified against the book’s answer key.

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