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Class 12 · Tamil Nadu State Board

Samacheer Kalvi 12th Maths Chapter 9 Exercise Solutions

Tamil Nadu Class 12 Maths Chapter 9 Applications of Integration exercise solutions with area under curves, area between curves, limits, and solved setup patterns.

Updated 25 May 202616 min read

How to Use This Page

This page covers all exercise patterns for Chapter 9 — Applications of Integration. Work through each section in order: start with the Key Concepts recap, attempt the MCQs and short answers on your own, then check against the solutions provided. For detail answers, write the full derivation before reading the model answer. Use the Common Mistakes and Exam Checklist sections as a final review before your exam.

Key Concepts (Quick Recap)

TermDefinition
Definite integral as areaThe area under y = f(x) from x = a to x = b equals the definite integral of f(x) from a to b
Area under a curveRegion bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b
Area between two curvesIntegral of (upper curve minus lower curve) between the intersection points
Upper and lower curvesThe curve with larger y-value is the upper curve; the one with smaller y-value is the lower curve
Vertical strip (dx)Used when integrating with respect to x; strip runs from lower curve to upper curve
Horizontal strip (dy)Used when integrating with respect to y; strip runs from left curve to right curve
Volume of solid of revolutionVolume generated by rotating a curve about the x-axis or y-axis
Disc methodV = pi * integral of [f(x)]² dx for rotation about the x-axis
Bernoulli's formulaA reduction formula for integrating x^n * e^x and similar products by repeated integration by parts
Gamma functionGamma(n) = integral from 0 to infinity of x^(n-1) e^(-x) dx; Gamma(n) = (n-1)! for positive integers
Beta functionB(m, n) = integral from 0 to 1 of x^(m-1) (1-x)^(n-1) dx; related to Gamma by B(m,n) = Gamma(m)Gamma(n)/Gamma(m+n)
Points of intersectionSolve the two curve equations simultaneously to find limits of integration
Symmetry in areaIf a region is symmetric about the y-axis, compute area for one half and double it

Chapter 9 — Exercise Solutions Guide

In area problems, the setup is more important than the integration itself. Always draw a rough graph.

Formula Recap

Area typeFormula
Area under y = f(x)integral from a to b f(x) dx
Area between curvesintegral from a to b [upper curve - lower curve] dx
Area with x as function of yintegral from c to d [right curve - left curve] dy

Standard Solution Pattern

1. Identify the curve or curves. 2. Find points of intersection if needed. 3. Decide whether to integrate with respect to x or y. 4. Write limits correctly. 5. Integrate and simplify.

I. Choose the Best Answer (MCQs)

1. The area under the curve y = x² from x = 0 to x = 2 is: (a) 4/3 (b) 8/3 (c) 2 (d) 16/3
Answer(b) 8/3 (Explanation: integral from 0 to 2 of x² dx = [x³/3] from 0 to 2 = 8/3 square units.)
2. The area enclosed between y = x and y = x² is: (a) 1/2 (b) 1/3 (c) 1/6 (d) 1/4
Answer(c) 1/6 (Explanation: The curves intersect at x = 0 and x = 1. Area = integral from 0 to 1 of (x - x²) dx = 1/2 - 1/3 = 1/6.)
3. The area under y = sin x from x = 0 to x = pi is: (a) 0 (b) 1 (c) 2 (d) pi
Answer(c) 2 (Explanation: integral from 0 to pi of sin x dx = [-cos x] from 0 to pi = -cos(pi) - (-cos 0) = 1 + 1 = 2.)
4. When finding the area between two curves, we always subtract: (a) Lower curve from upper curve (b) Upper curve from lower curve (c) x-values from y-values (d) Left curve from right curve
Answer(a) Lower curve from upper curve (Explanation: Area = integral of (upper curve - lower curve) dx so the result is always positive.)
5. The volume of the solid generated by revolving y = x² about the x-axis from x = 0 to x = 1 is: (a) pi/5 (b) pi/3 (c) 2pi/5 (d) pi/2
Answer(a) pi/5 (Explanation: V = pi integral from 0 to 1 of x⁴ dx = pi [x⁵/5] from 0 to 1 = pi/5.)
6. The value of Gamma(4) is: (a) 3 (b) 6 (c) 24 (d) 4
Answer(b) 6 (Explanation: Gamma(n) = (n-1)! so Gamma(4) = 3! = 6.)
7. Bernoulli's formula is used to evaluate: (a) Area between curves (b) Integrals of the form integral x^n e^x dx (c) Definite integrals with trigonometric functions (d) Partial fractions
Answer(b) Integrals of the form integral x^n e^x dx (Explanation: Bernoulli's formula provides a reduction formula for repeatedly applying integration by parts to x^n e^x.)
8. The area bounded by y² = 4x and x = 1 is: (a) 4/3 (b) 8/3 (c) 2 (d) 1
Answer(b) 8/3 (Explanation: y² = 4x gives y = ±2sqrt(x). Area = 2 integral from 0 to 1 of 2sqrt(x) dx = 4 [2x^(3/2)/3] from 0 to 1 = 8/3.)
9. If the area under y = f(x) from a to b is negative, then: (a) The area is negative (b) Take absolute value; area is always positive (c) The limits are wrong (d) The function is constant
Answer(b) Take absolute value; area is always positive (Explanation: When the curve lies below the x-axis, the definite integral is negative, but geometric area is taken as positive.)
10. The beta function B(1, 1) equals: (a) 0 (b) 1 (c) 2 (d) 1/2
Answer(b) 1 (Explanation: B(1,1) = integral from 0 to 1 of x⁰(1-x)⁰ dx = integral from 0 to 1 of 1 dx = 1.)

II. Fill in the Blanks

1. The area under y = f(x) from x = a to x = b is given by integral from a to b of f(x) dx. 2. To find the area between two curves, we integrate (upper curve minus lower curve) with respect to x. 3. The volume of the solid obtained by revolving y = f(x) about the x-axis is pi times integral of [f(x)]² dx. 4. The Gamma function Gamma(n) equals (n-1)! when n is a positive integer. 5. The curves y = x² and y = x intersect at x = 0 and x = 1. 6. If f(x) is below the x-axis on [a, b], the area equals the absolute value of the definite integral. 7. Bernoulli's formula reduces the power of x by 1 at each step of integration by parts. 8. B(m, n) = Gamma(m) Gamma(n) / Gamma(m+n).

10. When rotating about the y-axis, the volume formula uses x² inside the integral with dy.

9. The area bounded by y =xfrom x = -2 to x = 2 is 4 square units.

III. True or False

1. The definite integral always gives the geometric area. False. The definite integral can be negative when the curve lies below the x-axis. Geometric area is always taken as positive (absolute value).

2. The area between two curves is always computed as upper minus lower. True. We always subtract the lower curve from the upper curve to get a positive area value.

3. Gamma(1) = 1. True. Gamma(1) = integral from 0 to infinity of e^(-x) dx = 1.

4. The volume of revolution about the x-axis requires integration with respect to y. False. When rotating about the x-axis, we integrate with respect to x using V = pi * integral of [f(x)]² dx.

5. Points of intersection are needed only when the limits are not given. False. Even when limits seem obvious, you must verify intersection points to set correct boundaries.

6. Bernoulli's formula can be applied to any type of integral. False. Bernoulli's formula specifically applies to integrals of the form integral of x^n e^x dx using integration by parts repeatedly.

7. If y = f(x) is above the x-axis on [a, b], then the area equals the definite integral. True. When f(x) is non-negative, the area and the definite integral are the same.

8. The beta function B(m, n) is symmetric: B(m, n) = B(n, m). True. This follows from the substitution u = 1 - x in the definition.

IV. Match the Following

Column AColumn B
Area under y = f(x) from a to bintegral from a to b f(x) dx
Area between curvesintegral of (upper - lower) dx
Volume about x-axispi * integral [f(x)]² dx
Gamma(n)(n-1)! for positive integers
Beta(m, n)integral from 0 to 1 x^(m-1)(1-x)^(n-1) dx
Bernoulli's formulaReduction formula for integral x^n e^x dx

Worked Pattern

Area under y = x² from 0 to 2:

Area = integral 0 to 2 x² dx = [x³/3] 0 to 2 = 8/3 square units

Area Between Curves Pattern

If upper curve is y1 and lower curve is y2:

Area = integral from a to b (y1 - y2) dx

Never write lower minus upper; area must be positive.

Worked Examples

Example 1 — Area under a curve

Find the area under y = x² from x = 0 to x = 3.

Area = ∫₀³ x² dx = [x³/3]₀³ = 27/3 − 0 = 9 square units

Example 2 — Area under a trig curve

Find the area under y = sin x from x = 0 to x = π.

Area = ∫₀^π sin x dx = [−cos x]₀^π = (−cos π) − (−cos 0) = (1) − (−1) = 2 square units

Example 3 — Area between two curves

Find the area between y = x and y = x² from x = 0 to x = 1.

Here x ≥ x² on [0, 1], so upper = x, lower = x²: Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = (1/2 − 1/3) = 1/6 square units

V. Answer Briefly (2-Mark Questions)

1. State the formula for the area under a curve y = f(x) between x = a and x = b.
Answer

The area under y = f(x) from x = a to x = b is given by the definite integral of f(x) from a to b. If f(x) is non-negative on [a, b], the area equals the value of the integral. If f(x) dips below the x-axis, we take the absolute value of each part.

2. How do you decide whether to integrate with respect to x or y?
Answer

If the curves are easier to express as y = f(x) and the boundaries are vertical lines, integrate with respect to x. If the curves are easier to express as x = g(y) and the boundaries are horizontal lines, integrate with respect to y. Always choose the variable that avoids splitting the region into multiple parts.

3. What is the Gamma function? Give its integral definition.
Answer

The Gamma function is defined as Gamma(n) = integral from 0 to infinity of x^(n-1) e^(-x) dx. For positive integers, Gamma(n) = (n-1)! so Gamma(1) = 1, Gamma(2) = 1, Gamma(3) = 2, and Gamma(4) = 6. It extends the factorial function to non-integer values.

4. Write the formula for the volume of a solid of revolution about the x-axis.
Answer

The volume of the solid generated by revolving y = f(x) about the x-axis from x = a to x = b is V = pi * integral from a to b of [f(x)]² dx. This is called the disc method because each thin slice forms a disc of radius f(x) and thickness dx.

5. State the relationship between the Beta and Gamma functions.
Answer

The Beta and Gamma functions are related by B(m, n) = Gamma(m) Gamma(n) / Gamma(m + n). This allows us to compute Beta function values using known Gamma values. For example, B(1, 1) = Gamma(1) Gamma(1) / Gamma(2) = 1 * 1 / 1 = 1.

VI. Answer in Detail (5-Mark Questions)

1. Find the area bounded by the curve y = x² and the line y = 4. Show all steps clearly.
Answer

First, find the points of intersection by solving x² = 4, which gives x = -2 and x = 2. The line y = 4 is the upper curve and the parabola y = x² is the lower curve on this interval.

Area = ∫₋₂⁴ (4 − x²) dx Wait — the limits should be from x = −2 to x = 2: Area = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = (16/3) − (−16/3) = 32/3 square units

Since the region is symmetric about the y-axis, we can also compute area for x = 0 to x = 2 and double: 2 [4x − x³/3] from 0 to 2 = 2 (8/3) = 16/3... Wait, let me redo: 2 (8 − 8/3) = 2 16/3 = 32/3. Both methods agree. The final answer is 32/3 square units.

2. Derive the volume of a sphere of radius r using the method of solids of revolution.
Answer

A sphere of radius r is obtained by revolving the semicircle y = sqrt(r² − x²) about the x-axis from x = −r to x = r. Using the disc method:

V = π ∫₋ᵣʳ [√(r² − x²)]² dx = π ∫₋ᵣʳ (r² − x²) dx = π [r²x − x³/3]₋ᵣʳ = π [(r³ − r³/3) − (−r³ + r³/3)] = π [2r³/3 − (−2r³/3)] = π · 4r³/3 = 4πr³/3

This confirms the well-known formula V = 4/3 pi r³. The key steps are: express the semicircle as an explicit function, square it for the disc method, and evaluate the resulting polynomial integral.

3. State and prove Bernoulli's formula for integral of x^n e^x dx.
Answer

Bernoulli's formula states:

∫ xⁿ eˣ dx = eˣ [xⁿ − n·xⁿ⁻¹ + n(n−1)xⁿ⁻² − ... + (−1)ⁿ n!] + C

Proof using integration by parts: Let u = xⁿ, dv = eˣ dx. Then du = n·xⁿ⁻¹ dx and v = eˣ. Applying integration by parts:

∫ xⁿ eˣ dx = xⁿ eˣ − ∫ n·xⁿ⁻¹ eˣ dx

This is a reduction formula. Repeating the process n times, each step reduces the power of x by one. After n applications, we reach integral of eˣ dx = eˣ. Collecting terms with alternating signs gives Bernoulli's formula. This formula is particularly useful in evaluating definite integrals of the form integral from 0 to 1 of x^n e^x dx.

4. Find the area enclosed between the parabola y² = 4x and the line y = 2x − 4.
Answer

Step 1: Find points of intersection. From y² = 4x and y = 2x − 4, substitute: (2x − 4)² = 4x, so 4x² − 16x + 16 = 4x, giving 4x² − 20x + 16 = 0, or x² − 5x + 4 = 0. This factors as (x − 1)(x − 4) = 0, so x = 1 and x = 4. The corresponding y-values are y = −2 and y = 4.

Step 2: Since the parabola opens to the right, integrate with respect to y. Rewrite: x = y²/4 (parabola) and x = (y + 4)/2 (line).

Area = ∫₋₂⁴ [(y+4)/2 − y²/4] dy = [(y+4)²/4 − y³/12]₋₂⁴

Computing at y = 4: (8²/4) − (64/12) = 16 − 16/3 = 32/3. Computing at y = −2: (2²/4) − (−8/12) = 1 + 2/3 = 5/3. Area = 32/3 − 5/3 = 27/3 = 9 square units.

5. Evaluate the Beta function B(2, 3) using both the integral definition and the Gamma relation.
Answer

Method 1 — Using the integral definition:

B(2, 3) = ∫₀¹ x¹(1−x)² dx = ∫₀¹ (x − 2x² + x³) dx = [x²/2 − 2x³/3 + x⁴/4]₀¹ = 1/2 − 2/3 + 1/4 = 6/12 − 8/12 + 3/12 = 1/12

Method 2 — Using the Gamma relation:

B(2, 3) = Γ(2)·Γ(3) / Γ(5) = 1! · 2! / 4! = 1 · 2 / 24 = 2/24 = 1/12

Both methods give the same answer B(2, 3) = 1/12, which confirms the consistency of the Beta-Gamma relationship.

Common Mistakes to Avoid

1. ❌ Using lower curve minus upper curve in the area formula ✅ Always subtract lower curve from upper curve: Area = integral (upper - lower) dx

2. ❌ Forgetting to find intersection points before setting limits ✅ Solve the two equations simultaneously to find where curves intersect; these give your limits

3. ❌ Writing area as a negative number when the curve is below the x-axis ✅ Geometric area is always positive; take the absolute value of each part of the integral

4. ❌ Using dx when the region is better described with horizontal strips ✅ Choose dx or dy based on which orientation avoids splitting the region into multiple integrals

5. ❌ Omitting square units in the final answer for area problems ✅ Always state the unit: "square units" for area, "cubic units" for volume

6. ❌ Forgetting to multiply by pi in the volume of revolution formula ✅ The disc method formula is V = pi * integral [f(x)]² dx; pi is essential

7. ❌ Confusing Gamma(n) with n! directly ✅ Remember Gamma(n) = (n - 1)!, not n!; so Gamma(4) = 3! = 6

8. ❌ Dropping the constant of integration in indefinite integrals within reduction formulas ✅ Even in reduction formulas, track the constant carefully at each step

Exam Checklist

  • Rough diagram is drawn showing the region and the curves
  • Points of intersection are found algebraically and marked on the diagram
  • Upper and lower curves are correctly identified
  • Limits of integration match the intersection points or given boundaries
  • Upper minus lower is used (area is always positive)
  • The correct variable (dx or dy) is chosen based on the strip orientation
  • All integration steps are shown with proper notation
  • Definite integral limits are substituted correctly (upper limit minus lower limit)
  • Final answer has proper units (square units for area, cubic units for volume)
  • Pi is included in volume of revolution answers
  • Gamma and Beta function formulas are stated before substitution
  • Bernoulli's formula reduction steps are written out if used

Frequently Asked Questions

Setting correct limits and identifying the upper and lower curve are the most important steps.
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