Class 12 Mathematics Notes Chapter 8 (Application of Integrals) – Examplar Problems (English) Book

Detailed Notes with MCQs of Chapter 8, 'Application of Integrals'. This is a crucial chapter, not just for your board exams but also frequently tested in various government entrance exams. The core idea is simple: using the power of definite integration, which we learned previously, to calculate the area of regions bounded by curves. While the concept is straightforward, accuracy in sketching curves and setting up the correct integral is key, especially for the kind of problems you might see from the Exemplar book or in competitive exams.

Chapter 8: Application of Integrals - Detailed Notes

1. Fundamental Concept:

• Definite integration ∫[a, b] f(x) dx represents the algebraic sum of areas bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.
• 'Algebraic sum' means area above the x-axis is positive, and area below the x-axis is negative.
• For calculating the geometric area, we need to consider the absolute value.

2. Area Bounded by a Curve and the x-axis:

• Formula: The area A bounded by the curve y = f(x), the x-axis, and the lines x = a and x = b is given by:
  A = ∫[a, b] |f(x)| dx
• Procedure:
  • Sketch the curve y = f(x).
  • Identify the interval [a, b] on the x-axis.
  • Determine if f(x) is positive or negative in the interval [a, b].
  • If f(x) ≥ 0 for all x in [a, b], then A = ∫[a, b] f(x) dx.
  • If f(x) ≤ 0 for all x in [a, b], then A = |∫[a, b] f(x) dx| = -∫[a, b] f(x) dx.
  • If f(x) changes sign within [a, b] (say at x = c), split the integral:
    A = ∫[a, c] |f(x)| dx + ∫[c, b] |f(x)| dx. Evaluate each part based on the sign of f(x) in the respective sub-interval.
• Elemental Area: Think of the area as being composed of infinitely many thin vertical strips of width dx and height y (or |y|). The area of one such strip is dA = y dx (or |y| dx). Integrating dA from a to b gives the total area.

3. Area Bounded by a Curve and the y-axis:

• Formula: The area A bounded by the curve x = g(y), the y-axis, and the lines y = c and y = d is given by:
  A = ∫[c, d] |g(y)| dy
• Procedure: Similar to the x-axis case, but now we consider horizontal strips.
  • Sketch the curve x = g(y).
  • Identify the interval [c, d] on the y-axis.
  • Determine the sign of g(y) in [c, d].
  • If g(y) ≥ 0 (curve to the right of y-axis), A = ∫[c, d] g(y) dy.
  • If g(y) ≤ 0 (curve to the left of y-axis), A = |∫[c, d] g(y) dy| = -∫[c, d] g(y) dy.
  • If g(y) changes sign, split the integral at the points where g(y) = 0.
• Elemental Area: Think of horizontal strips of height dy and width x (or |x|). Area of one strip dA = x dy (or |x| dy). Integrate dA from c to d.

4. Area Between Two Curves:

• Case 1: Integrating with respect to x
  • Let y = f(x) and y = g(x) be two curves such that f(x) ≥ g(x) in the interval [a, b].
  • The area A bounded between these curves from x = a to x = b is:
    A = ∫[a, b] [f(x) - g(x)] dx (Area under Upper Curve - Area under Lower Curve)
  • Procedure:
    1. Sketch both curves.
    2. Find the points of intersection by solving f(x) = g(x). These often give you the limits a and b. If limits are given explicitly, use those.
    3. Identify which curve is the 'upper' curve (f(x)) and which is the 'lower' curve (g(x)) in the required interval.
    4. Set up and evaluate the integral.
• Case 2: Integrating with respect to y
  • Let x = f(y) and x = g(y) be two curves such that f(y) ≥ g(y) in the interval [c, d]. (f(y) is the curve to the right, g(y) is the curve to the left).
  • The area A bounded between these curves from y = c to y = d is:
    A = ∫[c, d] [f(y) - g(y)] dy (Area w.r.t Right Curve - Area w.r.t Left Curve)
  • Procedure: Similar to Case 1, but oriented horizontally. Find intersections, identify right/left curves, and integrate w.r.t y.

5. Essential Steps for Solving Problems:

1. Sketch: Draw a rough but clear sketch of the curves involved. This is the most critical step. Identify the region whose area is required.
2. Intersections: Find the points of intersection of the curves by solving their equations simultaneously. These points often determine the limits of integration.
3. Choose Variable: Decide whether to use vertical strips (dx) or horizontal strips (dy). Choose the one that makes the setup simpler (e.g., avoids splitting the integral if possible, or makes expressing functions easier).
4. Set up Integral:
   • If using dx: Identify upper curve y_U and lower curve y_L. The integral is ∫[x1, x2] (y_U - y_L) dx.
   • If using dy: Identify right curve x_R and left curve x_L. The integral is ∫[y1, y2] (x_R - x_L) dy.
   • If finding area with respect to an axis, use the formulas from sections 2 & 3, paying attention to signs/absolute values.
5. Evaluate: Calculate the definite integral using the fundamental theorem of calculus and standard integration formulas.
6. Check: The area must be positive. Does the value seem reasonable given your sketch?

6. Standard Curves (Know these shapes!):

• Lines: y = mx + c, x = k, y = k
• Parabolas: y² = 4ax, y² = -4ax, x² = 4ay, x² = -4ay. Also, shifted parabolas like (y-k)² = 4a(x-h).
• Circles: x² + y² = a² (centre (0,0), radius a), (x-h)² + (y-k)² = r² (centre (h,k), radius r).
• Ellipses: x²/a² + y²/b² = 1 (centre (0,0), intercepts (±a, 0), (0, ±b)).
• Modulus Function: y = |x| (V-shape at origin).
• Trigonometric Functions: y = sin x, y = cos x.

7. Important Tips for Exams:

• Symmetry: Use symmetry to simplify calculations. For example, the area of a circle x² + y² = a² is 4 * ∫[0, a] √(a² - x²) dx.
• Modulus: For regions involving |f(x)|, split the integral where f(x) changes sign.
• Multiple Curves: If a region is bounded by more than two curves, you might need to split the area into multiple parts and sum the corresponding integrals.
• Exemplar Focus: Expect problems involving intersections of parabolas, lines cutting circles/ellipses/parabolas, regions defined using inequalities, and possibly trigonometric or inverse trigonometric functions. Accuracy in finding intersection points and setting up the correct limits/integrand is vital.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts from Chapter 8, keeping the level relevant for government exams and touching upon Exemplar-type ideas:

1. The area of the region bounded by the curve y² = 8x and the line x = 2 is:
   (A) 16/3 sq. units
   (B) 32/3 sq. units
   (C) 64/3 sq. units
   (D) 8/3 sq. units

2. The area bounded by the curve y = x³, the x-axis, and the ordinates x = -1 and x = 2 is:
   (A) 17/4 sq. units
   (B) 15/4 sq. units
   (C) 0 sq. units
   (D) 16/4 sq. units

3. The area of the region bounded by the circle x² + y² = 16 is:
   (A) 4π sq. units
   (B) 8π sq. units
   (C) 16π sq. units
   (D) 32π sq. units

4. The area enclosed between the parabola y² = 4ax and the line y = mx is:
   (A) 8a² / (3m³)
   (B) 4a² / (3m³)
   (C) 8a³ / (3m²)
   (D) a² / (3m)

5. The area of the region bounded by the curve x = 2y - y² and the y-axis is:
   (A) 1/3 sq. units
   (B) 2/3 sq. units
   (C) 4/3 sq. units
   (D) 8/3 sq. units

6. The area bounded by y = |x - 1| and y = 1 is:
   (A) 1/2 sq. units
   (B) 1 sq. units
   (C) 2 sq. units
   (D) 4 sq. units

7. The area of the region bounded by the ellipse x²/9 + y²/4 = 1 is:
   (A) 6π sq. units
   (B) 9π sq. units
   (C) 4π sq. units
   (D) 36π sq. units

8. The area enclosed between the parabolas y² = x and x² = y is:
   (A) 1 sq. unit
   (B) 1/3 sq. units
   (C) 2/3 sq. units
   (D) 1/2 sq. units

9. The area of the region {(x, y) : x² ≤ y ≤ x} is:
   (A) 1/3 sq. units
   (B) 1/6 sq. units
   (C) 2/3 sq. units
   (D) 1/2 sq. units

10. The area bounded by the curve y = sin x between x = 0 and x = 2π is:
    (A) 0 sq. units
    (B) 2 sq. units
    (C) 4 sq. units
    (D) 2π sq. units

Answers to MCQs:

1. (C)
2. (A)
3. (C)
4. (A)
5. (C)
6. (B)
7. (A)
8. (B)
9. (B)
10. (C)

Remember to practice sketching these curves and setting up the integrals yourself. That's the best way to master this chapter. Good luck with your preparation!