Class 12 Mathematics Notes Chapter 2 (Application of Integrals) – Mathematics Part-II Book

Alright class, let's begin our focused session on Chapter 2: Application of Integrals from your NCERT Part II book. This chapter is conceptually straightforward but requires careful application of definite integration techniques, especially for competitive government exams where accuracy and speed are paramount. The core idea is to use the definite integral as a tool for calculating the area of regions bounded by curves.

Chapter 2: Application of Integrals - Detailed Notes

1. Introduction: Area as a Definite Integral

• Recall that the definite integral ∫_a^b f(x) dx represents the algebraic sum of areas of the regions bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.
• "Algebraic sum" means areas above the x-axis are positive, and areas below are negative.
• For calculating the geometric area, we need to ensure we consider the magnitude.

2. Area Under Simple Curves

• Case 1: Area bounded by the curve y = f(x), the x-axis, and lines x = a, x = b.

  • If f(x) ≥ 0 for all x in [a, b], then the area A is given by:
    A = ∫_a^b y dx = ∫_a^b f(x) dx
  • If f(x) ≤ 0 for all x in [a, b], the integral will be negative. Since area must be positive, the area A is:
    A = |∫_a^b f(x) dx| = - ∫_a^b f(x) dx
  • If f(x) changes sign in [a, b], say at x = c, then we split the integral:
    A = |∫_a^c f(x) dx| + |∫_c^b f(x) dx|
  • Visualisation: Imagine dividing the area into infinitesimally thin vertical strips of height 'y' and width 'dx'. The integral sums the areas (y * dx) of these strips from x = a to x = b.

• Case 2: Area bounded by the curve x = g(y), the y-axis, and lines y = c, y = d.

  • If g(y) ≥ 0 for all y in [c, d] (curve is to the right of y-axis), then the area A is:
    A = ∫_c^d x dy = ∫_c^d g(y) dy
  • If g(y) ≤ 0 for all y in [c, d] (curve is to the left of y-axis), the area A is:
    A = |∫_c^d g(y) dy| = - ∫_c^d g(y) dy
  • Visualisation: Imagine dividing the area into infinitesimally thin horizontal strips of width 'x' and height 'dy'. The integral sums the areas (x * dy) of these strips from y = c to y = d.

3. Area Between Two Curves

• Case 1: Integration with respect to x.

  • Let two curves be y = f(x) and y = g(x) 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 = ∫_a^b (Upper Curve - Lower Curve) dx
  • Finding Limits (a, b): Often, 'a' and 'b' are the x-coordinates of the points of intersection of the two curves. Solve f(x) = g(x) to find these points.

• Case 2: Integration with respect to y.

  • Let two curves be x = f(y) and x = g(y) such that f(y) ≥ g(y) in the interval [c, d]. (f(y) is the curve to the right).
  • The area A bounded between these curves from y = c to y = d is:
    A = ∫_c^d [f(y) - g(y)] dy = ∫_c^d (Right Curve - Left Curve) dy
  • Finding Limits (c, d): Often, 'c' and 'd' are the y-coordinates of the points of intersection. Solve f(y) = g(y) (or solve the original equations simultaneously) to find these points.

4. Key Steps for Solving Area Problems

1. Sketch the Curves: This is the MOST crucial step. Draw a rough but representative graph of the given curves and lines. Identify the required bounded region.
2. Find Points of Intersection: Determine the points where the bounding curves intersect. These points often define the limits of integration. Solve the equations simultaneously.
3. Choose the Integration Variable (dx or dy):
   • Use vertical strips (integrate w.r.t. x) if the region is easily defined by functions of the form y = f(x) and bounded by vertical lines or intersection points with clear x-coordinates. Use (Upper Curve - Lower Curve).
   • Use horizontal strips (integrate w.r.t. y) if the region is easily defined by functions of the form x = g(y) and bounded by horizontal lines or intersection points with clear y-coordinates. Use (Right Curve - Left Curve). Sometimes, one choice makes the integration much simpler than the other.
4. Set up the Definite Integral(s): Write the correct definite integral(s) based on the chosen variable, limits, and the functions defining the boundaries. Remember to subtract the lower curve from the upper curve (for dx) or the left curve from the right curve (for dy). You might need to split the area into multiple integrals if the upper/lower or right/left curve changes within the region.
5. Evaluate the Integral: Use the fundamental theorem of calculus and standard integration formulas. Be careful with calculations and signs.

5. Common Curves Encountered

• Lines: y = mx + c, x = k, y = k
• Parabolas:
  • y² = 4ax (opens right, vertex (0,0))
  • y² = -4ax (opens left, vertex (0,0))
  • x² = 4by (opens up, vertex (0,0))
  • x² = -4by (opens down, vertex (0,0))
  • Shifted parabolas: (y-k)² = 4a(x-h) etc.
• Circles: x² + y² = r² (centre (0,0), radius r), (x-h)² + (y-k)² = r² (centre (h,k), radius r)
• Ellipses: x²/a² + y²/b² = 1 (centre (0,0), semi-axes a, b)
• Modulus Function: y = |x| (V-shape at origin)

6. Important Standard Results (Useful for verification/MCQs)

• Area of circle x² + y² = r² is πr².
• Area of ellipse x²/a² + y²/b² = 1 is πab.
• Area bounded by parabola y² = 4ax and its latus rectum (x = a) is (8/3)a².
• Area bounded by parabola y² = 4ax and the line y = mx is (8a²)/(3m³).
• Area bounded by parabolas y² = 4ax and x² = 4by is (16ab)/3.

7. Tips for Government Exams

• Sketching is Key: Don't skip sketching; it prevents errors in identifying the region and limits.
• Symmetry: Look for symmetry. If the region is symmetric about the x-axis, y-axis, or origin, calculate the area of one part and multiply accordingly. This saves time.
• Standard Formulas: Memorize standard integral formulas, especially ∫√(a²-x²) dx, ∫√(a²+x²) dx, ∫√(x²-a²) dx.
• Practice: Practice problems involving intersections of lines, parabolas, circles, and ellipses. These are common question types.
• Check Units: Area is always positive and usually expressed in "square units".

Multiple Choice Questions (MCQs)

1. The area of the region bounded by the curve y² = 4x, the y-axis, and the lines y = 1 and y = 3 is:
   (a) 9 sq. units
   (b) 28/3 sq. units
   (c) 26/3 sq. units
   (d) 7 sq. units

2. The area bounded by the curve y = x³, the x-axis, and the ordinates x = -1 and x = 2 is:
   (a) 15/4 sq. units
   (b) 17/4 sq. units
   (c) 19/4 sq. units
   (d) 4 sq. units

3. The area of the region bounded by the ellipse x²/16 + y²/9 = 1 is:
   (a) 16π sq. units
   (b) 9π sq. units
   (c) 144π sq. units
   (d) 12π sq. units

4. The area bounded by the parabola y = x² and the line y = 4 is:
   (a) 16/3 sq. units
   (b) 32/3 sq. units
   (c) 64/3 sq. units
   (d) 8 sq. units

5. 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 sq. units
   (b) π/2 sq. units
   (c) π/3 sq. units
   (d) π/4 sq. units

6. The area of the region bounded by the curve y = |x - 1|, the x-axis, and the lines x = 0 and x = 2 is:
   (a) 1 sq. unit
   (b) 2 sq. units
   (c) 1/2 sq. unit
   (d) 3/2 sq. units

7. The area bounded by the parabola x² = 4y and its latus rectum (y = 1) is:
   (a) 4/3 sq. units
   (b) 8/3 sq. units
   (c) 16/3 sq. units
   (d) 2/3 sq. units

8. The area enclosed between the curves y = x² and y = x is:
   (a) 1/6 sq. unit
   (b) 1/3 sq. unit
   (c) 1/2 sq. unit
   (d) 5/6 sq. unit

9. Using integration, the area of the triangle whose vertices are (1, 0), (2, 2), and (3, 1) is:
   (a) 3/2 sq. units
   (b) 5/2 sq. units
   (c) 2 sq. units
   (d) 1 sq. unit

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

Answer Key for MCQs:

1. (c) 26/3
2. (b) 17/4
3. (d) 12π
4. (b) 32/3
5. (a) π
6. (a) 1
7. (b) 8/3
8. (a) 1/6
9. (a) 3/2
10. (c) 4

Remember to practice sketching these regions and setting up the integrals yourself. Good luck with your preparation!