Application Of Integrals Class 12 Maths: Chapter 8 Notes

Key Terminology

Area under a Curve

The area of the region bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b. It is calculated by the definite integral ∫[a to b] f(x) dx.

Elementary Area

An infinitesimally small area of a region, typically represented as a thin rectangle (strip). Its area is the product of its length and width (e.g., y dx or x dy).

Vertical Strip

An elementary rectangle with its height parallel to the y-axis and an infinitesimal width dx. Its area is y·dx. Used when integrating with respect to x.

Horizontal Strip

An elementary rectangle with its length parallel to the x-axis and an infinitesimal width dy. Its area is x·dy. Used when integrating with respect to y.

Points of Intersection

The points where two or more curves cross each other. Their coordinates are found by solving the equations of the curves simultaneously. These points often determine the limits of integration.

Bounded Region

A region in the Cartesian plane enclosed by one or more curves and/or lines, having a finite area.

Must-Remember Formulas & Concepts

• Area with Vertical Strips (dx): The area bounded by the curve y = f(x), the x-axis, and lines x = a and x = b is A = ∫[a to b] y dx = ∫[a to b] f(x) dx.
• Area with Horizontal Strips (dy): The area bounded by the curve x = g(y), the y-axis, and lines y = c and y = d is A = ∫[c to d] x dy = ∫[c to d] g(y) dy.
• Area Below x-axis: If f(x) ≤ 0 for x ∈ [a, b], the area is A = |∫[a to b] f(x) dx|. Geometrically, the integral will be negative, so we take the absolute value.
• Area Between Two Curves (dx): For y = f(x) and y = g(x) where f(x) ≥ g(x) from x = a to x = b, the area is A = ∫[a to b] [f(x) - g(x)] dx or ∫[a to b] (y_upper - y_lower) dx.
• Area Between Two Curves (dy): For x = f(y) and x = g(y) where f(y) ≥ g(y) from y = c to y = d, the area is A = ∫[c to d] [f(y) - g(y)] dy or ∫[c to d] (x_right - x_left) dy.
• Symmetry is Your Friend: If the area is symmetrical about the x-axis or y-axis, calculate the area of one part and multiply by the number of symmetrical parts (e.g., 2 or 4). This simplifies limits and calculations.
• Standard Curves: Be very familiar with sketching standard curves: Parabolas (y^2=4ax, x^2=4ay), Circles (x^2+y^2=a^2), and Ellipses (x^2/a^2 + y^2/b^2 = 1).
• Finding Limits: If limits are not given, they are the x or y coordinates of the points of intersection of the bounding curves.

Step-by-Step Method to Find Area of a Bounded Region

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Concept: Area Between Two Curves

Finding the area between two intersecting curves is a fundamental application of integration. Consider two curves, y = f(x) and y = g(x). The first and most critical step is to sketch the graphs to understand their relative positions. Let's assume f(x) is the upper curve and g(x) is the lower curve in the region of interest.

Next, you must find the points of intersection by setting f(x) = g(x) and solving for x. Let the solutions be x = a and x = b. These become your limits of integration.

The logic is simple: the area between the curves is the area under the upper curve minus the area under the lower curve. This is expressed as a single integral:
Area = ∫[a to b] f(x) dx - ∫[a to b] g(x) dx = ∫[a to b] [f(x) - g(x)] dx.

This formula, Area = ∫[a to b] (y_upper - y_lower) dx, is your primary tool. The elemental rectangle (vertical strip) has a height of (f(x) - g(x)) and a width of dx. We sum up these infinitesimally thin rectangles from a to b to get the total area. If using horizontal strips is easier (e.g., for parabolas like y^2=4x), the principle remains the same, but the integral becomes Area = ∫[c to d] (x_right - x_left) dy, where c and d are the y-coordinates of the intersection points.

Worked Examples

• Find the area of the region bounded by the curve y² = 9x and the line x = 4 in the first quadrant. 1. Curve: Parabola y² = 9x opening to the right. y = 3√x in the first quadrant. 2. Limits: Bounded by x=0 (y-axis) and x=4. 3. Setup: Using vertical strips (dx). Area = ∫[0 to 4] y dx. 4. Integration: Area = ∫[0 to 4] 3√x dx = 3 [x^(3/2) / (3/2)] from 0 to 4. 5. Calculation: Area = 3 (2/3) [4^(3/2) - 0] = 2 (2^2)^(3/2) = 2 2³ = 16 sq. units.
• Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1. 1. Curve: Ellipse with semi-major axis a=4 and semi-minor axis b=3. 2. Symmetry: The ellipse is symmetric about both axes. We can find the area in the first quadrant and multiply by 4. 3. Function: From the equation, y = (3/4)√(16 - x²). Limits for the first quadrant are x=0 to x=4. 4. Setup: Total Area = 4 ∫[0 to 4] y dx = 4 ∫[0 to 4] (3/4)√(16 - x²) dx. 5. Calculation: Area = 3 ∫[0 to 4] √(4² - x²) dx. Using the standard integral ∫√(a² - x²) dx, we get 3 [x/2 √(16-x²) + 16/2 sin⁻¹(x/4)] from 0 to 4. This evaluates to 3 [0 + 8 sin⁻¹(1)] = 3 8 (π/2) = 12π sq. units. (Matches formula πab = π4*3).

CBSE Exam Traps & Scoring Tips

1. Sketching is Mandatory: Even if not explicitly asked, always draw a rough sketch. It helps you identify the correct upper/lower curves and the limits of integration. Examiners often award marks for a correct sketch.

2. dx vs. dy Choice: Don't default to dx. For curves like y² = 4ax or x = f(y), integrating with respect to y (using horizontal strips) is often much easier. Look at the equations to decide.

3. The Modulus Trap: Area is always positive. If you calculate an area and get a negative answer, it means the region is below the x-axis. You should use the absolute value: Area = |∫ f(x) dx|. If a curve crosses the x-axis within the interval [a, b], say at c, you must split the integral: ∫[a to c] f(x) dx + |∫[c to b] f(x) dx|.

4. Symmetry Simplifies: For circles, ellipses, and even functions like y=cos(x) or y=x², use symmetry to your advantage. Calculating 2 * ∫[0 to a] is usually faster and less error-prone than ∫[-a to a].

Quick Revision Check

• What is the formula for the area bounded by the curve x = g(y), the y-axis, and the lines y = c and y = d? The area is given by A = ∫[c to d] x dy = ∫[c to d] g(y) dy.
• How do you find the limits of integration for the area enclosed between a parabola y² = 4x and a line y = x? By solving the two equations simultaneously. Substitute y=x into the parabola's equation: x² = 4x => x(x-4) = 0. So, the intersection points have x-coordinates x=0 and x=4, which are the limits.
• If a curve is symmetric about the y-axis, how can you simplify the area calculation from x = -a to x = a? You can calculate the area from x=0 to x=a and then multiply the result by 2. Area = 2 * ∫[0 to a] f(x) dx.
• What does the integral ∫[a to b] [f(x) - g(x)] dx represent, assuming f(x) ≥ g(x)? It represents the area of the region bounded between the upper curve y=f(x) and the lower curve y=g(x) from the vertical line x=a to x=b.

Frequently Asked Questions

Frequently Asked Questions