Application of Integrals Notes: Class 12 Mathematics Chapter 8
Class 12 Mathematics Chapter 8 Application of Integrals covers using definite integrals to find the area under a curve, the area enclosed between two curves, and the areas of standard geometric shapes such as circles and ellipses.
Overview of Topics & Subtopics (Click to Jump)
1. Area Under a Simple Curve
Short Note:
• The area enclosed between a curve y = f(x), the x-axis, and the ordinates x = a and x = b is given by the definite integral ∫[a to b] f(x)dx (taking the absolute value if the curve dips below the x-axis).
• When the required region lies along the y-axis, the area is instead computed by integrating x as a function of y between the corresponding limits.
2. Area Between Two Curves
Short Note:
• The area enclosed between two curves y = f(x) and y = g(x) over an interval is found by integrating the difference between the upper and lower functions, [f(x) - g(x)], over that interval.
• The points of intersection of the two curves, found by solving their equations simultaneously, determine the limits of integration for the enclosed region.
3. Areas of Standard Curves
Short Note:
• The area of a circle of radius r, given by x² + y² = r², can be derived using integration and matches the well-known geometric formula πr².
• Similarly, the area enclosed by an ellipse (x²/a² + y²/b² = 1) can be found using integration, giving the formula πab.
NCERT Textbook Exercise Questions
How is the area under a curve calculated using definite integration?
Answer: The area under a curve y = f(x), bounded by the x-axis and the vertical lines x = a and x = b, is calculated using the definite integral ∫[a to b] f(x)dx. If part of the curve lies below the x-axis, the corresponding integral value is negative, so its absolute value must be taken to represent the actual geometric area.
Find the area enclosed by the curve y = x² and the x-axis between x = 0 and x = 2.
Answer: The required area is given by the definite integral ∫[0 to 2] x² dx, which evaluates to [x³/3] from 0 to 2. Substituting the limits gives (2³/3) - (0³/3) = 8/3 square units.
How do we find the area between two intersecting curves?
Answer: First, the points of intersection of the two curves are determined by solving their equations simultaneously; these points give the limits of integration. The area is then found by integrating the difference between the upper curve and the lower curve, [f(x) - g(x)], over the interval defined by these intersection points.
Derive the area of a circle x² + y² = r² using integration.
Answer: The upper semicircle can be expressed as y = √(r² - x²), and integrating this from x = -r to x = r gives the area of the upper half of the circle. Doubling this result (to account for both the upper and lower halves) and evaluating the standard integral yields the familiar formula for the area of a circle, πr².
