Class 12 Mathematics (041) · Chapter 8 NotesApplication of Integrals
Comprehensive notes for Class 12 Maths Chapter 8: Application of Integrals. Learn to calculate areas under curves, circles, and ellipses using NCERT methods.
In previous mathematical studies, students learned to calculate the areas of standard geometric figures like triangles, rectangles, and circles using elementary formulas. However, these basic tools are insufficient for determining the area enclosed by irregular curves. Chapter 8 of Class 12 Mathematics introduces the powerful application of Integral Calculus to solve these complex spatial problems. By viewing a region as a collection of infinitely thin strips, integration allows us to sum these elementary areas to find the total space bounded by a curve and the coordinate axes. This chapter focuses on finding the area under simple curves, including lines, circles, parabolas, and ellipses in their standard forms. Students will learn to interpret the definite integral not just as a numerical value, but as a geometric representation of area. Mastery of this chapter is essential for understanding how calculus bridges the gap between abstract functions and physical dimensions, providing a foundation for advanced engineering and physics applications.
What you'll learn
1Define the area under a curve as a limit of a sum of elementary strips.
2Calculate the area bounded by a curve, the x-axis, and given vertical ordinates.
3Determine the area bounded by a curve, the y-axis, and given horizontal lines.
4Apply absolute values to find the total area when a curve lies below the x-axis.
5Utilize symmetry properties to simplify area calculations for circles and ellipses.
6Sketch and identify regions bounded by standard curves and lines.
Chapter at a glance
01Area Under Simple Curves
02Area Between Two Curves
03Application to Volume of Solids
04Differential Equations and Their Applications
05Physical Applications and Real-World Problems
Detailed chapter notes
01
The Concept of Elementary Area
The fundamental approach to finding the area under a curve involves dividing the region into a large number of very thin vertical or horizontal strips. For a curve defined by y = f(x), we consider an arbitrary vertical strip with a height 'y' and an infinitesimal width 'dx'. The area of this single strip, known as the elementary area (dA), is expressed as dA = y dx. By integrating this expression between the limits x = a and x = b, we effectively sum all these tiny rectangles to find the total area of the region bounded by the curve, the x-axis, and the vertical lines x = a and x = b.
dA = y dx (for vertical strips)
dA = x dy (for horizontal strips)
Integration acts as the summation of these elementary areas.
02
Area Bounded by the X-axis
When a region is bounded by a continuous curve y = f(x), the x-axis, and the ordinates x = a and x = b, the total area A is calculated using the definite integral of the function with respect to x. If the curve lies entirely above the x-axis, the integral yields a positive value representing the area. This method is intuitive and relies on the Fundamental Theorem of Calculus to evaluate the boundaries. It is the most common application of integration in geometry, allowing for the measurement of spaces that do not have straight edges.
FormulaArea = Integral from a to b of f(x) dx
The limits a and b represent the vertical boundaries (ordinates).
The function f(x) must be continuous over the interval [a, b].
03
Area Bounded by the Y-axis
In some cases, it is more convenient to calculate the area with respect to the y-axis. This occurs when the region is bounded by a curve x = g(y), the y-axis, and the horizontal lines y = c and y = d. Instead of vertical strips, we use horizontal elementary strips of width 'dy' and length 'x'. The area is then found by integrating the function g(y) with respect to y between the limits c and d. This dual approach provides flexibility, as some functions are easier to integrate when expressed in terms of y rather than x.
FormulaArea = Integral from c to d of g(y) dy
The limits c and d represent the horizontal boundaries.
Useful for curves that are functions of y.
04
Handling Negative Areas and Split Regions
A definite integral can result in a negative value if the curve lies below the x-axis (where f(x) < 0). However, since area is a physical quantity, it must always be positive. In such instances, we take the absolute numerical value of the integral. If a curve crosses the x-axis within the interval [a, b], the region is split into two parts: one above the axis and one below. To find the total area, we must calculate the integral for each section separately and add their absolute values. This ensures that the negative portion does not subtract from the positive portion.
Area = |Integral of f(x) dx| for regions below the axis.
Total Area = |A1| + A2 for curves crossing the axis.
Always sketch the curve to identify where it intersects the x-axis.
05
Symmetry in Circles and Ellipses
Standard curves like circles and ellipses exhibit symmetry about both the x-axis and the y-axis. For a circle defined by x² + y² = a², the total area is four times the area of the portion in the first quadrant. Similarly, for an ellipse x²/a² + y²/b² = 1, we calculate the area from x = 0 to x = a in the first quadrant and multiply by four. This symmetry significantly simplifies the integration process. For a circle, this derivation leads to the familiar formula πa², while for an ellipse, the total area is found to be πab.
Circle Area4 * Integral from 0 to a of sqrt(a² - x²) dx = πa²
Ellipse Area4 * (b/a) * Integral from 0 to a of sqrt(a² - x²) dx = πab
Symmetry reduces the complexity of integration limits.
06
Integration of Trigonometric Regions
The chapter also covers finding areas bounded by trigonometric curves like y = sin x or y = cos x. Because these functions are periodic and oscillate above and below the x-axis, calculating the area over a large interval (like 0 to 2π) requires careful attention. For example, the area of y = cos x from 0 to 2π is composed of three distinct regions. We must integrate from 0 to π/2, π/2 to 3π/2, and 3π/2 to 2π, taking the absolute value of the middle section to ensure the total area is correctly summed as 4 square units.
Identify roots of the trigonometric function to find split points.
Use absolute values for intervals where the function is negative.
Trigonometric areas often involve standard integral results like [sin x] or [-cos x].
Want the complete chapter resources?Topic notes, quizzes and flashcards for Application of Integrals.
The area under the curve y = x between x = 0 and x = 2 is:
A2 square units
B4 square units
C1 square unit
D8 square units
Show answer
Answer: (A) 2 square units
Area = ∫₀² x dx = [x²/2]₀² = 4/2 - 0 = 2 square units. This is a direct application of the fundamental theorem of calculus.
Question 02
The area between two curves y = f(x) and y = g(x) where f(x) ≥ g(x) between x = a and x = b is given by:
A∫[a to b] [f(x) - g(x)] dx
B∫[a to b] [g(x) - f(x)] dx
C∫[a to b] f(x) dx + ∫[a to b] g(x) dx
D∫[a to b] [f(x) + g(x)] dx
Show answer
Answer: (A) ∫[a to b] [f(x) - g(x)] dx
The area between two curves is the integral of the difference of the upper curve minus the lower curve over the given interval.
Question 03
The volume of a solid of revolution generated by rotating the curve y = f(x) about the x-axis between x = a and x = b is given by:
AV = π ∫[a to b] [f(x)]² dx
BV = ∫[a to b] [f(x)]² dx
CV = 2π ∫[a to b] f(x) dx
DV = π ∫[a to b] f(x) dx
Show answer
Answer: (A) V = π ∫[a to b] [f(x)]² dx
The disk method for volume of revolution about the x-axis uses the formula V = π ∫[a to b] [f(x)]² dx, where [f(x)]² represents the square of the radius of each circular disk.
Question 04
The order of a differential equation is defined as:
Athe highest power of the dependent variable
Bthe highest power of the derivative present in the equation
Cthe highest order of the derivative present in the equation
Dthe number of variables in the equation
Show answer
Answer: (C) the highest order of the derivative present in the equation
The order of a differential equation is determined by the order of the highest derivative. For example, d³y/dx³ + dy/dx = 0 has order 3.
Question 05
The area enclosed between the curve y = x² and the x-axis from x = 0 to x = 2 is:
A8/3 square units
B4 square units
C16/3 square units
D6 square units
Show answer
Answer: (A) 8/3 square units
The area is given by ∫₀² x² dx = [x³/3]₀² = 8/3 square units. This is a direct application of definite integration to find the area under a curve.
Ready for more practice?Unlock the full quiz for this chapter.
An elementary area is the area of an infinitesimally thin strip (either vertical or horizontal) within a region. For a curve y = f(x), a vertical elementary strip has an area dA = y dx. Integration sums these infinite strips to find the total area.
Why do we take the absolute value of an integral when finding area?
Definite integrals can be negative if the curve lies below the x-axis. Since geometric area is a non-negative physical quantity, we take the absolute value of the integral's result to represent the actual space occupied.
How do you find the area if a curve is symmetrical?
If a curve is symmetrical about an axis, you can calculate the area of one symmetrical part and multiply it by the number of parts. For example, the area of a circle is 4 times the area of its first-quadrant portion.
When should I use horizontal strips instead of vertical strips?
Horizontal strips (x dy) are used when the boundaries are given as horizontal lines (y = c, y = d) or when the curve is more easily expressed as a function of y, such as x = g(y).
How do you calculate the area of a region that crosses the x-axis?
You must find the points where the curve intersects the x-axis to divide the interval. Calculate the definite integral for each sub-interval separately, then add their absolute values to get the total area.