One of the most important geometrical applications of integration is the determination of the area enclosed by a curve, coordinate axes and straight lines. Integration provides a systematic mathematical method for calculating such areas, even when the boundaries are curved and cannot be measured using elementary geometrical formulae.
In differential calculus, we study the rate of change of a function, whereas integral calculus enables us to accumulate infinitely many infinitesimal quantities. The calculation of the area under a curve is a fundamental example of this accumulation process. It also establishes the connection between definite integration and geometrical measurement.
Consider a continuous function \(y=f(x)\) defined over the interval \([a,b]\). We wish to determine the area enclosed by the curve \(y=f(x)\), the x-axis, and the ordinates \(x=a\) and \(x=b\).
Chapter 8 · CBSE · Class XII
Table of Contents
22 sections
- 1 Overview ›
- 2 Geometrical Interpretation of Area Under a Curve ›
- 3 Elementary Area ›
- 4 Derivation of the Formula for Area ›
- 5 Why Do We Use Definite Integration? ›
- 6 Area Bounded by a Curve and the y-axis ›
- 7 Derivation of the Area Formula ›
- 8 Important Observation ›
- 9 Signed Area and Geometrical Area ›
- 10 Important Formula: Signed Area versus Total Area ›
- 11 General Formulae for Area Under Simple Curves ›
- 12 Example ›
- 13 Choosing Between Vertical and Horizontal Strips ›
- 14 Practical Rule ›
- 15 Applications and Significance of Area Under Curves ›
- 16 Significance ›
- 17 Applications Beyond Mathematics ›
- 18 Exam Tip ›
- 19 Common Mistakes ›
- 20 CBSE Competency-Based Case Study (HOTS) ›
- 21 Key Takeway ›
- 22 Quick Revision Formula Sheet ›
🗺️ Overview
🔍 Geometrical Interpretation of Area Under a Curve
Definition: Area Under a Curve
The area under a curve \(y=f(x)\) between \(x=a\) and \(x=b\) is the limiting sum of the areas of infinitely many thin vertical rectangular strips into which the region is divided.
To understand this concept, imagine dividing the region bounded by the curve, the x-axis and the vertical lines \(x=a\) and \(x=b\) into a large number of narrow vertical strips.
Each strip has a very small width \(dx\) and a height approximately equal to the ordinate \(y=f(x)\) at that position. Therefore, the area of an elementary rectangular strip is approximately the product of its height and width.
To understand this concept, imagine dividing the region bounded by the curve, the x-axis and the vertical lines \(x=a\) and \(x=b\) into a large number of narrow vertical strips.
Each strip has a very small width \(dx\) and a height approximately equal to the ordinate \(y=f(x)\) at that position. Therefore, the area of an elementary rectangular strip is approximately the product of its height and width.
📘 Elementary Area
🎨 SVG Diagram
Elementary Vertical Strip Under a Curve
📐 Derivation of the Formula for Area
🤔 Did You Know?
Why Do We Use Definite Integration?
Indefinite integration gives a family of antiderivatives, whereas definite integration evaluates the accumulated quantity over a specified interval. In the context of area, the limits \(a\) and \(b\) specify the horizontal boundaries of the region.
If \(F'(x)=f(x)\), the Fundamental Theorem of Calculus allows us to evaluate the area as:
If \(F'(x)=f(x)\), the Fundamental Theorem of Calculus allows us to evaluate the area as:
\[
\boxed{
A=F(b)-F(a)
}
\]
This result is valid as a signed integral. For a geometrical area, the sign of the function must also be considered.
📌 Area Bounded by a Curve and the y-axis
🎨 SVG Diagram
Elementary Horizontal Strip
📐 Derivation of the Area Formula
👁️ Important Observation
📌 Signed Area and Geometrical Area
🔢 Important Formula: Signed Area versus Total Area
🔢 General Formulae for Area Under Simple Curves
The following formulae are essential for solving NCERT exercises, CBSE Board questions and competitive examination problems involving areas bounded by curves and coordinate axes.
Formula 1: Area Under a Curve Above the x-axis
For a curve \(y=f(x)\geq0\) between \(x=a\) and \(x=b\):
\[\boxed{A=\int\limits_a^b f(x)\,dx}\]
Formula 2: Area Under a Curve Below the x-axis
For a curve \(y=f(x)\leq0\) between \(x=a\) and \(x=b\):
\[\boxed{A=-\int\limits_a^b f(x)\,dx}\]
Formula 3: Total Area When a Curve Crosses the x-axis
\[\boxed{A=\int\limits_a^b |f(x)|\,dx}\]
Formula 4: Area Bounded by a Curve and the y-axis
For a curve \(x=g(y)\geq0\), between \(y=c\) and \(y=d\):
\[\boxed{A=\int\limits_c^d g(y)\,dy}\]
Formula 5: Area Between Two Curves
If two curves are represented as \(y=f(x)\) and \(y=g(x)\), and \(f(x)\geq g(x)\) throughout \([a,b]\), then the area enclosed between them is:
\[\boxed{A=\int\limits_a^b [f(x)-g(x)]\,dx}\]
Similarly, if the curves are represented in terms of \(y\), and the right-hand curve is \(x=f(y)\), while the left-hand curve is \(x=g(y)\), then:
\[\boxed{A=\int\limits_c^d [f(y)-g(y)]\,dy}\]
\[\text{Area}=\int\limits_c^d(\text{Right curve}-\text{Left curve})\,dy\]
Formula 6: Area Using Symmetry
If \(f(x)\) is an even function, then \(f(-x)=f(x)\), and:
\[\boxed{\int\limits_{-a}^{a}f(x)\,dx=2\int\limits_0^a f(x)\,dx}\]
If \(f(x)\) is an odd function, then \(f(-x)=-f(x)\), and: \[\boxed{\int\limits_{-a}^{a}f(x)\,dx=0}\]
However, the total geometrical area for an odd function is not necessarily zero. It is:
\[\boxed{A=2\int\limits_0^a |f(x)|\,dx}\]
These symmetry properties are particularly useful in evaluating definite integrals and reducing lengthy area calculations.
✏️ Example
1
Question
Find the area bounded by the straight line \(y=2x\), the x-axis, and the ordinates \(x=0\) and \(x=3\).
The straight line lies above the x-axis in the given interval. Therefore, the required area is the definite integral of the function between the given limits. Geometrically, the enclosed region is also a triangle.
-
1
Identify the function and the integration limits.
-
2
Apply the area formula for a curve above the x-axis.
-
3
Evaluate the definite integral.
-
4
Verify the result using the geometrical area of a triangle.
-
\[y=2x,\quad a=0,\quad b=3\]
-
\[A=\int\limits_0^3 2x\,dx\]
-
\[A=2\left[\frac{x^2}{2}\right]_0^3\]
-
\[A=[x^2]_0^3\]
-
\[A=9-0\]
\[\boxed{A=9\text{ square units}}\]
2
Question
Find the area bounded by the parabola \(y=x^2\), the x-axis, and the ordinates \(x=0\) and \(x=2\).
The parabola \(y=x^2\) lies above the x-axis for all real values of \(x\), except at the origin where it touches the axis. Hence, the definite integral directly represents the geometrical area.
-
1
Express the required area as a definite integral.
-
2
Apply the power rule of integration.
-
3
Substitute the upper and lower limits.
-
\[A=\int\limits_0^2 x^2\,dx\]
-
\[A=\left[\frac{x^3}{3}\right]_0^2\]
-
\[A=\frac{2^3}{3}-\frac{0^3}{3}\]
\[\boxed{A=\frac83\text{ square units}}\]
3
Question
Find the geometrical area bounded by the curve \(y=x^2-4\), the x-axis, and the ordinates \(x=0\) and \(x=2\).
For \(0\leq x\leq2\), we have \(x^2\leq4\), so \(y=x^2-4\leq0\). The curve lies below the x-axis. Therefore, the negative value of the definite integral must be converted into a positive geometrical area.
-
\[A=\left|\int\limits_0^2(x^2-4)\,dx\right|\]
-
\[=\left|\left[\frac{x^3}{3}-4x\right]_0^2\right|\]
-
\[=\left|\frac83-8\right|\]
-
\[=\left|-\frac{16}{3}\right|\]
\[\boxed{A=\frac{16}{3}\text{ square units}}\]
4
Question
Find the total geometrical area enclosed between the curve \(y=x^2-1\), the x-axis, and the ordinates \(x=-2\) and \(x=2\).
The curve intersects the x-axis at:
\[x^2-1=0\]
\[x=\pm1\]
The function is positive on \([-2,-1]\) and \([1,2]\), and negative on \([-1,1]\). Hence, the interval must be divided at the points of intersection with the x-axis. -
1
Determine the x-intercepts of the curve.
-
2
Identify the intervals where the function is positive or negative.
-
3
Split the integral at the points where the sign changes.
-
4
Add the positive geometrical areas.
-
\[A=\int\limits_{-2}^{-1}(x^2-1)\,dx-\int\limits_{-1}^{1}(x^2-1)\,dx+\int\limits_1^2(x^2-1)\,dx\]
-
Since \(x^2-1\) is an even function, the two outer regions have equal areas, and the central region is symmetric about the y-axis.
\[ A=2\int\limits_1^2(x^2-1)\,dx +2\int\limits_0^1(1-x^2)\,dx \]\[ =2\left[\frac{x^3}{3}-x\right]_1^2 +2\left[x-\frac{x^3}{3}\right]_0^1 \]\[ =2\left(\frac53+\frac23\right) \]
\[\boxed{A=\frac{14}{3}\text{ square units}}\]
5
Question
Find the area bounded by the curve \(x=y^2\), the y-axis, and the lines \(y=0\) and \(y=2\).
The equation is expressed as \(x\) in terms of \(y\). Therefore, horizontal strips and integration with respect to \(y\) are more convenient.
-
\[A=\int\limits_0^2 y^2\,dy\]\[=\left[\frac{y^3}{3}\right]_0^2\]\[=\frac83\]
\[\boxed{A=\frac83\text{ square units}}\]
📌 Note
Choosing Between Vertical and Horizontal Strips
📜 Practical Rule
Rules
- For vertical strips, calculate the difference between the upper and lower boundaries.
- For horizontal strips, calculate the difference between the right and left boundaries.
- Use the method that gives simpler integration limits and a simpler integrand.
- If a strip changes its upper, lower, left or right boundary within the region, divide the region into separate parts.
🛠️ Applications and Significance of Area Under Curves
Importance in CBSE Class 12 Mathematics
The geometrical interpretation of definite integrals is an essential foundation for the NCERT Class 12 Mathematics Chapter 7, Integrals, particularly the section on applications of integrals. Students must understand the relationship between elementary area, definite integration, geometrical boundaries and the sign of a function before attempting more complex problems.
This topic supports the following learning objectives:
- Understanding definite integration as the limiting sum of elementary areas.
- Identifying the correct limits of integration from a geometrical figure.
- Distinguishing signed integrals from actual geometrical areas.
- Calculating areas enclosed by curves and coordinate axes.
- Choosing between vertical and horizontal strips.
- Applying symmetry to simplify definite integrals.
- Extending the method to areas bounded by two curves.
🌟 Significance
A strong understanding of area under curves is useful for JEE Main, JEE Advanced, CUET UG, BITSAT and other mathematics-based entrance examinations. Questions may require direct integration, graphical interpretation, symmetry, sign analysis or a combination of these techniques.
- JEE Main: Questions involving definite integrals, geometrical area, intersections of curves and symmetry.
- JEE Advanced: Multi-region area calculations, curve intersections, transformations, symmetry and non-trivial choices of integration variable.
- CUET UG: Conceptual and application-based questions involving definite integration and elementary area formulae.
- BITSAT: Direct applications of integration, geometrical interpretation and efficient evaluation of definite integrals.
🛠️ Applications Beyond Mathematics
The principle of accumulating infinitesimal quantities is used in several scientific and engineering disciplines:
- Physics: Displacement from velocity-time graphs, work done by a variable force and accumulated charge.
- Engineering: Cross-sectional areas, material quantities, fluid mechanics and structural calculations.
- Economics: Consumer surplus, producer surplus and accumulated marginal quantities.
- Statistics: Probabilities represented by areas under probability density curves.
- Data Science: Continuous probability distributions and cumulative distribution functions.
⚡ Exam Tip
❌ Common Mistakes
| Common mistake | Correct approach |
|---|---|
| Assuming every definite integral gives positive area. | Remember that a definite integral represents signed area. |
| Taking a negative area as the final geometrical answer. | Use the absolute value or reverse the sign when the entire curve is below the axis. |
| Not splitting the integral when a curve crosses the x-axis. | Find the intercepts and divide the integration interval wherever the sign changes. |
| Using \(dx\) when horizontal strips would simplify the region. | Consider both vertical and horizontal strips before choosing the integration variable. |
| Confusing upper and lower curves. | For vertical strips, subtract lower from upper. |
| Confusing right and left boundaries. | For horizontal strips, subtract left from right. |
| Applying symmetry to the signed area without considering the sign. | Distinguish between an algebraic integral and the total geometrical area. |
| Using the wrong integration limits. | Determine the intersection points and the actual extent of the region before integration. |
📋 Case Study
CBSE Competency-Based Case Study (HOTS)
Case Study: Designing a Curved Garden
An architect is designing a landscaped garden whose upper boundary is represented by the parabola \(y=4-x^2\). The garden extends from \(x=-2\) to \(x=2\), and its lower boundary is the x-axis. The architect wants to determine the area available for planting grass.
Based on this information, answer the following questions.
Question 1. At which points does the parabolic boundary intersect the x-axis?
(A) \((0,0)\) and \((2,0)\)
(B) \((-2,0)\) and \((2,0)\)
(C) \((-4,0)\) and \((4,0)\)
(D) \((0,2)\) and \((0,-2)\)
Answer: (B)
Explanation:
\[
4-x^2=0
\]
\[
x=\pm2
\]
Hence, the points of intersection are \((-2,0)\) and \((2,0)\).
Question 2. Which definite integral represents the area of the garden?
(A) \(\int\limits_{-2}^{2}(x^2-4)\,dx\)
(B) \(\int\limits_{-2}^{2}(4-x^2)\,dx\)
(C) \(\int\limits_0^2(x^2-4)\,dx\)
(D) \(\int\limits_{-2}^{2}x^2\,dx\)
Answer: (B)
Explanation: The parabola lies above the x-axis between \(x=-2\) and \(x=2\). Therefore, the required area is the integral of the upper boundary minus the lower boundary.
Question 3. Calculate the total area of the garden.
(A) \(\frac{16}{3}\) square units
(B) \(\frac{20}{3}\) square units
(C) \(\frac{32}{3}\) square units
(D) \(16\) square units
Answer: (C)
Solution:
\[
A=\int\limits_{-2}^{2}(4-x^2)\,dx
\]
Since \(4-x^2\) is an even function:
\[
A=2\int\limits_0^2(4-x^2)\,dx
\]
\[
=2\left[4x-\frac{x^3}{3}\right]_0^2
\]
\[
=2\left(8-\frac83\right)
\]
\[
=2\times\frac{16}{3}
\]
\[
\boxed{A=\frac{32}{3}\text{ square units}}
\]
Question 4. What is the significance of the symmetry of the garden about the y-axis?
(A) It makes the area zero.
(B) It allows the area on one side of the y-axis to be doubled.
(C) It changes the integration variable to \(y\).
(D) It makes the curve lie below the x-axis.
Answer: (B)
Explanation: The equation \(y=4-x^2\) is an even function. The left and right halves of the garden are identical, so the area can be calculated over \([0,2]\) and doubled.
Question 5 (HOTS). If the architect decides to use horizontal strips instead of vertical strips, which expression represents the same garden area?
(A) \(\int\limits_0^4 2\sqrt{4-y}\,dy\)
(B) \(\int\limits_0^4 \sqrt{4-y}\,dy\)
(C) \(\int\limits_{-2}^{2}(4-y^2)\,dy\)
(D) \(\int\limits_0^4 (4-y)\,dy\)
Answer: (A)
Solution:
The equation of the parabola is:
\[
y=4-x^2
\]
Expressing \(x\) in terms of \(y\):
\[
x=\pm\sqrt{4-y}
\]
For a horizontal strip, the right boundary is \(x=\sqrt{4-y}\), and the left boundary is \(x=-\sqrt{4-y}\). Thus, the horizontal length of the strip is:
\[
\sqrt{4-y}-(-\sqrt{4-y})
=2\sqrt{4-y}
\]
The vertical limits are \(y=0\) and \(y=4\). Therefore:
\[
A=\int\limits_0^4 2\sqrt{4-y}\,dy
\]
\[
=2\int\limits_0^4(4-y)^{1/2}\,dy
\]
Substituting \(u=4-y\), we get:
\[
A=2\int\limits_0^4 u^{1/2}\,du
\]
\[
=2\left[\frac{2u^{3/2}}{3}\right]_0^4
\]
\[
=\frac43(8)
\]
\[
\boxed{A=\frac{32}{3}\text{ square units}}
\]
Both vertical and horizontal strips give the same geometrical area. This demonstrates that the choice of integration variable is a matter of convenience, not a change in the area being calculated.
🗒️ Key Takeway
- The area under a curve can be obtained by summing the areas of infinitely many elementary strips.
- For a vertical strip, \(dA=y\,dx\).
- For a horizontal strip, \(dA=x\,dy\).
- If the curve lies above the x-axis, the definite integral directly gives its area.
- If the curve lies below the x-axis, the geometrical area is the negative of the definite integral.
- If the curve crosses the x-axis, divide the interval at its intercepts and add the absolute areas.
- The general geometrical area between a curve and the x-axis is \(\int\limits_a^b|f(x)|\,dx\).
- For two curves, use upper curve minus lower curve with vertical strips, or right curve minus left curve with horizontal strips.
- Symmetry can substantially reduce the calculations involved in definite integration.
⚡ Quick Revision Formula Sheet
| Concept | Formula |
|---|---|
| Elementary vertical strip | \(dA=y\,dx\) |
| Elementary horizontal strip | \(dA=x\,dy\) |
| Area above x-axis | \(\displaystyle\int\limits_a^b f(x)\,dx\) |
| Area below x-axis | \(\displaystyle-\int\limits_a^b f(x)\,dx\) |
| Total area with sign changes | \(\displaystyle\int\limits_a^b|f(x)|\,dx\) |
| Area to the right of y-axis | \(\displaystyle\int\limits_c^d g(y)\,dy\) |
| Area between two curves | \(\displaystyle\int\limits_a^b[\text{Upper}-\text{Lower}]\,dx\) |
| Area using horizontal strips | \(\displaystyle\int\limits_c^d[\text{Right}-\text{Left}]\,dy\) |
| Integral of an even function | \(\displaystyle\int\limits_{-a}^{a}f(x)\,dx=2\int\limits_0^a f(x)\,dx\) |
| Integral of an odd function | \(\displaystyle\int\limits_{-a}^{a}f(x)\,dx=0\) |