Class 12 Mathematics Chapter 8: Application of Integrals
CBSE 2026–27 | Detailed NCERT-aligned study resource
This chapter turns definite integration into a geometry tool. The key skill is not merely evaluating an integral; it is correctly identifying the region, finding intersections, choosing limits, and deciding whether to integrate with respect to x or y.
1. Area Under a Curve
If a continuous curve y=f(x) lies above the x-axis from x=a to x=b, its area with the x-axis is ∫ₐᵇ f(x)dx.
│ ╭──── curve y=f(x)
│ █████████ shaded region
│ █████████████
└────────────── x
a b
Worked Example 1
Find the area under y=x² from x=0 to x=2.
A=∫₀²x²dx=[x³/3]₀²=8/3. Therefore, the area is 8/3 square units.
2. Area When the Curve Is Below the x-axis
A definite integral may be negative because it represents signed area. Geometric area is always non-negative. If f(x)<0 throughout the interval, area is −∫ₐᵇf(x)dx.
Example
For y=−x² from 0 to 1, the integral is −1/3, but the geometric area is 1/3 square unit.
3. Area Between Two Curves
When two curves y=f(x) and y=g(x) enclose a region, first solve f(x)=g(x) to obtain intersection points. Then identify which curve is above the other.
──────────────
██████████████ ← area
──────────────
Lower curve
Area = ∫(upper − lower)dx
Worked Example 2: y=x and y=x²
Intersections satisfy x=x², so x=0 and x=1. Between 0 and 1, x is above x².
A=∫₀¹(x−x²)dx=[x²/2−x³/3]₀¹=1/2−1/3=1/6 square unit.
4. Area With Respect to y
Sometimes horizontal strips are simpler. If x=right boundary and x=left boundary, then A=∫(right−left)dy. Always sketch first.
5. Standard Curves in the CBSE Scope
Circle
For x²+y²=a², the upper semicircle is y=√(a²−x²). By symmetry, the area of the complete circle can be obtained by integrating the upper half and doubling it.
Parabola
For y²=4ax, write y=2√(ax) for the upper branch when appropriate. The limits must come from the geometry of the bounded region.
Ellipse
For x²/a²+y²/b²=1, the positive upper branch is y=b√(1−x²/a²). Symmetry can reduce the required integral.
6. How to Solve an Area Question
- Draw a rough labelled graph.
- Find intersection points.
- Decide whether vertical or horizontal strips are easier.
- Identify upper/lower or right/left boundaries.
- Write the limits.
- Set up the definite integral.
- Evaluate and state square units.
7. Exam-Level Example
Find the area enclosed between y=x and y=4−x². First solve x=4−x², giving x²+x−4=0. The roots provide the limits. On the interval between the intersections, determine the upper curve before writing ∫(4−x²−x)dx. The essential scoring steps are the intersection calculation, correct ordering of curves, limits, integration and final area.
8. Common Mistakes
- Writing limits without finding intersections.
- Subtracting lower curve from upper curve in the wrong order.
- Forgetting that geometric area cannot be negative.
- Using dx automatically when dy gives a simpler region.
- Failing to split the integral when the upper/lower curve changes.
9. Practice Set
- Find the area between y=x and y=x².
- Find the area between y=4−x² and the x-axis.
- Find the area enclosed by a circle and a coordinate axis using symmetry.
- Find the area enclosed by a suitable parabola and a line after determining their intersections.
- For an ellipse in standard form, set up an integral for one quadrant and use symmetry.
- Draw the region bounded by two given curves and decide whether dx or dy is more efficient.
10. Quick Revision
| Region | Integral idea |
|---|---|
| Under y=f(x) | ∫f(x)dx |
| Between two curves | ∫(upper−lower)dx |
| Horizontal strips | ∫(right−left)dy |
| Symmetric region | Use symmetry to reduce calculation |
