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Application of Integrals
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Class 12MathsChapter 8NCERTCBSE
Application of Integrals - Class 12 Maths Chapter 8
📌 Quick Overview: Chapter 8 applies definite integrals to find areas of geometric shapes bounded by curves. You will learn how to find the area under a curve, area between two curves, and areas of standard shapes like circles, parabolas, and ellipses using integration. This chapter is compact but very important — it almost always appears as a 5-mark question in CBSE board exams. The key skill is setting up the correct integral with correct limits.
Topics Covered:
- > Area Under a Curve (basic)
- > Area using Vertical Strips
- > Area using Horizontal Strips
- > Area of Standard Curves
- > Area - Circle
- > Area - Parabola
- > Area - Ellipse
- > Area Between Two Curves
- > Finding Intersection Points
- > Mixed Problems
- > Important Board Questions
- > MCQ Practice
1. Area Under a Curve - Basic Concept
Core Idea: The area of the region bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is given by the definite integral:
Area = Integral[a to b] |f(x)| dx
If f(x) >= 0 on [a, b] (curve is above x-axis):
Area = Integral[a to b] f(x) dx
If f(x) is partly above and partly below the x-axis:
Area = |Integral[a to c] f(x)dx| + |Integral[c to b] f(x)dx|
(split at point c where f(c) = 0)
Area = Integral[a to b] |f(x)| dx
If f(x) >= 0 on [a, b] (curve is above x-axis):
Area = Integral[a to b] f(x) dx
If f(x) is partly above and partly below the x-axis:
Area = |Integral[a to c] f(x)dx| + |Integral[c to b] f(x)dx|
(split at point c where f(c) = 0)
⭐ Key Sign Rule: Area is always POSITIVE. When the curve dips below the x-axis, the integral gives a negative value — take the absolute value (modulus) before adding.
2. Vertical Strips vs Horizontal Strips
VERTICAL STRIPS (integrate with respect to x):
Area = Integral[a to b] y dx = Integral[a to b] f(x) dx
Use when: curve is given as y = f(x), limits on x-axis
HORIZONTAL STRIPS (integrate with respect to y):
Area = Integral[c to d] x dy = Integral[c to d] g(y) dy
Use when: curve is given as x = g(y), limits on y-axis
OR when vertical strips are difficult (e.g. parabola opening sideways)
How to Choose:
> If the curve is y = f(x) and bounded by vertical lines x = a, x = b → use vertical strips.
> If the curve is x = g(y) or bounded by horizontal lines → use horizontal strips.
> When area is between a curve and the y-axis → use horizontal strips.
> If the curve is y = f(x) and bounded by vertical lines x = a, x = b → use vertical strips.
> If the curve is x = g(y) or bounded by horizontal lines → use horizontal strips.
> When area is between a curve and the y-axis → use horizontal strips.
3. Standard Curve Equations You Must Know ⭐
| Curve | Equation | Key Feature |
|---|---|---|
| Circle (centre at origin) | x^2 + y^2 = r^2 | Area = pi.r^2 |
| Ellipse (centre at origin) | x^2/a^2 + y^2/b^2 = 1 | Area = pi.a.b |
| Parabola (upward) | y^2 = 4ax | Opens right along x-axis |
| Parabola (downward) | x^2 = 4ay | Opens upward along y-axis |
| Line | y = mx + c | Straight line |
| Semicircle (upper) | y = sqrt(r^2 - x^2) | Upper half of circle |
⭐ Standard Results to memorise:
> Integral[0 to r] sqrt(r^2 - x^2) dx = pi.r^2/4 (area of quarter circle)
> Integral[-r to r] sqrt(r^2 - x^2) dx = pi.r^2/2 (area of semicircle)
> Integral[0 to a] sqrt(a^2 - x^2) dx = pi.a^2/4
> Integral[0 to r] sqrt(r^2 - x^2) dx = pi.r^2/4 (area of quarter circle)
> Integral[-r to r] sqrt(r^2 - x^2) dx = pi.r^2/2 (area of semicircle)
> Integral[0 to a] sqrt(a^2 - x^2) dx = pi.a^2/4
4. Area of a Circle using Integration ⭐
Find the area of circle x^2 + y^2 = a^2 using integration.
The circle has equation x^2 + y^2 = a^2.
Upper semicircle: y = sqrt(a^2 - x^2)
Area of circle = 4 × (area of one quadrant)
= 4 × Integral[0 to a] sqrt(a^2 - x^2) dx
Using the formula Integral sqrt(a^2-x^2) dx = (x/2)sqrt(a^2-x^2) + (a^2/2)sin⁻¹(x/a):
= 4 × [(x/2)sqrt(a^2-x^2) + (a^2/2)sin⁻¹(x/a)] from 0 to a
= 4 × [(0 + (a^2/2).sin⁻¹(1)) - (0 + 0)]
= 4 × (a^2/2).(pi/2)
= pi.a^2 ✓
The circle has equation x^2 + y^2 = a^2.
Upper semicircle: y = sqrt(a^2 - x^2)
Area of circle = 4 × (area of one quadrant)
= 4 × Integral[0 to a] sqrt(a^2 - x^2) dx
Using the formula Integral sqrt(a^2-x^2) dx = (x/2)sqrt(a^2-x^2) + (a^2/2)sin⁻¹(x/a):
= 4 × [(x/2)sqrt(a^2-x^2) + (a^2/2)sin⁻¹(x/a)] from 0 to a
= 4 × [(0 + (a^2/2).sin⁻¹(1)) - (0 + 0)]
= 4 × (a^2/2).(pi/2)
= pi.a^2 ✓
5. Area of an Ellipse using Integration
Find the area of ellipse x^2/a^2 + y^2/b^2 = 1.
Upper half: y = (b/a)sqrt(a^2 - x^2)
Area = 4 × Integral[0 to a] (b/a)sqrt(a^2 - x^2) dx
= (4b/a) × Integral[0 to a] sqrt(a^2 - x^2) dx
= (4b/a) × (pi.a^2/4)
= pi.a.b ✓
Upper half: y = (b/a)sqrt(a^2 - x^2)
Area = 4 × Integral[0 to a] (b/a)sqrt(a^2 - x^2) dx
= (4b/a) × Integral[0 to a] sqrt(a^2 - x^2) dx
= (4b/a) × (pi.a^2/4)
= pi.a.b ✓
6. Area Related to Parabola ⭐
Example 1: Find the area bounded by y^2 = 4x, x = 1 and x = 4.
y = sqrt(4x) = 2sqrt(x) (upper half of parabola)
By symmetry about x-axis:
Area = 2 × Integral[1 to 4] 2sqrt(x) dx
= 4 × Integral[1 to 4] x^(1/2) dx
= 4 × [x^(3/2) / (3/2)] from 1 to 4
= 4 × (2/3)[x^(3/2)] from 1 to 4
= (8/3)[8 - 1]
= 56/3 sq. units
y = sqrt(4x) = 2sqrt(x) (upper half of parabola)
By symmetry about x-axis:
Area = 2 × Integral[1 to 4] 2sqrt(x) dx
= 4 × Integral[1 to 4] x^(1/2) dx
= 4 × [x^(3/2) / (3/2)] from 1 to 4
= 4 × (2/3)[x^(3/2)] from 1 to 4
= (8/3)[8 - 1]
= 56/3 sq. units
Example 2: Find the area bounded by the parabola y = x^2 and y = x.
Find intersection: x^2 = x → x(x-1) = 0 → x = 0, x = 1
In [0,1]: y = x is above y = x^2 (check at x = 0.5: 0.5 > 0.25)
Area = Integral[0 to 1] (x - x^2) dx
= [x^2/2 - x^3/3] from 0 to 1
= (1/2 - 1/3) - 0
= 3/6 - 2/6 = 1/6 sq. units
Find intersection: x^2 = x → x(x-1) = 0 → x = 0, x = 1
In [0,1]: y = x is above y = x^2 (check at x = 0.5: 0.5 > 0.25)
Area = Integral[0 to 1] (x - x^2) dx
= [x^2/2 - x^3/3] from 0 to 1
= (1/2 - 1/3) - 0
= 3/6 - 2/6 = 1/6 sq. units
7. Area Between Two Curves ⭐ (Most Important)
Formula: Area between y = f(x) (upper curve) and y = g(x) (lower curve) from x = a to x = b:
Area = Integral[a to b] [f(x) - g(x)] dx
where f(x) >= g(x) on [a, b].
Area = Integral[a to b] [f(x) - g(x)] dx
where f(x) >= g(x) on [a, b].
Steps for Area Between Two Curves:
Step 1: Find intersection points — solve f(x) = g(x) to get limits a and b.
Step 2: Determine which curve is on top (substitute a test x value).
Step 3: Integrate [upper curve - lower curve] between the limits.
Step 4: If curves cross in between, split the integral at crossing point.
Step 1: Find intersection points — solve f(x) = g(x) to get limits a and b.
Step 2: Determine which curve is on top (substitute a test x value).
Step 3: Integrate [upper curve - lower curve] between the limits.
Step 4: If curves cross in between, split the integral at crossing point.
Example 1: Find the area between y = x^2 and y = sqrt(x).
Intersection: x^2 = sqrt(x) → x^4 = x → x(x^3 - 1) = 0 → x = 0 or x = 1
Check at x = 0.5: sqrt(0.5) = 0.707 > (0.5)^2 = 0.25
So y = sqrt(x) is above y = x^2 on [0,1].
Area = Integral[0 to 1] (sqrt(x) - x^2) dx
= [x^(3/2)/(3/2) - x^3/3] from 0 to 1
= [(2/3)(1) - (1/3)] - 0
= 2/3 - 1/3 = 1/3 sq. units
Intersection: x^2 = sqrt(x) → x^4 = x → x(x^3 - 1) = 0 → x = 0 or x = 1
Check at x = 0.5: sqrt(0.5) = 0.707 > (0.5)^2 = 0.25
So y = sqrt(x) is above y = x^2 on [0,1].
Area = Integral[0 to 1] (sqrt(x) - x^2) dx
= [x^(3/2)/(3/2) - x^3/3] from 0 to 1
= [(2/3)(1) - (1/3)] - 0
= 2/3 - 1/3 = 1/3 sq. units
Example 2: Find the area enclosed by the circle x^2 + y^2 = 4 and the line x + y = 2 in the first quadrant.
Circle: radius 2, line x + y = 2 → y = 2 - x
Intersection: x^2 + (2-x)^2 = 4 → x^2 + 4 - 4x + x^2 = 4
2x^2 - 4x = 0 → 2x(x-2) = 0 → x = 0 or x = 2
In first quadrant from x=0 to x=2:
Upper: y = sqrt(4-x^2) (circle), Lower: y = 2-x (line)
Area = Integral[0 to 2] [sqrt(4-x^2) - (2-x)] dx
= [Integral of sqrt(4-x^2) dx] - [2x - x^2/2] from 0 to 2
= [(x/2)sqrt(4-x^2) + 2sin⁻¹(x/2)] from 0 to 2 - [(4-2) - 0]
= [0 + 2.sin⁻¹(1) - 0 - 0] - 2
= 2.(pi/2) - 2
= (pi - 2) sq. units
Circle: radius 2, line x + y = 2 → y = 2 - x
Intersection: x^2 + (2-x)^2 = 4 → x^2 + 4 - 4x + x^2 = 4
2x^2 - 4x = 0 → 2x(x-2) = 0 → x = 0 or x = 2
In first quadrant from x=0 to x=2:
Upper: y = sqrt(4-x^2) (circle), Lower: y = 2-x (line)
Area = Integral[0 to 2] [sqrt(4-x^2) - (2-x)] dx
= [Integral of sqrt(4-x^2) dx] - [2x - x^2/2] from 0 to 2
= [(x/2)sqrt(4-x^2) + 2sin⁻¹(x/2)] from 0 to 2 - [(4-2) - 0]
= [0 + 2.sin⁻¹(1) - 0 - 0] - 2
= 2.(pi/2) - 2
= (pi - 2) sq. units
Example 3: Find the area between the parabola y = x^2 - 4x and the line y = 2x - x^2 - 4? No — let us do a standard one. Area between y^2 = 4x and x^2 = 4y.
Intersection: From y^2 = 4x → x = y^2/4. Substitute in x^2 = 4y:
(y^2/4)^2 = 4y → y^4/16 = 4y → y^4 = 64y → y(y^3 - 64) = 0 → y = 0 or y = 4
When y = 4: x = 16/4 = 4. Points: (0,0) and (4,4).
For x in [0,4]: upper curve is y = 2sqrt(x) (from y^2=4x), lower is y = x^2/4 (from x^2=4y)
Area = Integral[0 to 4] [2sqrt(x) - x^2/4] dx
= [2.(x^(3/2))/(3/2) - x^3/12] from 0 to 4
= [(4/3)x^(3/2) - x^3/12] from 0 to 4
= [(4/3)(8) - 64/12] - 0
= 32/3 - 16/3
= 16/3 sq. units
Intersection: From y^2 = 4x → x = y^2/4. Substitute in x^2 = 4y:
(y^2/4)^2 = 4y → y^4/16 = 4y → y^4 = 64y → y(y^3 - 64) = 0 → y = 0 or y = 4
When y = 4: x = 16/4 = 4. Points: (0,0) and (4,4).
For x in [0,4]: upper curve is y = 2sqrt(x) (from y^2=4x), lower is y = x^2/4 (from x^2=4y)
Area = Integral[0 to 4] [2sqrt(x) - x^2/4] dx
= [2.(x^(3/2))/(3/2) - x^3/12] from 0 to 4
= [(4/3)x^(3/2) - x^3/12] from 0 to 4
= [(4/3)(8) - 64/12] - 0
= 32/3 - 16/3
= 16/3 sq. units
8. Area Between Curve and Y-axis
Area bounded by x = g(y), the y-axis, and lines y = c and y = d:
Area = Integral[c to d] |g(y)| dy
Area = Integral[c to d] |g(y)| dy
Example: Find the area bounded by x^2 = y, y = 1, y = 4 and the y-axis.
x = sqrt(y) (right half)
Area = Integral[1 to 4] sqrt(y) dy
= [y^(3/2)/(3/2)] from 1 to 4
= (2/3)[4^(3/2) - 1^(3/2)]
= (2/3)[8 - 1]
= 14/3 sq. units
x = sqrt(y) (right half)
Area = Integral[1 to 4] sqrt(y) dy
= [y^(3/2)/(3/2)] from 1 to 4
= (2/3)[4^(3/2) - 1^(3/2)]
= (2/3)[8 - 1]
= 14/3 sq. units
9. Important Board Exam Questions
Q1. Find the area of the region bounded by the curve y^2 = x, the lines x = 1, x = 4 and the x-axis in the first quadrant.
y = sqrt(x) in first quadrant
Area = Integral[1 to 4] sqrt(x) dx
= [x^(3/2) / (3/2)] from 1 to 4
= (2/3)[4^(3/2) - 1]
= (2/3)[8 - 1]
= 14/3 sq. units
Area = Integral[1 to 4] sqrt(x) dx
= [x^(3/2) / (3/2)] from 1 to 4
= (2/3)[4^(3/2) - 1]
= (2/3)[8 - 1]
= 14/3 sq. units
Q2. Find the area enclosed between the parabola y = x^2 and the line y = x + 2.
Intersection: x^2 = x + 2 → x^2 - x - 2 = 0 → (x-2)(x+1) = 0 → x = -1, x = 2
Check at x = 0: line gives y = 2, parabola gives y = 0 → line is above.
Area = Integral[-1 to 2] [(x+2) - x^2] dx
= [x^2/2 + 2x - x^3/3] from -1 to 2
= (2 + 4 - 8/3) - (1/2 - 2 + 1/3)
= (18/3 - 8/3) - (3/6 - 12/6 + 2/6)
= 10/3 - (-7/6)
= 10/3 + 7/6 = 20/6 + 7/6
= 27/6 = 9/2 sq. units
Check at x = 0: line gives y = 2, parabola gives y = 0 → line is above.
Area = Integral[-1 to 2] [(x+2) - x^2] dx
= [x^2/2 + 2x - x^3/3] from -1 to 2
= (2 + 4 - 8/3) - (1/2 - 2 + 1/3)
= (18/3 - 8/3) - (3/6 - 12/6 + 2/6)
= 10/3 - (-7/6)
= 10/3 + 7/6 = 20/6 + 7/6
= 27/6 = 9/2 sq. units
Q3. Find the area of the region {(x,y): x^2 + y^2 <= 1 <= x + y}.
This is the area inside the circle x^2 + y^2 = 1 but above the line x + y = 1.
Line x+y=1 intersects circle at: x^2+(1-x)^2=1 → 2x^2-2x=0 → x=0 or x=1.
Points: (0,1) and (1,0).
Area = Integral[0 to 1] [sqrt(1-x^2) - (1-x)] dx
= [(x/2)sqrt(1-x^2) + (1/2)sin⁻¹x] from 0 to 1 - [x - x^2/2] from 0 to 1
= [(0 + pi/4) - 0] - [(1 - 1/2) - 0]
= pi/4 - 1/2
= (pi - 2)/4 sq. units
Line x+y=1 intersects circle at: x^2+(1-x)^2=1 → 2x^2-2x=0 → x=0 or x=1.
Points: (0,1) and (1,0).
Area = Integral[0 to 1] [sqrt(1-x^2) - (1-x)] dx
= [(x/2)sqrt(1-x^2) + (1/2)sin⁻¹x] from 0 to 1 - [x - x^2/2] from 0 to 1
= [(0 + pi/4) - 0] - [(1 - 1/2) - 0]
= pi/4 - 1/2
= (pi - 2)/4 sq. units
Q4. Using integration, find the area of the triangle with vertices A(2,0), B(4,5) and C(6,3).
Find equations of sides AB, BC, CA.
AB: slope = (5-0)/(4-2) = 5/2 → y = (5/2)(x-2) = 5x/2 - 5
BC: slope = (3-5)/(6-4) = -1 → y - 5 = -(x-4) → y = -x + 9
CA: slope = (0-3)/(2-6) = 3/4 → y - 3 = (3/4)(x-6) → y = 3x/4 - 9/2
Area = Integral[2 to 4](AB line)dx + Integral[4 to 6](BC line)dx - Integral[2 to 6](CA line)dx
= Integral[2 to 4](5x/2-5)dx + Integral[4 to 6](-x+9)dx - Integral[2 to 6](3x/4-9/2)dx
= [5x^2/4 - 5x] from 2 to 4 + [-x^2/2 + 9x] from 4 to 6 - [3x^2/8 - 9x/2] from 2 to 6
= (20-20)-(5-10) + (-18+54)-(-8+36) - (27/2-27)-(3/2-9)
= 5 + 8 - (27/2 - 27 - 3/2 + 9)
= 13 - (24/2 - 18) = 13 - (12-18) = 13 - (-6) = 6 sq. units
AB: slope = (5-0)/(4-2) = 5/2 → y = (5/2)(x-2) = 5x/2 - 5
BC: slope = (3-5)/(6-4) = -1 → y - 5 = -(x-4) → y = -x + 9
CA: slope = (0-3)/(2-6) = 3/4 → y - 3 = (3/4)(x-6) → y = 3x/4 - 9/2
Area = Integral[2 to 4](AB line)dx + Integral[4 to 6](BC line)dx - Integral[2 to 6](CA line)dx
= Integral[2 to 4](5x/2-5)dx + Integral[4 to 6](-x+9)dx - Integral[2 to 6](3x/4-9/2)dx
= [5x^2/4 - 5x] from 2 to 4 + [-x^2/2 + 9x] from 4 to 6 - [3x^2/8 - 9x/2] from 2 to 6
= (20-20)-(5-10) + (-18+54)-(-8+36) - (27/2-27)-(3/2-9)
= 5 + 8 - (27/2 - 27 - 3/2 + 9)
= 13 - (24/2 - 18) = 13 - (12-18) = 13 - (-6) = 6 sq. units
Q5. Find the area bounded by the ellipse x^2/4 + y^2/9 = 1.
a = 2, b = 3
Area = pi.a.b = pi.(2).(3) = 6pi sq. units
Verification using integration:
y = (3/2)sqrt(4-x^2)
Area = 4 × Integral[0 to 2] (3/2)sqrt(4-x^2) dx
= 6 × Integral[0 to 2] sqrt(4-x^2) dx
= 6 × [pi(2)^2/4] = 6 × pi = 6pi ✓
Area = pi.a.b = pi.(2).(3) = 6pi sq. units
Verification using integration:
y = (3/2)sqrt(4-x^2)
Area = 4 × Integral[0 to 2] (3/2)sqrt(4-x^2) dx
= 6 × Integral[0 to 2] sqrt(4-x^2) dx
= 6 × [pi(2)^2/4] = 6 × pi = 6pi ✓
Q6. Find the smaller area enclosed by the circle x^2 + y^2 = 4 and the line x + y = 2.
Circle radius = 2. Line: y = 2-x.
Intersection with circle: x=0,y=2 and x=2,y=0 (first quadrant).
Smaller area = Area under circle arc - Area under line (from x=0 to x=2)
= Integral[0 to 2] sqrt(4-x^2) dx - Integral[0 to 2] (2-x) dx
= [(x/2)sqrt(4-x^2) + 2sin⁻¹(x/2)] from 0 to 2 - [2x - x^2/2] from 0 to 2
= [0 + 2.(pi/2) - 0] - [4-2]
= pi - 2
= (pi - 2) sq. units
Intersection with circle: x=0,y=2 and x=2,y=0 (first quadrant).
Smaller area = Area under circle arc - Area under line (from x=0 to x=2)
= Integral[0 to 2] sqrt(4-x^2) dx - Integral[0 to 2] (2-x) dx
= [(x/2)sqrt(4-x^2) + 2sin⁻¹(x/2)] from 0 to 2 - [2x - x^2/2] from 0 to 2
= [0 + 2.(pi/2) - 0] - [4-2]
= pi - 2
= (pi - 2) sq. units
10. Standard Results Summary ⭐
| Region | Area Formula |
|---|---|
| Full circle x^2 + y^2 = r^2 | pi.r^2 |
| Semicircle (upper half) | pi.r^2 / 2 |
| Quarter circle (first quadrant) | pi.r^2 / 4 |
| Full ellipse x^2/a^2 + y^2/b^2 = 1 | pi.a.b |
| Region between y^2=4ax and x^2=4ay | 16a^2/3 |
| Triangle with vertices (x1,y1), (x2,y2), (x3,y3) | Use integration of side equations |
| Area between y=f(x) and x-axis [a,b] | Integral[a,b] f(x) dx |
| Area between two curves f(x) and g(x) | Integral[a,b] [f(x)-g(x)] dx, f above g |
11. Key Formulas at a Glance
BASIC AREA FORMULAS:
Area under curve (above x-axis):
A = Integral[a to b] f(x) dx
Area to the left of curve (right of y-axis):
A = Integral[c to d] g(y) dy
Area between two curves:
A = Integral[a to b] [f(x) - g(x)] dx (f above g)
STANDARD INTEGRALS USED:
Integral[0 to a] sqrt(a^2-x^2) dx = pi.a^2 / 4
Integral sqrt(a^2-x^2) dx = (x/2)sqrt(a^2-x^2) + (a^2/2)sin⁻¹(x/a) + C
STANDARD AREAS:
Circle x^2+y^2=r^2 : pi.r^2
Ellipse x^2/a^2+y^2/b^2=1 : pi.a.b
Between y^2=4ax, x^2=4ay : 16a^2/3
STEPS FOR EVERY PROBLEM:
1. Sketch the curves (rough diagram always helps)
2. Find intersection points (limits of integration)
3. Identify upper and lower curves
4. Set up integral: Integral [upper - lower] dx
5. Evaluate and write answer with "sq. units"
12. MCQ Practice (1 Mark)
1. The area bounded by y = x^2 and y = x is:
(a) 1/6 sq. units (b) 1/3 sq. units (c) 1/2 sq. units (d) 1 sq. unit
Answer: (a) 1/6 sq. units
2. Area of region bounded by curve y = sin x between x = 0 and x = 2pi is:
(a) 0 (b) 2 sq. units (c) 4 sq. units (d) 1 sq. unit
Answer: (c) 4 sq. units — area from 0 to pi is 2, area from pi to 2pi is 2 (take modulus), total = 4
3. Area bounded by y^2 = 4x, x = 1 and x-axis (first quadrant) is:
(a) 2/3 sq. units (b) 4/3 sq. units (c) 8/3 sq. units (d) 1 sq. unit
Answer: (b) 4/3 sq. units — Integral[0,1] 2sqrt(x) dx = 2.[x^(3/2)/(3/2)] from 0 to 1 = 4/3
4. Area of the circle x^2 + y^2 = 16 is:
(a) 4pi (b) 8pi (c) 16pi (d) 32pi
Answer: (c) 16pi — r = 4, area = pi.r^2 = 16pi
5. The area between the curves y = sqrt(x) and y = x^2 is:
(a) 1/2 (b) 1/3 (c) 1/4 (d) 2/3
Answer: (b) 1/3 sq. units
13. Exam Tips
- Always draw a rough sketch first — even a simple diagram helps you identify upper/lower curves and correct limits.
- Find intersection points carefully — these are your limits of integration.
- For area between two curves: always subtract (lower from upper), never the other way.
- If a curve goes below the x-axis, compute each part separately and add their absolute values — do NOT integrate across the x-axis directly.
- For parabola y^2 = 4ax: use symmetry about x-axis — compute upper half and double it.
- Remember the shortcut: Integral[0 to a] sqrt(a^2-x^2) dx = pi.a^2/4 — this saves time for circle and ellipse problems.
- Write the answer with proper units: "sq. units" — missing units may cost marks.
- For triangle area by integration: find equations of all three sides first, then integrate each side equation over the correct x range.
✅ Summary: Chapter 8 is compact but guaranteed in board exams as a 5-mark question. The key skill is setting up the integral correctly — sketch the curves, find the intersection, identify which is above which, then integrate. Most problems involve circles, parabolas, or area between two standard curves. Master 3-4 problem types from this chapter and you will confidently score full marks on this question.
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