Class 12 Mathematics - NCERT Mathematics Part II
NCERT Solutions for Class 12 Maths Chapter 8: Application of Integrals
Chapter 8 covers the application of definite integrals to find areas under curves and areas between curves. Students learn to compute the area bounded by a curve and the x-axis (or y-axis), the area enclosed between two curves, and the area of regions bounded by lines and standard curves like parabolas, circles, and ellipses. This chapter is part of the Calculus unit which carries 35 marks in the CBSE Class 12 board exam 2027. Application of Integrals typically contributes 3-5 marks.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 8 has 2 exercises plus a Miscellaneous Exercise. Exercise 8.1 (13 Qs) covers area under simple curves (parabolas, circles, ellipses). Exercise 8.2 (7 Qs) covers area between two curves. The Miscellaneous Exercise has 19 questions. Total ~39 questions.
Table of Contents
8.1Exercise 8.1 - Area Under Simple Curves
8 key questions shown below covering area under parabolas, circles, ellipses, and lines.
Q1Find the area of the region bounded by the curve y² = x and the lines x = 1 and x = 4 and the x-axis in the first quadrant.
In the first quadrant, y = √x (positive root since y ≥ 0).
Area = ∫(1 to 4) √x dx = ∫(1 to 4) x^(1/2) dx
= [x^(3/2)/(3/2)](from 1 to 4)
= (2/3)[x^(3/2)](from 1 to 4)
= (2/3)[4^(3/2) − 1^(3/2)]
= (2/3)[8 − 1]
= (2/3)(7)
= 14/3 sq units.
Q2Find the area of the region bounded by y² = 9x, x = 2, x = 4, and the x-axis in the first quadrant.
In the first quadrant, y = 3√x.
Area = ∫(2 to 4) 3√x dx = 3∫(2 to 4) x^(1/2) dx
= 3 × (2/3)[x^(3/2)](from 2 to 4)
= 2[4^(3/2) − 2^(3/2)]
= 2[8 − 2√2]
= 16 − 4√2 sq units.
Q3Find the area of the region bounded by x² = 4y, y = 2, y = 4, and the y-axis in the first quadrant.
From x² = 4y, in the first quadrant x = 2√y.
Area = ∫(2 to 4) 2√y dy = 2∫(2 to 4) y^(1/2) dy
= 2 × (2/3)[y^(3/2)](from 2 to 4)
= (4/3)[4^(3/2) − 2^(3/2)]
= (4/3)[8 − 2√2]
= (32 − 8√2)/3 sq units.
Q4Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1.
The ellipse has a = 4 and b = 3.
By symmetry, area = 4 × (area in first quadrant).
From the equation: y = 3√(1 − x²/16) = (3/4)√(16 − x²).
Area = 4∫(0 to 4) (3/4)√(16 − x²) dx
= 3∫(0 to 4) √(16 − x²) dx.
Using the formula ∫(0 to a) √(a² − x²) dx = πa²/4:
∫(0 to 4) √(16 − x²) dx = π(4)²/4 = 4π.
Area = 3 × 4π = 12π sq units.
(Alternatively, area of ellipse = πab = π(4)(3) = 12π.)
Q5Find the area of the region bounded by the ellipse x²/4 + y²/9 = 1.
Here a = 2 and b = 3.
Area of ellipse = πab = π(2)(3) = 6π sq units.
Verification by integration:
y = 3√(1 − x²/4) = (3/2)√(4 − x²).
Area = 4∫(0 to 2) (3/2)√(4 − x²) dx = 6∫(0 to 2) √(4 − x²) dx.
∫(0 to 2) √(4 − x²) dx = π(2)²/4 = π.
Area = 6π. ✓
Q6Find the area of the region in the first quadrant enclosed by the x-axis, the line y = x, and the circle x² + y² = 32.
The line y = x intersects the circle x² + y² = 32 when:
x² + x² = 32 → 2x² = 32 → x² = 16 → x = 4 (first quadrant).
The region is bounded by:
- y = x from x = 0 to x = 4
- y = √(32 − x²) from x = 4 to x = 4√2
Area = ∫(0 to 4) x dx + ∫(4 to 4√2) √(32 − x²) dx
First integral:
∫(0 to 4) x dx = [x²/2](from 0 to 4) = 16/2 = 8.
Second integral:
∫(4 to 4√2) √(32 − x²) dx = [(x/2)√(32 − x²) + (32/2)sin⁻¹(x/(4√2))](from 4 to 4√2)
At x = 4√2: (4√2/2)√(32 − 32) + 16 sin⁻¹(1) = 0 + 16(π/2) = 8π.
At x = 4: (4/2)√(32 − 16) + 16 sin⁻¹(4/(4√2)) = 2(4) + 16 sin⁻¹(1/√2) = 8 + 16(π/4) = 8 + 4π.
Second integral = 8π − (8 + 4π) = 4π − 8.
Total area = 8 + (4π − 8) = 4π sq units.
Q9Find the area of the region bounded by the parabola y = x² and y = |x|.
By symmetry about the y-axis, area = 2 × (area in first quadrant).
In the first quadrant, y = |x| = x.
Intersection of y = x and y = x²: x² = x → x(x − 1) = 0 → x = 0 or x = 1.
In the first quadrant, x > x² for 0 < x < 1.
Area = 2∫(0 to 1) (x − x²) dx
= 2[x²/2 − x³/3](from 0 to 1)
= 2[1/2 − 1/3]
= 2[1/6]
= 1/3 sq units.
Q13Find the area bounded by the curves y = x and y = x² between x = 0 and x = 2.
Note: y = x and y = x² intersect at x = 0 and x = 1.
For 0 ≤ x ≤ 1: x ≥ x² (line is above parabola).
For 1 ≤ x ≤ 2: x² ≥ x (parabola is above line).
Area = ∫(0 to 1) (x − x²) dx + ∫(1 to 2) (x² − x) dx
First part:
∫(0 to 1) (x − x²) dx = [x²/2 − x³/3](from 0 to 1) = 1/2 − 1/3 = 1/6.
Second part:
∫(1 to 2) (x² − x) dx = [x³/3 − x²/2](from 1 to 2)
= (8/3 − 4/2) − (1/3 − 1/2)
= (8/3 − 2) − (1/3 − 1/2)
= (8/3 − 6/3) − (2/6 − 3/6)
= 2/3 − (−1/6)
= 2/3 + 1/6 = 5/6.
Total area = 1/6 + 5/6 = 1 sq unit.
8.2Exercise 8.2 - Area Between Two Curves
7 questions shown below covering area between standard curves.
Q1Find the area of the circle x² + y² = 4 not included in the parabola y² = 3x (i.e., the area of the circle minus the area common to both).
Circle: x² + y² = 4, radius = 2.
Parabola: y² = 3x.
Intersection: x² + 3x = 4 → x² + 3x − 4 = 0 → (x + 4)(x − 1) = 0 → x = 1 (taking positive value).
When x = 1: y² = 3, so y = ±√3.
Area of circle = π(2)² = 4π.
Common area (by symmetry, 2 × area in first quadrant):
= 2[∫(0 to 1) √(3x) dx + ∫(1 to 2) √(4 − x²) dx]
First integral:
∫(0 to 1) √3 · √x dx = √3 × (2/3)[x^(3/2)](from 0 to 1) = 2√3/3.
Second integral:
∫(1 to 2) √(4 − x²) dx = [(x/2)√(4 − x²) + 2 sin⁻¹(x/2)](from 1 to 2)
At x = 2: 0 + 2 sin⁻¹(1) = 2(π/2) = π.
At x = 1: (1/2)√3 + 2 sin⁻¹(1/2) = √3/2 + 2(π/6) = √3/2 + π/3.
Second integral = π − √3/2 − π/3 = 2π/3 − √3/2.
Common area = 2[2√3/3 + 2π/3 − √3/2]
= 2[4√3/6 − 3√3/6 + 2π/3]
= 2[√3/6 + 2π/3]
= √3/3 + 4π/3.
Required area = 4π − (√3/3 + 4π/3) = 4π − 4π/3 − √3/3 = 8π/3 − √3/3 = (8π − √3)/3 sq units.
Q2Find the area bounded by the curves (x − 1)² + y² = 1 and x² + y² = 1.
Circle 1: x² + y² = 1 (centre O(0,0), radius 1).
Circle 2: (x−1)² + y² = 1 (centre A(1,0), radius 1).
Intersection: x² + y² = (x−1)² + y²
x² = x² − 2x + 1 → 2x = 1 → x = 1/2.
When x = 1/2: y² = 1 − 1/4 = 3/4, y = ±√3/2.
By symmetry about the x-axis, area = 2 × (area above x-axis).
For the region common to both circles (above x-axis), split at x = 1/2:
- From x = 0 to x = 1/2: bounded above by Circle 2: y = √(1 − (x−1)²)
- From x = 1/2 to x = 1: bounded above by Circle 1: y = √(1 − x²)
Area = 2[∫(0 to 1/2) √(1 − (x−1)²) dx + ∫(1/2 to 1) √(1 − x²) dx].
By the symmetry of the configuration (both circles have radius 1), these two integrals are equal.
Using ∫√(a² − u²) du = (u/2)√(a² − u²) + (a²/2)sin⁻¹(u/a):
For ∫(1/2 to 1) √(1 − x²) dx:
= [(x/2)√(1 − x²) + (1/2)sin⁻¹(x)](from 1/2 to 1)
At x = 1: 0 + (1/2)(π/2) = π/4.
At x = 1/2: (1/4)(√3/2) + (1/2)(π/6) = √3/8 + π/12.
= π/4 − √3/8 − π/12 = 3π/12 − π/12 − √3/8 = 2π/12 − √3/8 = π/6 − √3/8.
By symmetry, ∫(0 to 1/2) √(1 − (x−1)²) dx = π/6 − √3/8.
Common area = 2[π/6 − √3/8 + π/6 − √3/8] = 2[π/3 − √3/4] = 2π/3 − √3/2 sq units.
Q3Find the area of the region bounded by the curves y = x² + 2, y = x, x = 0, and x = 3.
For 0 ≤ x ≤ 3: x² + 2 ≥ x (since x² − x + 2 = (x − 1/2)² + 7/4 > 0 always).
So the parabola is above the line.
Area = ∫(0 to 3) [(x² + 2) − x] dx
= ∫(0 to 3) (x² − x + 2) dx
= [x³/3 − x²/2 + 2x](from 0 to 3)
= 27/3 − 9/2 + 6
= 9 − 4.5 + 6
= 21/2 sq units.
Q4Using integration, find the area of the region bounded by the triangle whose vertices are (−1, 0), (1, 3), and (3, 2).
The three sides of the triangle:
Side 1: (−1, 0) to (1, 3):
Slope = (3−0)/(1−(−1)) = 3/2.
y − 0 = (3/2)(x + 1) → y = (3/2)(x + 1) = 3x/2 + 3/2.
Side 2: (1, 3) to (3, 2):
Slope = (2−3)/(3−1) = −1/2.
y − 3 = (−1/2)(x − 1) → y = −x/2 + 1/2 + 3 = −x/2 + 7/2.
Side 3: (−1, 0) to (3, 2):
Slope = (2−0)/(3−(−1)) = 2/4 = 1/2.
y − 0 = (1/2)(x + 1) → y = x/2 + 1/2.
Area = ∫(−1 to 1) [y_top − y_bottom] dx + ∫(1 to 3) [y_top − y_bottom] dx
From x = −1 to x = 1: top = Side 1, bottom = Side 3.
= ∫(−1 to 1) [(3x/2 + 3/2) − (x/2 + 1/2)] dx
= ∫(−1 to 1) (x + 1) dx
= [x²/2 + x](from −1 to 1)
= (1/2 + 1) − (1/2 − 1) = 3/2 − (−1/2) = 2.
From x = 1 to x = 3: top = Side 2, bottom = Side 3.
= ∫(1 to 3) [(−x/2 + 7/2) − (x/2 + 1/2)] dx
= ∫(1 to 3) (−x + 3) dx
= [−x²/2 + 3x](from 1 to 3)
= (−9/2 + 9) − (−1/2 + 3) = 9/2 − 5/2 = 2.
Total area = 2 + 2 = 4 sq units.
Q5Using integration, find the area of the region bounded by the triangle with vertices (1, 0), (2, 2), and (3, 1).
The three sides:
Side 1: (1, 0) to (2, 2):
Slope = (2−0)/(2−1) = 2. y = 2(x − 1) = 2x − 2.
Side 2: (2, 2) to (3, 1):
Slope = (1−2)/(3−2) = −1. y − 2 = −1(x − 2) → y = −x + 4.
Side 3: (1, 0) to (3, 1):
Slope = (1−0)/(3−1) = 1/2. y = (1/2)(x − 1) = x/2 − 1/2.
Area = ∫(1 to 2) [(2x − 2) − (x/2 − 1/2)] dx + ∫(2 to 3) [(−x + 4) − (x/2 − 1/2)] dx
First integral:
= ∫(1 to 2) (3x/2 − 3/2) dx = (3/2)∫(1 to 2) (x − 1) dx
= (3/2)[(x−1)²/2](from 1 to 2) = (3/2)(1/2) = 3/4.
Second integral:
= ∫(2 to 3) (−3x/2 + 9/2) dx = (3/2)∫(2 to 3) (3 − x) dx
= (3/2)[(3x − x²/2)](from 2 to 3) = (3/2)[(9 − 9/2) − (6 − 2)] = (3/2)[9/2 − 4] = (3/2)(1/2) = 3/4.
Total area = 3/4 + 3/4 = 3/2 sq units.
Q6Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is: (A) 2(π − 2) (B) π − 2 (C) 2π − 1 (D) 2(π + 2)
Circle: x² + y² = 4 (radius 2). Line: x + y = 2 → y = 2 − x.
Intersection: x² + (2−x)² = 4 → x² + 4 − 4x + x² = 4 → 2x² − 4x = 0 → 2x(x−2) = 0.
So x = 0 or x = 2. Points: (0, 2) and (2, 0).
Smaller area = ∫(0 to 2) [√(4 − x²) − (2 − x)] dx.
∫(0 to 2) √(4 − x²) dx = [(x/2)√(4−x²) + 2 sin⁻¹(x/2)](from 0 to 2)
= [0 + 2(π/2)] − [0 + 0] = π.
∫(0 to 2) (2 − x) dx = [2x − x²/2](from 0 to 2) = 4 − 2 = 2.
Smaller area = π − 2.
Answer: (B) π − 2.
Q7Area lying between the curves y² = 4x and y = 2x is: (A) 2/3 (B) 1/3 (C) 1/4 (D) 3/4
Intersection of y² = 4x and y = 2x:
(2x)² = 4x → 4x² = 4x → 4x(x − 1) = 0.
x = 0 or x = 1. Points: (0, 0) and (1, 2).
For 0 ≤ x ≤ 1:
y_parabola = 2√x (upper branch) and y_line = 2x.
Since 2√x ≥ 2x for 0 ≤ x ≤ 1 (because √x ≥ x for 0 ≤ x ≤ 1).
Area = ∫(0 to 1) (2√x − 2x) dx
= 2∫(0 to 1) (√x − x) dx
= 2[(2/3)x^(3/2) − x²/2](from 0 to 1)
= 2[2/3 − 1/2]
= 2[1/6]
= 1/3.
Answer: (B) 1/3.
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Using integration, find the area of the region bounded by the parabola y² = 4x and the circle x² + y² = 9.
Intersection: x² + 4x = 9 → x² + 4x − 9 = 0.
x = (−4 + √(16 + 36))/2 = (−4 + √52)/2 = (−4 + 2√13)/2 = −2 + √13.
Since √13 ≈ 3.606, x = −2 + 3.606 ≈ 1.606.
At this x-value: y² = 4(−2 + √13), y = ±2√(−2 + √13).
By symmetry about x-axis, area = 2 × (area in upper half).
Area = 2[∫(0 to −2+√13) 2√x dx + ∫(−2+√13 to 3) √(9 − x²) dx]
First integral: 2∫(0 to −2+√13) 2√x dx = 4[(2/3)x^(3/2)](from 0 to −2+√13)
= (8/3)(−2 + √13)^(3/2).
Second integral: 2∫(−2+√13 to 3) √(9 − x²) dx
= 2[(x/2)√(9−x²) + (9/2)sin⁻¹(x/3)](from −2+√13 to 3)
= 2[0 + (9/2)(π/2)] − 2[((−2+√13)/2)√(9−(−2+√13)²) + (9/2)sin⁻¹((−2+√13)/3)].
The exact evaluation requires careful computation. The final answer is:
Area = (4/3)(√13 − 2)^(3/2) + 9π/2 − 9 sin⁻¹((√13 − 2)/3) − (√13 − 2)√(9 − (√13 − 2)²) sq units.
For a cleaner CBSE version, this is often expressed as (9π/2 − 9sin⁻¹(1/3) − √8)/2 or evaluated numerically as approximately 11.5 sq units.
Find the area of the region {(x, y) : x² + y² ≤ 1 ≤ x + y}, using integration.
Region: inside the circle x² + y² = 1 AND above the line x + y = 1.
Intersection of x² + y² = 1 and x + y = 1:
x² + (1−x)² = 1 → x² + 1 − 2x + x² = 1 → 2x² − 2x = 0 → 2x(x−1) = 0.
x = 0 or x = 1. Points: (0, 1) and (1, 0).
The required area is above the line y = 1 − x and below the circle y = √(1 − x²) from x = 0 to x = 1.
Area = ∫(0 to 1) [√(1 − x²) − (1 − x)] dx
= ∫(0 to 1) √(1 − x²) dx − ∫(0 to 1) (1 − x) dx.
∫(0 to 1) √(1 − x²) dx = π(1)²/4 = π/4. (Quarter circle of radius 1.)
∫(0 to 1) (1 − x) dx = [x − x²/2](from 0 to 1) = 1 − 1/2 = 1/2.
Area = π/4 − 1/2 = (π − 2)/4 sq units.
Find the area of the region enclosed between the two circles x² + y² = 4 and (x − 2)² + y² = 4.
Circle 1: x² + y² = 4 (centre O(0,0), r = 2).
Circle 2: (x−2)² + y² = 4 (centre A(2,0), r = 2).
Intersection: x² + y² = (x−2)² + y² → x² = x² − 4x + 4 → 4x = 4 → x = 1.
At x = 1: y² = 4 − 1 = 3, y = ±√3.
By symmetry about x-axis, area = 2 × (upper half area).
The common region from x = −something, but by symmetry of the configuration about x = 1:
Area = 2[∫(0 to 1) √(4 − (x−2)²) dx + ∫(1 to 2) √(4 − x²) dx]
By the symmetry of both circles (same radius), these two integrals are equal.
∫(1 to 2) √(4 − x²) dx = [(x/2)√(4−x²) + 2 sin⁻¹(x/2)](from 1 to 2)
= [0 + 2(π/2)] − [(1/2)√3 + 2 sin⁻¹(1/2)]
= π − [√3/2 + 2(π/6)]
= π − √3/2 − π/3
= 2π/3 − √3/2.
By symmetry, each integral = 2π/3 − √3/2.
Total area = 2 × 2 × (2π/3 − √3/2) = 4(2π/3 − √3/2) = 8π/3 − 2√3 sq units.
Using integration, find the area of the region bounded by the line y − 1 = x, the x-axis, and the ordinates x = −2 and x = 3.
Line: y = x + 1. This crosses the x-axis when y = 0: x = −1.
For −2 ≤ x ≤ −1: y = x + 1 ≤ 0 (below x-axis).
For −1 ≤ x ≤ 3: y = x + 1 ≥ 0 (above x-axis).
Area = |∫(−2 to −1) (x+1) dx| + ∫(−1 to 3) (x+1) dx
∫(−2 to −1) (x+1) dx = [x²/2 + x](from −2 to −1) = (1/2 − 1) − (2 − 2) = −1/2.
∫(−1 to 3) (x+1) dx = [x²/2 + x](from −1 to 3) = (9/2 + 3) − (1/2 − 1) = 15/2 − (−1/2) = 8.
Area = 1/2 + 8 = 17/2 sq units.
Using integration, find the area bounded by the curve y = √(3x + 4), x-axis, and the lines x = 0 and x = 4.
Area = ∫(0 to 4) √(3x + 4) dx.
Let u = 3x + 4, du = 3 dx → dx = du/3.
When x = 0: u = 4. When x = 4: u = 16.
= (1/3)∫(4 to 16) √u du = (1/3) × (2/3)[u^(3/2)](from 4 to 16)
= (2/9)[16^(3/2) − 4^(3/2)]
= (2/9)[64 − 8]
= (2/9)(56)
= 112/9 sq units.
Find the area of the region bounded by the curve y² = 4x, the line y = 2x − 4, and the y-axis.
Parabola: y² = 4x → x = y²/4.
Line: y = 2x − 4 → x = (y + 4)/2.
Intersection: y²/4 = (y + 4)/2 → y² = 2(y + 4) → y² − 2y − 8 = 0 → (y − 4)(y + 2) = 0.
y = 4 or y = −2.
When y = 4: x = 16/4 = 4. When y = −2: x = 4/4 = 1.
Integrating with respect to y (easier since x is expressed as function of y):
For −2 ≤ y ≤ 4: x_line = (y+4)/2 ≥ x_parabola = y²/4.
Area = ∫(−2 to 4) [(y+4)/2 − y²/4] dy
= ∫(−2 to 4) [y/2 + 2 − y²/4] dy
= [y²/4 + 2y − y³/12](from −2 to 4)
At y = 4: 16/4 + 8 − 64/12 = 4 + 8 − 16/3 = 12 − 16/3 = 20/3.
At y = −2: 4/4 − 4 − (−8)/12 = 1 − 4 + 2/3 = −3 + 2/3 = −7/3.
Area = 20/3 − (−7/3) = 20/3 + 7/3 = 27/3 = 9 sq units.
Find the area of the region bounded by y² = 4ax and x² = 4ay, where a > 0.
Parabola 1: y² = 4ax → y = 2√(ax) (upper branch).
Parabola 2: x² = 4ay → y = x²/(4a).
Intersection: (2√(ax))² = x²/(4a) × 4a? No, set equal:
2√(ax) = x²/(4a) → squaring: 4ax = x⁴/(16a²) → 64a³x = x⁴ → x³ = 64a³ → x = 4a.
Alternatively: from y² = 4ax and x² = 4ay:
Substitute x = y²/(4a) into x² = 4ay: y⁴/(16a²) = 4ay → y³ = 64a³ → y = 4a.
So x = 4a. Intersection at (4a, 4a) and (0, 0).
For 0 ≤ x ≤ 4a: y₁ = 2√(ax) ≥ y₂ = x²/(4a).
Area = ∫(0 to 4a) [2√(ax) − x²/(4a)] dx
= 2√a × (2/3)[x^(3/2)](from 0 to 4a) − (1/(4a))[x³/3](from 0 to 4a)
= (4√a/3)(4a)^(3/2) − (4a)³/(12a)
= (4√a/3)(8a√a) − 64a³/(12a)
= 32a²/3 − 16a²/3
= 16a²/3 sq units.
Find the area bounded by y = x², the line y = x and x = 0 to x = 2.
y = x and y = x² intersect at x = 0 and x = 1.
For 0 ≤ x ≤ 1: x ≥ x². For 1 ≤ x ≤ 2: x² ≥ x.
Area = ∫(0 to 1)(x − x²) dx + ∫(1 to 2)(x² − x) dx
= [x²/2 − x³/3](from 0 to 1) + [x³/3 − x²/2](from 1 to 2)
= (1/2 − 1/3) + [(8/3 − 2) − (1/3 − 1/2)]
= 1/6 + [2/3 + 1/6]
= 1/6 + 5/6
= 1 sq unit.
Using integration, find the area of the triangle ABC with vertices A(2, 5), B(4, 7), and C(6, 2).
Equations of sides:
AB: slope = (7−5)/(4−2) = 1. y − 5 = 1(x−2) → y = x + 3.
BC: slope = (2−7)/(6−4) = −5/2. y − 7 = (−5/2)(x−4) → y = −5x/2 + 17.
AC: slope = (2−5)/(6−2) = −3/4. y − 5 = (−3/4)(x−2) → y = −3x/4 + 13/2.
Verify: AC at x=2: 13/2−3/2 = 5 ✓. At x=6: 13/2−18/4 = 13/2−9/2 = 2 ✓.
Area = ∫(2 to 4) [y_AB − y_AC] dx + ∫(4 to 6) [y_BC − y_AC] dx
First integral:
= ∫(2 to 4) [(x + 3) − (−3x/4 + 13/2)] dx
= ∫(2 to 4) [7x/4 − 7/2] dx
= (7/4)∫(2 to 4) (x − 2) dx
= (7/4)[(x−2)²/2](from 2 to 4)
= (7/4)(2) = 7/2.
Second integral:
= ∫(4 to 6) [(−5x/2 + 17) − (−3x/4 + 13/2)] dx
= ∫(4 to 6) [−7x/4 + 21/2] dx
= (−7/4)∫(4 to 6) (x − 6) dx
= (−7/4)[(x−6)²/2](from 4 to 6)
= (−7/4)(−2) = 7/2.
Total area = 7/2 + 7/2 = 7 sq units.
Find the area of the region bounded by the parabola y = x² and the line y = |x|.
By symmetry about y-axis, area = 2 × (area in first quadrant).
In first quadrant: y = x and y = x². Intersection: x² = x → x = 0, 1.
For 0 ≤ x ≤ 1: x ≥ x².
Area = 2∫(0 to 1) (x − x²) dx = 2[x²/2 − x³/3](from 0 to 1)
= 2(1/2 − 1/3) = 2(1/6) = 1/3 sq units.
Find the area of the region {(x, y) : 0 ≤ y ≤ x² + 1, 0 ≤ y ≤ x + 1, 0 ≤ x ≤ 2}.
The region satisfies BOTH y ≤ x² + 1 AND y ≤ x + 1, so the upper boundary is min(x² + 1, x + 1).
Intersection of y = x² + 1 and y = x + 1: x² + 1 = x + 1 → x² = x → x(x−1) = 0 → x = 0 or x = 1.
For 0 ≤ x ≤ 1: x² + 1 ≤ x + 1 (since x² ≤ x), so min = x² + 1.
For 1 ≤ x ≤ 2: x + 1 ≤ x² + 1 (since x ≤ x²), so min = x + 1.
Area = ∫(0 to 1) (x² + 1) dx + ∫(1 to 2) (x + 1) dx
= [x³/3 + x](from 0 to 1) + [x²/2 + x](from 1 to 2)
= (1/3 + 1) + [(2 + 2) − (1/2 + 1)]
= 4/3 + [4 − 3/2]
= 4/3 + 5/2
= 8/6 + 15/6
= 23/6 sq units.
Find the area of the region bounded by the parabola y² = 2x and the line x − y = 4.
Parabola: y² = 2x → x = y²/2.
Line: x = y + 4.
Intersection: y²/2 = y + 4 → y² = 2y + 8 → y² − 2y − 8 = 0 → (y − 4)(y + 2) = 0.
y = 4 or y = −2.
For −2 ≤ y ≤ 4: x_line = y + 4 ≥ x_parabola = y²/2.
Area = ∫(−2 to 4) [(y + 4) − y²/2] dy
= [y²/2 + 4y − y³/6](from −2 to 4)
At y = 4: 8 + 16 − 64/6 = 24 − 32/3 = 72/3 − 32/3 = 40/3.
At y = −2: 2 − 8 − (−8)/6 = −6 + 4/3 = −18/3 + 4/3 = −14/3.
Area = 40/3 − (−14/3) = 40/3 + 14/3 = 54/3 = 18 sq units.
Using integration, find the area of the triangle with vertices (−1, 2), (1, 5), and (3, 4).
Equations of sides:
Side AB (−1,2) to (1,5): slope = (5−2)/(1+1) = 3/2. y − 2 = (3/2)(x+1) → y = 3x/2 + 7/2.
Side BC (1,5) to (3,4): slope = (4−5)/(3−1) = −1/2. y − 5 = (−1/2)(x−1) → y = −x/2 + 11/2.
Side AC (−1,2) to (3,4): slope = (4−2)/(3+1) = 1/2. y − 2 = (1/2)(x+1) → y = x/2 + 5/2.
Area = ∫(−1 to 1) [y_AB − y_AC] dx + ∫(1 to 3) [y_BC − y_AC] dx
First:
= ∫(−1 to 1) [(3x/2 + 7/2) − (x/2 + 5/2)] dx = ∫(−1 to 1) (x + 1) dx
= [x²/2 + x](from −1 to 1) = (1/2 + 1) − (1/2 − 1) = 3/2 + 1/2 = 2.
Second:
= ∫(1 to 3) [(−x/2 + 11/2) − (x/2 + 5/2)] dx = ∫(1 to 3) (−x + 3) dx
= [−x²/2 + 3x](from 1 to 3) = (−9/2 + 9) − (−1/2 + 3) = 9/2 − 5/2 = 2.
Total area = 2 + 2 = 4 sq units.
Find the area enclosed by the parabola 4y = 3x² and the line 2y = 3x + 12.
Parabola: y = 3x²/4. Line: y = (3x + 12)/2.
Intersection: 3x²/4 = (3x + 12)/2 → 3x² = 2(3x + 12) → 3x² = 6x + 24 → 3x² − 6x − 24 = 0 → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0.
x = 4 or x = −2.
For −2 ≤ x ≤ 4: line is above parabola.
Area = ∫(−2 to 4) [(3x + 12)/2 − 3x²/4] dx
= ∫(−2 to 4) [3x/2 + 6 − 3x²/4] dx
= [3x²/4 + 6x − x³/4](from −2 to 4)
At x = 4: 12 + 24 − 16 = 20.
At x = −2: 3 − 12 + 2 = −7.
Area = 20 − (−7) = 27 sq units.
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The area enclosed by the circle x² + y² = a² is πa².
Reason (R): ∫(−a to a) √(a² − x²) dx = πa²/2.
Answer: (A)
Both are TRUE and the reason explains the assertion. The circle has upper half y = √(a²−x²) and lower half y = −√(a²−x²). Total area = 2∫(−a to a) √(a²−x²) dx = 2 × πa²/2 = πa². The reason gives the value of the integral for the upper semicircle, which when doubled gives the full area.
AR2
Assertion (A): The area bounded by y = x² and y = x from x = 0 to x = 1 is 1/6.
Reason (R): For 0 ≤ x ≤ 1, x ≥ x², so area = ∫(0 to 1) (x − x²) dx.
Answer: (A)
Both are TRUE and the reason correctly explains the assertion. ∫(0 to 1) (x − x²) dx = [x²/2 − x³/3](from 0 to 1) = 1/2 − 1/3 = 1/6.
AR3
Assertion (A): The area of the region {(x,y): x² + y² ≤ 1 ≤ x + y} is (π − 2)/4.
Reason (R): The area is ∫(0 to 1) [√(1−x²) − (1−x)] dx.
Answer: (A)
Both are TRUE and the reason explains the assertion. The region is inside the unit circle and above the line x + y = 1, from x = 0 to x = 1. ∫(0 to 1) √(1−x²) dx = π/4 (quarter circle). ∫(0 to 1)(1−x) dx = 1/2. Difference = π/4 − 1/2 = (π−2)/4.
AR4
Assertion (A): Area between y² = 4x and y = 2x is 1/3.
Reason (R): The curves intersect at (0,0) and (1,2), and 2√x ≥ 2x for 0 ≤ x ≤ 1.
Answer: (A)
Both are TRUE and the reason explains the assertion. ∫(0 to 1)(2√x − 2x) dx = 2[2x^(3/2)/3 − x²/2](from 0 to 1) = 2[2/3 − 1/2] = 2(1/6) = 1/3.
AR5
Assertion (A): The area of the ellipse x²/9 + y²/4 = 1 is 6π.
Reason (R): Area of ellipse x²/a² + y²/b² = 1 is πab.
Answer: (A)
Both are TRUE and the reason explains the assertion. Here a = 3 and b = 2, so area = π(3)(2) = 6π.
AR6
Assertion (A): ∫(0 to 2) |x − 1| dx = 1.
Reason (R): |x − 1| = 1 − x for x < 1 and x − 1 for x ≥ 1.
Answer: (A)
Both are TRUE and the reason explains the assertion. ∫(0 to 1)(1−x) dx + ∫(1 to 2)(x−1) dx = [x − x²/2](from 0 to 1) + [x²/2 − x](from 1 to 2) = (1−1/2) + [(2−2) − (1/2−1)] = 1/2 + 1/2 = 1.
Case Study Questions
A landscape architect is designing a garden in the shape of the region bounded by the parabola y = 8 − x² and the line y = 2x. The garden will have different types of plants in different sections. All measurements are in metres.
(i) Find the points where the parabola y = 8 − x² and the line y = 2x intersect. (1 mark)
Set 8 − x² = 2x → x² + 2x − 8 = 0 → (x + 4)(x − 2) = 0.
x = −4 or x = 2.
When x = −4: y = −8. When x = 2: y = 4.
Intersection points: (−4, −8) and (2, 4).
(ii) For the interval −4 ≤ x ≤ 2, which curve is above the other? (1 mark)
Compare: (8 − x²) − 2x = −x² − 2x + 8 = −(x² + 2x − 8) = −(x+4)(x−2).
For −4 < x < 2: (x+4) > 0 and (x−2) < 0, so −(x+4)(x−2) > 0.
Therefore y = 8 − x² is above y = 2x for −4 < x < 2.
(iii)(a) Find the total area of the garden (area between the parabola and the line). (2 marks)
Area = ∫(−4 to 2) [(8 − x²) − 2x] dx
= ∫(−4 to 2) (8 − x² − 2x) dx
= [8x − x³/3 − x²](from −4 to 2)
At x = 2: 16 − 8/3 − 4 = 12 − 8/3 = 28/3.
At x = −4: −32 + 64/3 − 16 = −48 + 64/3 = −80/3.
Area = 28/3 − (−80/3) = 28/3 + 80/3 = 108/3 = 36 sq m.
(iii)(b) If flowering plants are placed in the region from x = 0 to x = 2, find the area of the flowering section. (2 marks)
Flowering area = ∫(0 to 2) [(8 − x²) − 2x] dx
= ∫(0 to 2) (8 − x² − 2x) dx
= [8x − x³/3 − x²](from 0 to 2)
= 16 − 8/3 − 4 − 0
= 12 − 8/3
= 36/3 − 8/3
= 28/3 sq m (approximately 9.33 sq m).
A factory has a curved roof modelled by the equation y = 12 − x²/3 (in metres), where x ranges from −6 to 6. The factory floor is at y = 0. Solar panels are to be installed on the roof. The cross-section of the roof is the area between the curve and the x-axis.
(i) Find the points where the roof curve meets the floor (y = 0). (1 mark)
Set 12 − x²/3 = 0 → x² = 36 → x = ±6.
The roof meets the floor at (−6, 0) and (6, 0).
(ii) Find the cross-sectional area of the factory (area under the roof curve from x = −6 to x = 6). (2 marks)
By symmetry about the y-axis:
Area = 2∫(0 to 6) (12 − x²/3) dx
= 2[12x − x³/9](from 0 to 6)
= 2[72 − 216/9]
= 2[72 − 24]
= 2 × 48
= 96 sq m.
(iii)(a) If solar panels can only be installed on the portion where the roof height is at least 8 m (y ≥ 8), find the width of this portion. (2 marks)
12 − x²/3 ≥ 8 → x²/3 ≤ 4 → x² ≤ 12 → |x| ≤ 2√3.
The width = 2 × 2√3 = 4√3 m (approximately 6.93 m).
(iii)(b) Find the cross-sectional area of the region where y ≥ 8. (2 marks)
Area = ∫(−2√3 to 2√3) [(12 − x²/3) − 8] dx = ∫(−2√3 to 2√3) (4 − x²/3) dx.
By symmetry:
= 2∫(0 to 2√3) (4 − x²/3) dx
= 2[4x − x³/9](from 0 to 2√3)
= 2[8√3 − (2√3)³/9]
= 2[8√3 − 24√3/9]
= 2[8√3 − 8√3/3]
= 2[24√3/3 − 8√3/3]
= 2 × 16√3/3
= 32√3/3 sq m (approximately 18.48 sq m).
MCQ Practice
MCQ1. Area lying in the first quadrant and bounded by the circle x² + y² = 4 and the lines x = 0 and x = 2 is:
(A) π
(B) π/2
(C) π/3
(D) π/4
Answer: (A) - This is the area of a quarter circle of radius 2 = π(2)²/4 = **π**.
MCQ2. Area of the region bounded by the curve y = cos x between x = 0 and x = π is:
(A) 0
(B) 1
(C) 2
(D) π
Answer: (C) - cos x ≥ 0 on [0, π/2] and cos x ≤ 0 on [π/2, π]. Area = ∫(0 to π/2) cos x dx + |∫(π/2 to π) cos x dx| = [sin x](0 to π/2) + |[sin x](π/2 to π)| = 1 + |0 − 1| = 1 + 1 = **2**.
MCQ3. The area bounded by the curve y = x|x|, x-axis, and the ordinates x = −1 and x = 1 is:
(A) 0
(B) 1/3
(C) 2/3
(D) 4/3
Answer: (C) - y = x|x| means y = x² for x ≥ 0 and y = −x² for x < 0. Area = ∫(−1 to 0) |−x²| dx + ∫(0 to 1) x² dx = ∫(−1 to 0) x² dx + ∫(0 to 1) x² dx = 1/3 + 1/3 = **2/3**.
MCQ4. The area enclosed by the ellipse x²/a² + y²/b² = 1 is:
(A) πab
(B) π(a + b)
(C) πa²b²
(D) πab/2
Answer: (A) - Area of ellipse = **πab**. This is derived using integration: 4∫(0 to a) (b/a)√(a²−x²) dx = 4(b/a)(πa²/4) = πab.
MCQ5. Area bounded by y² = 4x and the line y = 2x is:
(A) 1/6
(B) 1/3
(C) 1/2
(D) 2/3
Answer: (B) - Intersection at (0,0) and (1,2). Area = ∫(0 to 1)(2√x − 2x) dx = 2[2/3 − 1/2] = 2(1/6) = **1/3**.
MCQ6. The area of the region bounded by y = |x − 1| and y = 1 is:
(A) 1/2
(B) 1
(C) 2
(D) 4
Answer: (B) - y = |x−1| meets y = 1 at |x−1| = 1, so x = 0 or x = 2. The region is a triangle with base 2 and height 1. Area = (1/2)(2)(1) = **1**. By integration: ∫(0 to 2) [1 − |x−1|] dx = ∫(0 to 1)(1−(1−x)) dx + ∫(1 to 2)(1−(x−1)) dx = ∫(0 to 1) x dx + ∫(1 to 2)(2−x) dx = 1/2 + 1/2 = 1.
MCQ7. Area bounded by the curve y = sin x, x-axis, and between x = 0 and x = 2π is:
(A) 0
(B) 2
(C) 4
(D) 2π
Answer: (C) - sin x ≥ 0 on [0,π] and sin x ≤ 0 on [π,2π]. Area = ∫(0 to π) sin x dx + |∫(π to 2π) sin x dx| = [−cos x](0 to π) + |[−cos x](π to 2π)| = 2 + 2 = **4**.
MCQ8. The area between the curve y = 4 + 3x − x² and x-axis is:
(A) 125/6
(B) 125/3
(C) 125/4
(D) 125/12
Answer: (A) - Roots: 4 + 3x − x² = 0 → x² − 3x − 4 = 0 → (x−4)(x+1) = 0, so x = −1 and x = 4. Area = ∫(−1 to 4)(4+3x−x²) dx = [4x + 3x²/2 − x³/3](from −1 to 4). At x=4: 16+24−64/3 = (120−64)/3 = 56/3. At x=−1: −4+3/2+1/3 = (−24+9+2)/6 = −13/6. Area = 56/3 − (−13/6) = 112/6 + 13/6 = **125/6**.
MCQ9. Smaller area enclosed by the circle x² + y² = 4 and the line x + y = 2 is:
(A) 2(π − 2)
(B) π − 2
(C) 2π − 1
(D) 2(π + 2)
Answer: (B) - Intersection at (0,2) and (2,0). Area = ∫(0 to 2)[√(4−x²) − (2−x)] dx = π − 2 = **π − 2**.
MCQ10. Area bounded by the curves y = x², y = x, x = 0 and x = 2 is:
(A) 1
(B) 1/6
(C) 5/6
(D) 11/6
Answer: (A) - Curves meet at x=1. For [0,1]: x ≥ x². For [1,2]: x² ≥ x. Area = ∫(0 to 1)(x−x²)dx + ∫(1 to 2)(x²−x)dx = 1/6 + 5/6 = **1**.
MCQ11. The area of the region {(x,y): x² ≤ y ≤ x} is:
(A) 1/2
(B) 1/3
(C) 1/6
(D) 1/12
Answer: (C) - x² ≤ y ≤ x means x² ≤ x, so 0 ≤ x ≤ 1. Area = ∫(0 to 1)(x − x²)dx = 1/2 − 1/3 = **1/6**.
MCQ12. The area enclosed between y = √x and y = x is:
(A) 1/2
(B) 1/3
(C) 1/6
(D) 1
Answer: (C) - Intersection at x=0 and x=1 (√x = x → x = x² → x(x−1) = 0). For 0 ≤ x ≤ 1: √x ≥ x. Area = ∫(0 to 1)(√x − x)dx = [2x^(3/2)/3 − x²/2](from 0 to 1) = 2/3 − 1/2 = **1/6**.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the area enclosed by the circle x² + y² = a².
Area = πa². By integration: 4∫(0 to a) √(a²−x²) dx = 4 × πa²/4 = **πa²**.
Find the area bounded by y = x², x-axis, x = 1 and x = 2.
∫(1 to 2) x² dx = [x³/3](from 1 to 2) = 8/3 − 1/3 = **7/3 sq units**.
What is the area of the ellipse x²/25 + y²/16 = 1?
a = 5, b = 4. Area = πab = π(5)(4) = **20π sq units**.
Find ∫(0 to 2) |x − 1| dx.
∫(0 to 1)(1−x) dx + ∫(1 to 2)(x−1) dx = 1/2 + 1/2 = **1**.
Area enclosed between y² = 4x and y = 2x is:
Intersection at (0,0) and (1,2). Area = ∫(0 to 1)(2√x − 2x) dx = **1/3 sq units**.
2-3 Mark Questions
Find the area bounded by y = √(3x+4), x-axis, x = 0 and x = 4.
Let u = 3x+4. Area = (2/9)[16^(3/2) − 4^(3/2)] = (2/9)(56) = 112/9 sq units.
Find the area of the region bounded by y = x² and y = |x|.
By symmetry: 2∫(0 to 1)(x − x²) dx = 2(1/6) = 1/3 sq units.
Find the area bounded by y = cos x, x-axis, x = 0, x = π.
∫(0 to π/2) cos x dx + |∫(π/2 to π) cos x dx| = 1 + 1 = 2 sq units.
Find the smaller area enclosed by circle x² + y² = 4 and line x + y = 2.
∫(0 to 2)[√(4−x²) − (2−x)] dx = π − 2 = (π − 2) sq units.
3-4 Mark Questions
Find the area bounded by y² = 2x and x − y = 4.
Integrating w.r.t. y from −2 to 4: ∫[(y+4) − y²/2] dy = 18 sq units.
Using integration, find the area of the triangle with vertices (−1, 2), (1, 5), (3, 4).
Find side equations, integrate in two parts: 4 sq units.
Find the area enclosed by the parabolas y² = 4ax and x² = 4ay.
Intersection at (0,0) and (4a,4a). Area = ∫(0 to 4a)[2√(ax) − x²/(4a)] dx = 16a²/3 sq units.
Key Concepts
Important Formulas
Study Plan - Application of Integrals
Area Under Simple Curves
Read Section 8.1. Understand the concept of area under a curve using definite integrals. Complete Exercise 8.1 (all 13 Qs). Focus on area under parabolas (y = x², y² = 4ax), circles (x² + y² = r²), and ellipses (x²/a² + y²/b² = 1). Practice using symmetry to simplify calculations. Master the formula ∫(0 to a)√(a²−x²) dx = πa²/4.
Area Between Two Curves
Read Section 8.2. Complete Exercise 8.2 (all 7 Qs). Focus on finding intersection points, determining which curve is on top, and integrating with respect to the correct variable. Practice area between parabola and line, two circles, and parabola and parabola. Always sketch the region first.
Triangle Area and Miscellaneous Problems
Complete the Miscellaneous Exercise (all 19 Qs). Focus on triangle area using integration (find side equations, split at vertices). Practice absolute value integrals (∫|f(x)| dx). Solve problems involving regions defined by set notation like {(x,y): x² + y² ≤ 1 ≤ x + y}.
PYQs and Practice
Solve all 14 PYQs from 2019-2026. Practice MCQs, assertion-reason, and case study questions. Focus on the most common patterns: parabola-line area (appears in 80% of papers), circle-line area, and triangle area. Time yourself: target 8 minutes for a 5-mark area problem.
Revision and Mixed Practice
Revise all formulas and key concepts. Practice 5 mixed problems combining different curve types. Create a quick reference sheet of standard areas: circle (πr²), ellipse (πab), area between y² = 4ax and line, area between two parabolas (16a²/3). Attempt one full-length practice paper.
Common Mistakes to Avoid
- ✗Not sketching the region before integrating. A rough sketch helps identify which curve is on top, the limits of integration, and whether the integral needs to be split at intersection points
- ✗Forgetting to take the absolute value when a curve goes below the x-axis. The integral ∫f(x) dx can be negative, but area is always positive. Split the integral where f(x) changes sign
- ✗Not splitting the integral when curves cross. If y = f(x) is above y = g(x) on [a,c] but below on [c,b], you must compute two separate integrals with the correct order of subtraction
- ✗Confusing integration with respect to x vs y. Sometimes integrating w.r.t. y is much easier (e.g., area between y² = 4x and a line). Choose the variable that gives simpler limits and integrands
- ✗Wrong limits of integration - not finding intersection points correctly. Always solve the curves simultaneously to find exact intersection points before setting up the integral
- ✗Forgetting symmetry: if the region is symmetric about the x-axis or y-axis, compute area in one part and multiply. This saves time and reduces errors in the board exam
Tips for Board Exam
- ✓Always draw a rough sketch of the region before integrating. Even a quick diagram helps you identify the correct setup and avoid errors. Examiners award marks for correct diagrams
- ✓For area between a parabola and a line (most common 5-mark question), find intersection points first, determine which curve is on top, then integrate. Practice at least 5 different parabola-line combinations
- ✓When the boundary involves a circle or ellipse, use the formula ∫(0 to a)√(a²−x²) dx = πa²/4 directly instead of computing the integral from scratch. This saves significant time
- ✓For triangle area problems using integration, find equations of all three sides using two-point form, then carefully split the integral at the appropriate x-values corresponding to vertices
- ✓Consider integrating with respect to y when dealing with horizontal parabolas (y² = 4ax). Express both curves as x = f(y), find y-limits from intersection, and integrate x_right − x_left
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Start Application of Integrals - Choose Your ModeAbout Application of Integrals - Class 12 Mathematics
Chapter 8 Application of Integrals covers the use of definite integrals to compute areas of plane regions. The chapter begins with the fundamental concept that the area under a curve y = f(x) between x = a and x = b is given by ∫(a to b) f(x) dx when f(x) ≥ 0, and |∫(a to b) f(x) dx| when f(x) ≤ 0.
Students learn to find areas of regions bounded by standard curves: parabolas (y = x², y² = 4ax), circles (x² + y² = r²), and ellipses (x²/a² + y²/b² = 1). The area of an ellipse πab and the quarter-circle integral ∫(0 to a)√(a²−x²) dx = πa²/4 are key results.
The chapter then covers area between two curves: A = ∫(a to b)[f(x) − g(x)] dx where f(x) ≥ g(x). Students learn to handle cases where curves cross by splitting the integral at intersection points. Integrating with respect to y is introduced as an alternative when curves are expressed more naturally as x = g(y).
For the CBSE 2027 board exam, this chapter typically contributes one 5-mark question (area between parabola and line, or area of triangle using integration) or one 3-mark question (area under a standard curve). Combined with Chapter 7 (Integrals), the total contribution to the 35-mark Calculus unit is about 15-18 marks. The most frequently tested problems involve area between y² = 4ax and a line, area enclosed by two curves, and area of triangles using integration.
Frequently Asked Questions
How many exercises are in NCERT Class 12 Maths Chapter 8 Application of Integrals?
There are 2 exercises plus a Miscellaneous Exercise, totalling approximately 39 questions. Exercise 8.1 has 13 questions on area under simple curves, Exercise 8.2 has 7 questions on area between two curves, and the Miscellaneous Exercise has 19 questions. This is a relatively short chapter compared to Chapter 7 (Integrals).
How many marks does Application of Integrals carry in the CBSE board exam?
Application of Integrals typically carries 3-5 marks in the board exam. Usually one 5-mark question on area between curves appears, or one 3-mark question. Combined with Chapter 7, the total integrals-related marks are 15-18 out of the 35-mark Calculus unit.
What is the most important topic in Chapter 8 for the board exam?
Area between a parabola and a line is the most frequently tested topic (appears in about 80% of board papers). The typical 5-mark question asks to find the area bounded by y² = 4ax and a line like y = mx or x − y = c. Practice at least 5 different parabola-line combinations. Triangle area using integration is the second most common topic.
Should I integrate with respect to x or y?
Choose the variable that gives simpler expressions and easier limits. For vertical parabolas (y = x²), integrate w.r.t. x. For horizontal parabolas (y² = 4ax), it is often easier to integrate w.r.t. y (express both curves as x = f(y)). For circles and ellipses, integrate w.r.t. x using the upper half equation y = √(r²−x²).
How do I find the area of a triangle using integration?
Step 1: Find equations of all three sides using the two-point form. Step 2: Identify the leftmost and rightmost x-values among vertices. Step 3: Split the integral at the middle vertex x-value. Step 4: In each interval, integrate (upper side − lower side). The two sub-areas add up to the total triangle area.
What is the formula for the area of an ellipse?
Area = πab, where a is the semi-major axis and b is the semi-minor axis. For x²/a² + y²/b² = 1: area = πab. This can be derived by integration: 4∫(0 to a) (b/a)√(a²−x²) dx = 4(b/a)(πa²/4) = πab. For the circle (special case a = b = r), area = πr².
Is NCERT enough for Application of Integrals in the CBSE board exam?
Yes, NCERT is more than sufficient. Since the chapter has only 39 questions, attempt all of them. The board questions closely follow NCERT examples and exercises. Pay special attention to Examples 5-10 in the textbook (area between curves) and the Miscellaneous Exercise. Supplement with PYQs for practice.
What are the standard area results I should memorize?
Key results to memorize: (1) Circle x² + y² = r²: area = πr². (2) Ellipse x²/a² + y²/b² = 1: area = πab. (3) ∫(0 to a)√(a²−x²) dx = πa²/4. (4) Area between y² = 4ax and x² = 4ay: 16a²/3. (5) Area between y² = 4ax and line y = mx: 8a²/(3m³). These save computation time in the exam.