📘 Calculus · Chapter MN16🎯 NDA Level : High Priority
Area under curves is the most direct geometric application of definite integration. Instead of just computing a number, the definite integral here answers the question: how much space is enclosed? For NDA, this chapter tests area bounded by a single curve and the axes, area between two curves, and a few standard shapes like circles, ellipses, and parabolas whose areas are derived directly.
📌 What to expect in NDA (based on 2022–2024 papers): (1) Area between a curve y = f(x) and the x-axis from x = a to x = b; (2) Area bounded by a curve and the y-axis using ∫|x| dy; (3) Area of a parabolic region (e.g., y = x² or y² = 4ax); (4) Area enclosed between two curves y = f(x) and y = g(x); (5) Finding the intersection point of two curves before setting up the integral; (6) Using symmetry to simplify — area of circle, ellipse by doubling first-quadrant area; (7) Area of regions where the curve goes below the x-axis (modulus needed); (8) Standard results: circle πr², ellipse πab, parabolic segment.
Topics at a Glance
① Area: Curve & x-Axis
∫ab |y| dx — horizontal strips
② Area: Curve & y-Axis
∫cd |x| dy — vertical strips
③ Area Between Two Curves
∫[f(x)−g(x)] dx
④ Standard Area Results
Circle, ellipse, parabola
⑤ Symmetry in Area
Doubling first-quadrant area
⑥ Mixed & NDA Problems
Intersection finding, split intervals
1. Area Between a Curve and the Axes
1.1
Area Under y = f(x) and the x-Axis
Add |y| — if curve dips below x-axis, split at the zeros and use modulus
The area bounded by the curve y = f(x), the x-axis, and the vertical lines x = a and x = b is computed using a definite integral. The absolute value ensures we always get a positive area even when the curve is below the x-axis.
⚡ Area Formulae — Curve and Axes
AREA BETWEEN y = f(x), x-AXIS, x = a and x = b:
Case 1: f(x) ≥ 0 throughout [a, b] (curve above x-axis):
A = ∫ab f(x) dx = ∫ab y dx
Case 2: f(x) ≤ 0 throughout [a, b] (curve below x-axis):
A = −∫ab f(x) dx = |∫ab f(x) dx|
Case 3: Mixed (curve crosses x-axis at x = c inside [a, b]):
A = |∫ac f(x) dx| + |∫cb f(x) dx|
(Split at every zero of f(x) and add magnitudes separately!)
AREA BETWEEN x = g(y), y-AXIS, y = c and y = d:
A = ∫cd |x| dy = ∫cd |g(y)| dy
(Use horizontal strips; integrate with respect to y)
The single most common error in NDA area problems is forgetting to split the integral when the curve crosses the x-axis. Never blindly integrate over the full interval — always sketch or check the sign of f(x) first.
Fig 1: Left — curve above x-axis: simple integral gives area. Right — curve crosses at c: split and sum absolute values of each part.
Worked Example — Area Above x-Axis
Find the area bounded by y = x², the x-axis, x = 0 and x = 3.
y = x² ≥ 0 on [0, 3]. No sign change → no splitting needed.
Area Using Horizontal Strips (Integrating w.r.t. y)
When the curve is naturally expressed as x = g(y), integrate along y
⚡ Area Between Curve and y-Axis
When x = g(y) is the curve and we want area between curve, y-axis,
y = c and y = d:
A = ∫cd |x| dy = ∫cd |g(y)| dy
This is the same concept as the x-integration but with x and y roles swapped.
Use this when:
(a) The curve is given as x = f(y) (e.g., x = y², x = √y)
(b) The region is bounded on left/right by the curve and by the y-axis
(c) Integration w.r.t. x would require awkward inverse functions
Example: Area between x = y² and y-axis, y = 0 to y = 2:
A = ∫02 y² dy = [y³/3]02 = 8/3 sq. units
The choice between dx and dy integration depends on which is simpler. If the curve is y=f(x), integrate dx. If the curve is x=g(y), integrate dy. Both give the same area, just different setups.
📝 TOPIC-WISE PYQ
Area Under Curve — NDA-Pattern Questions
Q1. The area bounded by y = x³, x-axis, x = 1 and x = 2 is:
Q2. Area bounded by y = sin x and x-axis from x = 0 to x = π is:
(a) 0 (b) 1 (c) 2 (d) π
Answer: (c) 2
sin x ≥ 0 on [0, π]. A = ∫0π sin x dx = [−cos x]0π = −cos π + cos 0 = 1+1 = 2.
Q3. Area bounded by y = sin x between x = 0 and x = 2π is:
(a) 0 (b) 2 (c) 4 (d) π
Answer: (c) 4
sin x ≥ 0 on [0, π] and ≤ 0 on [π, 2π]. Must split:
A = ∫0π sin x dx + |∫π2π sin x dx| = 2 + |−2| = 4.
Trap: ∫02π sin x dx = 0 (net area) ≠ total area = 4.
Q4. The area bounded by x = y², the y-axis and y = 3 (above y = 0) is:
(a) 9 (b) 3 (c) 18 (d) 6
Answer: (a) 9
A = ∫03 y² dy = [y³/3]03 = 27/3 = 9.
🔥 TRICKY QUESTIONS
Area Basics — Splitting & Sign Traps
🤯 T1. Find the area bounded by y = x² − 4 and the x-axis.
Zeros of x²−4 = 0: x = ±2. On (−2, 2): y = x²−4 < 0 (curve below x-axis).
A = |∫−22 (x²−4) dx|.
By symmetry (even function): = 2|∫02 (x²−4) dx|.
∫02 (x²−4) dx = [x³/3−4x]02 = 8/3−8 = −16/3.
A = 2|−16/3| = 32/3 sq. units. Trap: Computing ∫−22(x²−4)dx = −32/3 and reporting a negative area. Area is always positive.
🤯 T2. Find the area bounded by y = |x| and the x-axis from x = −2 to x = 3.
|x| is always ≥ 0. A = ∫−23 |x| dx.
Split at x=0 (where |x| changes definition): ∫−20 (−x) dx + ∫03 x dx.
= [−x²/2]−20 + [x²/2]03
= (0−(−2)) + (9/2−0) = 2 + 9/2 = 13/2 sq. units. For |x| type: always split at x=0. Left side: −x (since x<0 makes |x|=−x); Right side: x.
2. Standard Area Results
2.1
Areas of Circle, Ellipse & Parabola by Integration
These are derived once and memorised — used directly in NDA problems
● Circle
A = πr²
x²+y²=r². First quadrant: ∫0r √(r²−x²) dx = πr²/4. Full: ×4.
◯ Ellipse
A = πab
x²/a²+y²/b²=1. First quadrant: (b/a)∫0a √(a²−x²) dx = πab/4. Full: ×4.
▶ Parabola
A = (2/3)base × height
y²=4ax (standard parabola). Area from x=0 to x=t: (4/3)√(at³) = (8/3)a ⋅ √(at).
⚡ Standard Area Results — Derivation Summary
CIRCLE x² + y² = r²:
First quadrant area = ∫0r √(r²−x²) dx = [x√(r²−x²)/2 + r²/2 ⋅ sin⁻¹(x/r)]0r
= πr²/4. Full circle = 4 × πr²/4 = πr².
ELLIPSE x²/a² + y²/b² = 1:
y = (b/a)√(a²−x²). First quadrant = (b/a) ⋅ πa²/4 = πab/4.
Full ellipse = πab.
PARABOLA y² = 4ax (opens right):
Area from vertex to x = h: 2 ∫0h √(4ax) dx = 2 ⋅ 2√a ⋅ [x^(3/2)/(3/2)]0h
= (8√a/3) h^(3/2) = (8/3)√(ah³).
PARABOLA y = ax² (opens up):
Area under parabola to height h from x = −√(h/a) to √(h/a):
= (4/3) × base × height where base = 2√(h/a).
SEMI-CIRCLE (upper half):
Area = πr²/2.
For NDA: memorise the four results (πr², πab, and the two parabola forms). Verify a circle result by symmetry: 4 × (first quadrant area) and confirm the integral gives πr²/4.
Fig 2: Standard shapes and their areas. Circle (green), Ellipse (amber), Parabola y²=4ax (blue) with shaded area to x=h.
📝 TOPIC-WISE PYQ
Standard Shapes — NDA-Pattern Questions
Q5. The area enclosed by the parabola y² = 4x and the line x = 4 is:
Q6. Area of the region bounded by the ellipse x²/9 + y²/4 = 1 is:
(a) 6π (b) 12π (c) 9π (d) 4π
Answer: (a) 6π
a² = 9 → a = 3. b² = 4 → b = 2. Area = πab = π ⋅ 3 ⋅ 2 = 6π.
Q7. The area of the smaller portion enclosed by the circle x²+y²=1 and x+y=1 is:
(a) (π−2)/4 (b) π/2 (c) π−2 (d) π/4
Answer: (a) (π−2)/4
The line x+y=1 cuts the unit circle. Area of the smaller segment:
= Area of circular sector − Area of triangle.
= π(1)²/4 − (1/2)(1)(1) = π/4 − 1/2 = (π−2)/4.
3. Area Between Two Curves
3.1
Area Enclosed Between y = f(x) and y = g(x)
Subtract lower curve from upper curve — find intersection limits first
When two curves intersect at x = a and x = b, the region between them has area equal to the integral of the vertical gap between the curves.
⚡ Area Between Two Curves — Complete Method
If f(x) ≥ g(x) on [a, b] (f is above g):
A = ∫ab [f(x) − g(x)] dx = ∫ab [upper curve − lower curve] dx
Steps:
1. Find intersection points: solve f(x) = g(x) → gives limits a and b.
2. Determine which curve is on top in the interval.
3. Integrate [f(x) − g(x)] from a to b.
If curves interchange (g becomes upper in part of [a,b]):
A = ∫ac [f−g] dx + ∫cb [g−f] dx
(split at crossing point c and always subtract smaller from larger)
Also valid with horizontal strips (x = f(y), x = g(y)):
A = ∫cd [f(y) − g(y)] dy (f is right curve, g is left curve)
Always identify the intersection points first — they are the limits of integration. Then verify which curve is above the other by testing a point in the interval. The formula A = |∫[f−g]dx| also works if you’re unsure which is upper.
Fig 3: Area between curves y=f(x) (upper, green) and y=g(x) (lower, red). The shaded area = ∫ab [f(x)−g(x)] dx where a, b are the x-coordinates of the intersection points.
1
Find intersections: Solve f(x) = g(x). These give the limits a and b.
2
Identify upper and lower: Pick a test point between a and b. If f(test) > g(test), then f is upper.
3
Set up the integral: A = ∫ab [f(x) − g(x)] dx (or [g(x)−f(x)] if g is upper).
4
Evaluate: Find the antiderivative, substitute limits, compute F(b) − F(a).
Worked Example — Area Between Two Curves
Find the area enclosed between y = x² and y = x.
Intersections: x² = x → x(x−1) = 0 → x = 0 and x = 1.
Test x = 1/2: f(1/2) = 1/4, g(1/2) = 1/2. Since 1/2 > 1/4, y=x is upper.
A = ∫01 (x − x²) dx = [x²/2 − x³/3]01 = 1/2 − 1/3 = 1/6 sq. unit.
Worked Example — Parabola and Line
Find the area between y = x² and y = 2x + 3.
Intersections: x² = 2x+3 → x²−2x−3 = 0 → (x−3)(x+1) = 0 → x = −1 and x = 3.
Upper curve: y = 2x+3 (line) on (−1, 3). [Test x=0: line gives 3, parabola gives 0. Line > parabola.]
A = ∫−13 [(2x+3) − x²] dx = [x²+3x − x³/3]−13.
At x=3: 9+9−9 = 9. At x=−1: 1−3+1/3 = −5/3.
A = 9 − (−5/3) = 9 + 5/3 = 32/3 = 32/3 sq. units.
📝 TOPIC-WISE PYQ
Area Between Two Curves — NDA-Pattern Questions
Q8. Area bounded between y = x² and y = √x is:
(a) 1/3 (b) 1/6 (c) 2/3 (d) 1/2
Answer: (a) 1/3
Intersections: x² = √x → x⁴ = x → x(x³−1)=0 → x=0 or x=1.
On (0,1): √x > x². A = ∫01 (√x − x²) dx = [2x^(3/2)/3 − x³/3]01 = 2/3 − 1/3 = 1/3.
Q9. The area between the parabola y = x² and the straight line y = x + 2 is:
Answer: (c) (2π/3−√3/2)
By symmetry about x=1/2, the total enclosed area = 2 × (area of one circular segment).
Intersection: x=1/2. Sector angle = 2π/3 (120°). Area = 2(π/3 − √3/4) = 2π/3 − √3/2.
🔥 TRICKY QUESTIONS
Area Between Curves — Multiple Intersection & Switching Traps
🤯 T3. Find the area enclosed between y = sin x and y = cos x in [0, π/2].
Intersection in [0, π/2]: sin x = cos x → tan x = 1 → x = π/4.
On (0, π/4): cos x > sin x. On (π/4, π/2): sin x > cos x. Curves switch!
A = ∫0π/4 (cos x − sin x) dx + ∫π/4π/2 (sin x − cos x) dx.
First: [sin x + cos x]0π/4 = (1/√2+1/√2) − (0+1) = √2 − 1.
Second: [−cos x − sin x]π/4π/2 = (0−1) − (−1/√2−1/√2) = −1+√2 = √2−1.
Total = (√2−1) + (√2−1) = 2(√2 − 1) = 2√2 − 2. Key trap: Not splitting at the crossing point π/4. Without splitting, the two parts cancel partially and give an incorrect smaller value.
🤯 T4. Find the area of the region {(x,y): y² ≤ 4x, 4x² + 4y² ≤ 9}.
Region inside both: y² = 4x (parabola) and x²+y² = 9/4 (circle of radius 3/2).
Find intersection: substitute y²=4x into circle: x²+4x=9/4 → 4x²+16x−9=0 → x=(1/2) (taking positive root).
Split the area into two parts:
Part A: From x=0 to x=1/2 — bounded between the parabola arms.
A = 2∫01/2 2√x dx + 2∫1/23/2 √(9/4−x²) dx.
= (8/3)(1/2)^(3/2) + [area of circular sector + segment]
= (8/3)(√2/4) + π(9/4)⋅(angle)/(2π) = standard result.
Final area = 2√2/3 + π(9/4)⋅(1/3) − √2/4 ⋅ (correction term). This composite region problem requires careful identification of which curve bounds the region at each x-interval. Always sketch the region first.
🤯 T5. The area bounded by y = x|x| and y = x² is:
y = x|x| means: y = x² for x ≥ 0 and y = −x² for x < 0.
The other curve is y = x² always.
For x ≥ 0: both are x² → same curve, no area between them.
For x < 0: upper is y = x² (positive); lower is y = −x² (negative).
Intersection: only at x = 0 (for x < 0).
But we need boundaries — say from x = −a to x = 0:
A = ∫−a0 [x² − (−x²)] dx = ∫−a0 2x² dx = [2x³/3]−a0 = 0 − (−2a³/3) = 2a³/3. If a=1: Area = 2/3 sq. units. The key insight: analyse y=x|x| piecewise before computing any area involving it.
4. Using Symmetry to Simplify Area Problems
4.1
Symmetry About Axes & Doubling Technique
Recognise symmetry to halve or quarter the calculation
⚡ Symmetry Rules for Area
1. SYMMETRIC ABOUT x-AXIS (even in y):
If curve gives same shape above and below x-axis:
Total area = 2 × (area for y ≥ 0)
Example: y² = 4ax — symmetric about x-axis
A from x=0 to x=h: 2 ∫0h √(4ax) dx
2. SYMMETRIC ABOUT y-AXIS (even in x):
If f(−x) = f(x) on [−a, a]:
∫−aa f(x) dx = 2 ∫0a f(x) dx
Example: x²+y²=r² — full area = 4 × (first quadrant area)
3. ORIGIN SYMMETRIC (odd function):
If f(−x) = −f(x): ∫−aa f(x)dx = 0 (for area, use |f| or split)
4. FOUR-QUADRANT SYMMETRY:
Ellipse x²/a²+y²/b²=1: area = 4 ∫0a (b/a)√(a²−x²) dx = 4 ⋅ πab/4 = πab
Symmetry saves enormous time in NDA. Always check: is the curve symmetric about x-axis, y-axis, or both? If yes, compute only one portion and multiply.
Common Symmetric Curves in NDA
x²+y²=r²: symmetric about both axes and origin
y²=4ax: symmetric about x-axis only
x²/a²+y²/b²=1: symmetric about both axes
y=x²: symmetric about y-axis
y=|x|: symmetric about y-axis
Quick Area Results (NDA-ready)
Circle radius r: πr²
Ellipse semi-axes a, b: πab
Parabola y²=4ax, 0 to h: (8/3)√(ah³)
Semi-circle: πr²/2
Quarter-circle: πr²/4
📝 Master Formula Sheet — MN16 Area Under Curves
All critical formulae for rapid pre-exam revision.
▒ Area: Curve & x-Axis
A = ∫ab f(x) dx (if f ≥ 0)
A = |∫ab f(x) dx| (if f ≤ 0)
Mixed: split at zeros, sum |each part|
Never just integrate and take absolute value of total!
▓ Area: Curve & y-Axis
A = ∫cd |g(y)| dy
Use when x = g(y) form is natural
Horizontal strips integrate w.r.t. y
x = y², 0 to 2: ∫02y²dy = 8/3
▔ Area Between Two Curves
A = ∫ab[f(x)−g(x)]dx (f upper, g lower)
Find a, b by solving f(x) = g(x)
Test which is upper with a point
If curves cross inside: split and sum
⚬ Standard Areas
Circle r: πr²
Ellipse a,b: πab
Parabola y²=4ax to x=h: (8/3)√(ah³)
Semi-circle: πr²/2
Quarter-circle: πr²/4
◆ Symmetry
About x-axis: double y≥0 part
About y-axis: double x≥0 part
Both axes: quadruple first quadrant
sin x on [0,2π]: total area = 4, NOT 0
📌 Key Results
∫01 (√x − x²) dx = 1/3
∫−12[(x+2)−x²] dx = 9/2
y=x and y=x²: enclosed area = 1/6
Always sketch before integrating!
⚡ Quick Revision Booster — MN16 Area Under Curves
▒ Single Curve Steps
Find zeros of f(x) in [a,b]
Split at each zero
Integrate each piece separately
Take absolute value of each result
Sum all absolute values
▔ Two Curves Steps
Solve f(x)=g(x) → limits a, b
Test which is upper in (a,b)
∫[upper − lower] dx
If curves cross, split and sum
Always positive result
⚬ Standard Areas
Circle: πr²
Ellipse: πab
Parabola to x=h: (8/3)√(ah³)
Derive using first-quadrant + symmetry
◆ Symmetry Shortcuts
Parabola y²=4ax: use 2×upper half
Circle/Ellipse: 4×first quadrant
Even function on [−a,a]: 2×[0,a]
sin/cos over full period: handle signs
📈 Common Results
y=x vs y=x²: area = 1/6
sin x, [0,π]: area = 2
sin x, [0,2π]: area = 4 (NOT 0)
y=x and y=x+2 vs y=x²: area = 9/2
🚨 Critical Exam Traps
sin x on [0,2π]: integrate blindly gives 0; actual area = 4
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