Application of Integrals

🟢 Lite — Quick Review

Rapid summary for last-minute revision before your exam.

Application of Integrals — Key Facts for JEE Main Area under curve y = f(x) from x = a to x = b: ∫_a^b f(x) dx (above x-axis only; below x-axis gives negative) Area between curves y = f(x) and y = g(x): ∫_a^b |f(x) − g(x)| dx where a, b are intersection points For region bounded by curves in polar coordinates: Area = ½∫ r² dθ For parametric curves: Area = ∫ y dx with x as parameter ⚡ Exam tip: Always find intersection points first — they give the limits of integration. Sketch the curves mentally!

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🟡 Standard — Core Study

Standard content for students with a few days to months.

Application of Integrals — JEE Main Study Guide

Area under curve (Cartesian):

• Region above x-axis: A = ∫_a^b f(x) dx
• Region below x-axis: A = −∫_a^b f(x) dx (take absolute)
• Total area between curve and x-axis: A = ∫_a^b |f(x)| dx

Area between two curves: When f(x) ≥ g(x) on [a, b]: A = ∫_a^b [f(x) − g(x)] dx When curves cross, split into intervals where one is above the other

Steps to find area:

1. Draw rough sketch of curves
2. Find intersection points (solve equations simultaneously)
3. Determine which curve is above in each interval
4. Set up integral with correct limits
5. Evaluate

Area bounded by curves:

• If curves are y = f(x) and y = g(x): use horizontal strips or vertical strips
• For curves given as x = f(y) and x = g(y): integrate with respect to y: A = ∫_c^d [f(y) − g(y)] dy

When to use vertical vs horizontal strips:

• Vertical strips: when it’s easier to express y as function of x
• Horizontal strips: when curves are better expressed as x = f(y) or when bounded region is tall and narrow

Area of region bounded by:

• Parametric curve x = f(t), y = g(t): A = ∫ y (dx/dt) dt with t-limits
• Polar curve r = f(θ): A = ½ ∫ r² dθ

Standard areas:

• Circle x² + y² = r²: total area = πr²; quarter area = πr²/4
• Ellipse x²/a² + y²/b² = 1: total area = πab
• Parabola y² = 4ax and x = a: area = 4a²/3

Area in polar coordinates: A = ½ ∫_{θ₁}^{θ₂} r² dθ for region bounded by r = f(θ) between angles θ₁ and θ₂

Area of sector: In polar coordinates, sector from θ = 0 to θ = α has area = (1/2)r²α (if r is constant) For variable r, use the integral formula

• Key formula: Area = ∫_a^b |f(x) − g(x)| dx; polar: A = ½∫ r² dθ
• Common trap: When finding area between curve and x-axis, if curve dips below axis, you must split the integral and add absolute values
• Exam weight: 1 question per year (4 marks); often combined with differentiation or differential equations

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🔴 Extended — Deep Dive

Comprehensive coverage for students on a longer study timeline.

Application of Integrals — Comprehensive JEE Main Notes

Area using horizontal strips: When region is bounded by curves x = f(y) and x = g(y): A = ∫{c}^{d} |f(y) − g(y)| dy Example: region bounded by y² = 4x and x = 4 → use y-limits from y = −4 to y = 4 x = y²/4 from parabola, x = 4 from line; A = ∫{-4}^{4} (4 − y²/4) dy

Volume of solid of revolution:

• About x-axis: V = π∫ y² dx
• About y-axis: V = π∫ x² dy
• Shell method (about y-axis using vertical strip): V = 2π∫ x·y dx

Volume using cylindrical shells:

• Strip parallel to axis of rotation: use washers/discs
• Strip perpendicular to axis: use shells For region rotated about y-axis using vertical strip: V = 2π∫ x·f(x) dx

Volume of solid with known cross-section: If area of cross-section perpendicular to x-axis is A(x), then V = ∫ A(x) dx

Area between polar curves: For two curves r₁(θ) and r₂(θ): A = ½∫ (r₁² − r₂²) dθ with appropriate limits Need to find where r₁ = r₂ (intersection angles)

Important polar curves and areas:

• Cardioid r = a(1 + cos θ): area = 3πa²/2
• Lemniscate r² = a² cos 2θ: area = a²
• Circle r = 2a cos θ: area = πa²

Area bounded by parametric curve: For x = f(t), y = g(t): A = ∫ y (dx/dt) dt with t from t₁ to t₂ If curve is traced more than once, account for that

Symmetry in area calculation:

• About x-axis: if region is symmetric, double the area for y ≥ 0
• About y-axis: if symmetric, double the area for x ≥ 0
• About origin: if traced symmetrically, may need to multiply appropriately

Area using double integration: A = ∫∫ dx dy over region For region bounded by curves: set up limits carefully

Wronskian and area: Not directly relevant, but used in differential equations context

Surface area of revolution: About x-axis: S = 2π∫ y √(1 + (dy/dx)²) dx About y-axis: S = 2π∫ x √(1 + (dx/dy)²) dy

Arc length:

• Cartesian: s = ∫ √(1 + (dy/dx)²) dx
• Parametric: s = ∫ √((dx/dt)² + (dy/dt)²) dt
• Polar: s = ∫ √(r² + (dr/dθ)²) dθ

Bounded region in polar: For curve r = f(θ), area = (1/2)∫ r² dθ from θ₁ to θ₂ If curve goes to origin, find angle where r = 0 for limits

Area common to two curves: For intersection of circle and parabola, etc.: Find intersection points, determine which curve is inside in each region

Volume of solid generated by rotating area between curves:

• About x-axis: washers method: π∫ (y₁² − y₂²) dx where y₁ > y₂
• About y-axis: washers: π∫ (x₁² − x₂²) dy

Integration by parts for area: Sometimes integration by parts needed to evaluate ∫ x·f(x) dx

Proving area formulas: Using integration, we can derive area of circle = πr² by integrating y = √(r² − x²) from −r to r

Area under rate curve: If dy/dx gives rate, area under rate curve gives total change Example: velocity-time graph, area under curve = displacement

Mean value via integration: Average value of f(x) on [a, b] = (1/(b−a))∫_a^b f(x) dx

Using substitution in area: When region is transformed, Jacobian may be needed for double integral

• Remember: For area between curves y₁(x) and y₂(x): A = ∫_a^b |y₁ − y₂| dx; find intersection points for limits; polar area = (1/2)∫ r² dθ
• Previous years: “Find area bounded by y = x² and y = x” [2023]; “Find area of region bounded by y = sin x from x=0 to x=2π and x-axis” [2024]; “Find area of circle r = 2a cos θ” [2024]

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📊 JEE Main Exam Essentials

Detail	Value
Questions	90 (30 per subject)
Time	3 hours
Marks	300 (90 per subject)
Section	Physics (30), Chemistry (30), Mathematics (30)
Negative	−1 for wrong answer
Mode	Computer-based

🎯 High-Yield Topics for JEE Main Mathematics

• Calculus (Differentiation + Integration) — ~35 marks combined
• Coordinate Geometry (straight lines, circles, conics) — ~20 marks
• Algebra (Complex Numbers, Quadratics, P&C, Probability) — ~25 marks
• Trigonometry + Inverse Trigonometry — ~15 marks
• Vector + 3D — ~15 marks

📝 Previous Year Question Patterns

• Application of Integrals: 1 question per year, 4 marks
• Common patterns: area between curve and x-axis, area between two curves, area in polar coordinates
• Weight: medium frequency, high scoring

💡 Pro Tips

• Always find intersection points first — they define the integration limits
• Sketch the curves mentally to understand which is above/below
• For area between y = f(x) and y = g(x), integrate |f(x) − g(x)| with respect to x
• If the curve goes below x-axis, the integral gives negative value — take absolute for area
• For polar curves, sketch the curve first to understand how r changes with θ
• When in doubt about which method to use (vertical/horizontal strips), try to set up both and pick whichever gives simpler integrals

🔗 Official Resources

• NTA Official JEE Main
• JEE Main Syllabus PDF

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