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MathsMediumClass 12

Area Under Curves — Simple regions

Definite Integration & Area

17

JEE Qs

8%

Hard

75

min

Always sketch the region accurately and correctly identify points of intersection to set up the integral for the area precisely.

🧮 Key Formulas

Area = |∫[a to b] f(x) dx| (for area bounded by y=f(x), x-axis, x=a, x=b)
Area = |∫[c to d] g(y) dy| (for area bounded by x=g(y), y-axis, y=c, y=d)
Area = ∫[a to b] |f(x) - g(x)| dx (for area between y=f(x) and y=g(x) from x=a to x=b)
Area = ∫[c to d] |f(y) - g(y)| dy (for area between x=f(y) and x=g(y) from y=c to y=d)

✅ Key Points for JEE

  • 1Always draw an accurate sketch of the region bounded by the given curves/lines to correctly identify the upper/lower or right/left functions and the limits of integration.
  • 2Identify all points of intersection between the curves. These points often define the limits of integration or indicate where the 'upper' or 'lower' function changes.
  • 3Choose the variable of integration (dx or dy) judiciously. Integrate with respect to x (vertical strips) if y can be easily expressed as a function of x, and with respect to y (horizontal strips) if x can be easily expressed as a function of y.
  • 4Area is always a positive quantity. If the definite integral evaluates to a negative number, take its absolute value. This often happens if the region lies below the x-axis or to the left of the y-axis.
  • 5Utilize symmetry if the region is symmetric about an axis or the origin to simplify calculations (e.g., calculate area in one quadrant and multiply).

⚠️ Common Mistakes

  • ✕Failing to sketch the region properly, leading to incorrect identification of limits or the 'upper/lower' function.
  • ✕Incorrectly finding points of intersection between curves, which results in wrong integration limits.
  • ✕Assuming the area is simply ∫f(x)dx, without considering parts of the curve below the x-axis or switching 'upper' and 'lower' functions where they intersect, leading to cancellation and incorrect net area.
  • ✕Choosing the wrong variable of integration (dx vs dy) which makes the problem unnecessarily complex or impossible to solve easily.
  • ✕Calculating the definite integral but forgetting to take its absolute value when the region lies below the x-axis or to the left of the y-axis.

📝 Practice Questions

See all

Q1. Let f(x) = ∫t0 (1) 253 (2) 154 (3) 125 (4) 157 →

2025·NumericalHard

Q11.Let the area enclosed between the curves |y| = 1 −x2 and x2 + y2 = 1 be α. If 9α = βπ + γ; β, γ are integers, then the value of |β −γ| equals. (1) 27 (2) 33 (3) 15 (4) 18

2025·MCQMedium

Q21.If 24 ∫ 0 4 (sin 4x − 12π + [2 sin x])dx = 2π + α, where [⋅] denotes the greatest integer function, then α is equal to _______.

2025·NumericalMedium

Q6. Let for f(x) = 7 tan8 x + 7 tan6 x −3 tan4 x −3 tan2 x, I1 = ∫π/40 f(x)dx and I2 = ∫π/40 xf(x)dx. Then 7I1 + 12I2 is equal to : (1) 2 (2) 1 (3) 2π (4) π

2025·MCQMedium

Q13.The area of the region, inside the circle (x −2√3)2 + y2 = 12 and outside the parabola y2 = 2√3x is : (1) 3π + 8 (2) 6π −16 (3) 3π −8 (4) 6π −8

2025·MCQHard

Q7. The area of the region enclosed by the curves y = x2 −4x + 4 and y2 = 16 −8x is : (1) 8 (2) 4 3 3 (3) 8 (4) 5 x ∈R. Then the numbers of local maximum and local minimum points of f ,

2025·MCQMedium

NCERT Chapters

  • Class 12 Mathematics Ch 7: Integrals
  • Class 12 Mathematics Ch 8: Application of Integrals
  • Class 11 Mathematics Ch 10: Straight Lines
  • Class 11 Mathematics Ch 11: Conic Sections