Area Under Curves for JEE Main: Complete Guide

Area under curves is the applied culmination of integration in JEE Main mathematics, contributing 1–2 questions per session. The chapter tests students' ability to set up definite integrals for areas, identify the correct limits of integration, handle area between two curves (where the "upper" and "lower" functions may switch), and apply standard area results for common geometric shapes like circles, parabolas, and ellipses. This guide provides complete coverage with worked examples from the easiest (single curve, clear limits) to the hardest (area between two conics with intersection point calculation).

Fundamental Area Formulas and Setup

Area under a curve y = f(x) between x = a and x = b (where f(x) ≥ 0 in [a,b]): A = integral(a to b) f(x) dx. If f(x) ≤ 0 in [a,b]: A = |integral(a to b) f(x) dx| = −integral(a to b) f(x) dx. If f(x) changes sign in [a,b]: split the interval at the zeros of f and sum the absolute values of each sub-interval integral. Area between x = g(y) and y-axis between y = c and y = d: A = integral(c to d) g(y) dy. Setting up the limits correctly is the most critical skill — sketch the curve, identify where it intersects the x-axis or y-axis, and determine the positive/negative regions. A quick sketch (even rough) before computing the integral saves 3–5 minutes of algebraic confusion in every area problem. For the integration techniques used to evaluate these area integrals, see our Integration Techniques Guide.

Standard area results for common curves: circle x² + y² = r² has area pi·r². The first-quadrant area is pi·r²/4 = integral(0 to r) sqrt(r²−x²) dx. Ellipse x²/a² + y²/b² = 1 has area pi·a·b (or 4×(first quadrant area) = 4 × integral(0 to a) (b/a)·sqrt(a²−x²) dx = pi·a·b). Parabola y² = 4ax and line x = h: area enclosed between the parabola and the line is (2/3)·base×height analogy — specifically, area from x=0 to x=h is (4/3)·h·sqrt(h·a/a) ... more precisely, area = integral(−sqrt(4ah) to sqrt(4ah)) (h − y²/(4a)) dy = (8/3)·sqrt(ah³)... JEE Main directly tests the standard result: area enclosed by y² = 4ax and x = h is (4h/3)·sqrt(4ah)/2 = (8/3)·h^(3/2)/sqrt(a) — derive once and use. These standard results significantly speed up JEE Main area problems.

Area Between Two Curves

Area between y = f(x) (upper curve) and y = g(x) (lower curve) between x = a and x = b: A = integral(a to b) [f(x) − g(x)] dx. The "upper" and "lower" roles may switch at intersection points — always find intersection points first and split the integral if needed. Finding intersection points: solve f(x) = g(x) to get x-coordinates of intersections. JEE Main standard problem: area between parabola y = x² and line y = x. Intersection: x² = x → x = 0 or x = 1. For 0 ≤ x ≤ 1: y = x is above y = x² (since x > x² in (0,1)). Area = integral(0 to 1) (x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6. This specific result (area between y=x and y=x² is 1/6) is a standard JEE Main answer. Practise area under curves problems on our JEE Main math mock tests — area problems with intersection points are among the most frequently tested definite integral applications.

Area between two parabolas: y = x² and y² = x (i.e., y = sqrt(x) for the upper branch). Intersection: x² = sqrt(x) → x^4 = x → x(x³−1) = 0 → x = 0 or x = 1. For 0 ≤ x ≤ 1: sqrt(x) > x² (since x^(1/2) > x^2 for x ∈ (0,1)). Area = integral(0 to 1) (sqrt(x) − x²) dx = [2x^(3/2)/3 − x³/3] from 0 to 1 = 2/3 − 1/3 = 1/3. This is another standard JEE Main result worth memorising. The general pattern: for y = x^(1/n) and y = x^n (in the region where they intersect at 0 and 1), area = 1/n − 1/(n+1) + ... the formula simplifies based on the specific exponents — derive for each case.

Area with Absolute Value Functions and Inequalities

Area problems involving |f(x)|: split the domain at zeros of f. For A = integral(−2 to 3) |x² − 4| dx: zeros at x = ±2. In [−2, 2]: x²−4 ≤ 0, so |x²−4| = 4−x². In [2, 3]: x²−4 ≥ 0, so |x²−4| = x²−4. Area = integral(−2 to 2) (4−x²) dx + integral(2 to 3) (x²−4) dx = [4x − x³/3] from −2 to 2 + [x³/3 − 4x] from 2 to 3. Evaluate: = (8 − 8/3) − (−8 + 8/3) + (9 − 12) − (8/3 − 8) = (16/3 + 16/3) + (−3 + 8 − 8/3) = 32/3 + 5 − 8/3 = 24/3 + 5 = 8 + 5 = 13. This type of |f(x)| area problem is a JEE Main favourite and requires careful case-by-case integration after zero identification.

Region defined by inequalities: some JEE Main problems ask for the area of the region satisfying multiple inequalities (e.g., y ≥ 0, y ≤ x², x ≤ 2). Sketch the region, identify its boundary curves and intersection points, then integrate. The key skill is translating the inequality constraints into a definite integral setup — the region boundary defines the limits and the integrand. For linear constraint regions, the area is often a polygon or a combination of triangle and curved region that can be computed by splitting into sub-regions.

Tricky Area Problems and Exam Strategy

Three tricky area problem types in JEE Main: (1) Area of region between a curve and its tangent/normal: find the tangent line, identify the enclosed region, and integrate. (2) Area swept by a moving line or curve (dynamic areas): parametric area formulas using Green's theorem — more JEE Advanced territory but occasionally appears in JEE Main numerical type. (3) Area of region defined by modulus inequalities: |y| ≤ 2, |x| ≤ 3 defines a rectangle; |y| ≤ |x| defines a region between two lines through origin; combining them requires careful sketch. For JEE Main, area problems are almost always solvable by standard definite integral setup — there are rarely tricks that require advanced theory. The difficulty lies in: (a) correctly identifying the upper and lower curves, (b) finding intersection points accurately, (c) computing the integral without arithmetic errors. Build these three skills specifically, and area problems become reliable 4-mark pickups. Create a free account on our platform to access 120+ area under curves problems sorted by difficulty. Our premium subscription includes definite integrals and area chapter tests with full solutions. For the definite integral properties that make area calculations faster (king's rule, periodicity), see our Definite Integrals Guide.

Exam time allocation for area problems: 3–4 minutes for problems with a single curve and clear limits; 4–5 minutes for two-curve area with intersection calculation; 6+ minutes for complex multi-region or absolute value area problems. Flag any area problem that takes more than 3 minutes in the first pass and return to it during the review period. Do not sacrifice 8 minutes on a complex area problem when simpler questions elsewhere in the paper remain unsolved.