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Area questions are some of the most predictable in JEE maths, and some of the easiest to get wrong. The integration is usually simple. The marks go on the step before: working out which region is meant, where it crosses the axis, and which curve is on top. That is why the most useful single habit for this chapter is to sketch before you integrate. This post shows why, with worked examples and a practice routine.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists "determining areas of the regions bounded by simple curves in standard forms" under Integral Calculus. The NCERT Class 12 chapter on application of integrals covers areas under simple curves, including circles and ellipses. Check the syllabus for your exam year.
Why does a definite integral not always give the area?
A definite integral adds up f(x) dx. Where the curve is below the x-axis, f(x) is negative, so those pieces subtract. The integral is a signed area. Area itself is never negative.
This idea trips up many students. In interviews with 100 students aged 16 to 22, A. Orton (1983) found that items on integration as the limit of a sum were difficult even for good students. Steven Jones (2013) found that seeing the integral as an addition over many small pieces was one of the more productive ways students thought about it. If you picture thin strips of height f(x), it is clear why strips below the axis come out negative.
Worked example 1: the NCERT trap
NCERT's miscellaneous exercise asks for the area bounded by y = x³, the x-axis and the lines x = -2 and x = 1, with options including -15/4, 15/4 and 17/4.
Sketch. y = x³ is below the axis for x < 0 and above for x > 0. It crosses at x = 0, inside the interval.
Split at the crossing. From -2 to 0: ∫ x³ dx = [x⁴/4] from -2 to 0 = 0 - 4 = -4. Area of this part: 4. From 0 to 1: ∫ x³ dx = 1/4.
Add the areas: 4 + 1/4 = 17/4.
If you integrate straight from -2 to 1 you get 1/4 - 4 = -15/4, which is an option, and wrong. Its absolute value, 15/4, is also an option, and also wrong. The sketch is what tells you to split.
Worked example 2: a full sine wave
Find the area between y = sin x and the x-axis from 0 to 2π.
∫ from 0 to 2π of sin x dx = 0, because the part above the axis and the part below cancel. The area is 2 × ∫ from 0 to π of sin x dx = 2 × [-cos x] from 0 to π = 2 × 2 = 4. NCERT includes this exact question in the same exercise.
How do you find the area between two curves?
Use (upper curve - lower curve), integrated between the intersection points. The sketch tells you which is upper.
Worked example 3: y = x and y = x².
Intersections: x = x² gives x = 0 and x = 1.
Which is on top? At x = 1/2, y = x gives 0.5 and y = x² gives 0.25, so the line is above the parabola between 0 and 1.
Integrate: ∫ from 0 to 1 of (x - x²) dx = [x²/2 - x³/3] from 0 to 1 = 1/2 - 1/3 = 1/6.
If the curves cross again inside the interval, the upper and lower curves swap there, and you must split.
When should you use horizontal strips?
Some regions are easier with strips parallel to the x-axis. NCERT's summary gives both forms: ∫ y dx for vertical strips and ∫ x dy for horizontal ones.
Worked example 4: the region bounded by y² = 4x and the line x = 1.
The parabola opens to the right and is symmetric about the x-axis. Using vertical strips, the upper half is y = 2√x, so the area is 2 × ∫ from 0 to 1 of 2√x dx = 2 × [4x^(3/2)/3] from 0 to 1 = 2 × 4/3 = 8/3. With horizontal strips, x runs from y²/4 to 1 and y from -2 to 2: ∫ from -2 to 2 of (1 - y²/4) dy = 4 - 16/12 = 8/3. Same answer, and a good check.
Symmetry helps for closed curves too. NCERT finds the area of the ellipse x²/a² + y²/b² = 1 as four times the first-quadrant area, which gives πab. For a circle of radius a, that becomes πa².
Common slips and how to avoid them
Integrating across an axis crossing without splitting.
Taking the absolute value of the whole integral instead of each part.
Subtracting lower minus upper, which gives a negative area.
Missing an intersection point because you solved the equation without sketching.
Using vertical strips where the region needs two different top curves, when horizontal strips need only one integral.
A practice routine
Sketch-only drill: for ten area questions, sketch, mark crossings and shade the region. Do not integrate yet.
Split drill: five single-curve areas over intervals that cross the axis.
Two-curve set: five regions between a line and a parabola, or two parabolas.
Strip choice: three regions done both ways, vertical and horizontal, to check the answer.
The properties in our post on definite integrals often shorten these calculations. To check a sketch, a graphing tool helps if you use it after drawing by hand; see our post on graphing tools for maths. If the idea of an integral as a sum of strips is still unclear, the Fermi Concept Library covers every concept in mathematics, each with its own lesson.
Today, find the area bounded by y = x³ - x and the x-axis from -1 to 1, sketching first; the answer is 1/2, not 0.
