# Area Under Curves > Area under curves for JEE Mathematics; strip selection, modulus handling, parabola and ellipse standard areas with common sign traps. - Canonical URL: https://prepelephant.com/topics/jee/mathematics/area-under-curves - Exam / course: JEE · Subject: Mathematics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Area Under Curves", PrepElephant, https://prepelephant.com/topics/jee/mathematics/area-under-curves ## Direct answer The area bounded by y = f(x), the x-axis and the lines x = a, x = b is the integral of |f(x)| dx from a to b, split at the zeros of f; the area between two curves is the integral of (upper curve - lower curve) dx between their intersection abscissae. Choosing the strip direction — vertical dx or horizontal dy — before integrating is what keeps the algebra small. ## What you must remember - Curve with the x-axis: area = integral of |y| dx; find every intersection with the axis first and split the integral wherever f changes sign. - Between two curves: integrate (y upper - y lower) dx between consecutive intersection x-values; when unsure which curve is upper, test a sample point in the interval. - Horizontal strips: for curves better written as x = g(y), area = integral of (x right - x left) dy between suitable y-values — y^2 = 4ax regions usually fall to this in one step. - Symmetry: even functions and quadrant symmetry let you compute one piece and multiply; use the geometry before the algebra. - Standard areas: ellipse x^2/a^2 + y^2/b^2 = 1 encloses pi a b; circle of radius r encloses pi r^2; between the parabola y^2 = 4ax and its latus rectum x = a the area is (8/3) a^2. - Modulus curves: y = |f(x)| reflects the negative lobes above the axis; sketch the sign-changing pieces first, then integrate the positive parts. - Parametric curves: substitute x(t), y(t) with dx = x'(t) dt and convert the limits to t values before integrating. ## Common confusion Integrating y dx across a sign change without the modulus computes net signed area, not geometric area — the classic half-answer or, for sin x over [0, 2pi], an answer of zero for a region that plainly has area. The second recurring slip is unbounded enthusiasm: setting limits before finding all intersection points, which misses a lobe or an extra crossing. Sketch first, mark every intersection, then let the integrals follow the picture; area questions punish students who integrate before they look. ## Exam-focused takeaway JEE Main asks standard regions — line with parabola, curve with axis, ellipse sectors, modulus sketches — as numerical-value questions where the sketch is most of the solution. JEE Advanced prefers asymmetric regions, curves given implicitly or parametrically, and areas defined by inequalities where you must shade the feasible region piecewise before integrating. Strip selection is the real skill: if a vertical strip would cross different curve pairs in different subintervals, switch to dy or split the region before doing any integration. ## Frequently asked questions ### Why did my area computation return zero? You integrated a function whose positive and negative parts cancel — sin x over [0, 2pi] is the standard example; use |f(x)| and split at the zeros. ### When should I integrate using dy instead of dx? When the curve is naturally x in terms of y — such as y^2 = 4ax — or when vertical strips would need different upper curves on different subintervals. ### What is the area between the parabola y^2 = 4ax and its latus rectum? (8/3) a^2, computed with horizontal strips as the integral of (a - y^2/(4a)) dy from y = -2a to 2a, or by doubling the first-quadrant piece with vertical strips. ### How do I find the area between two curves? Locate every intersection, then integrate (upper - lower) dx across each interval between consecutive intersections and add the pieces. ### Is the area of an ellipse pi a b? Yes, where a and b are the semi-major and semi-minor axes; it reduces to pi r^2 when a = b.