Area under curves is a direct application of definite integration and is one of the most reliably tested topics in JEE Main, with 2-3 questions appearing annually. The fundamental principle is that the definite integral from a to b of f(x) dx computes the signed area between y = f(x) and the x-axis. When f(x) is positive, the integral gives the area directly; when negative, it gives the negative of the area. For geometric (unsigned) area, we integrate |f(x)|, splitting at the zeros of f.

For area between two curves y = f(x) and y = g(x), the formula is integral of (f - g) dx where f is the upper curve. The critical step is finding ALL intersection points to determine limits and which curve is on top in each subinterval. If curves cross within the interval, split the integral at each crossing point.

Horizontal strips (integrating with respect to y) are often simpler for curves expressed as x = g(y), such as parabolas opening sideways. The formula becomes integral of ($x_{right}$ - $x_{left}$) dy.

Symmetry is a powerful shortcut: even functions allow doubling the half-integral; odd functions integrate to zero (signed). Four-fold symmetry (about both axes) allows quadrupling the first-quadrant integral. The area of an ellipse $x^2$/$a^2$ + $y^2$/$b^2$ = 1 is piab.

Key standard results to memorize: parabola $y^2$ = 4ax cut by latus rectum gives 8$a^2$/3; between y = $x^2$ and y = x gives 1/6; between two parabolas $y^2$ = 4ax and $x^2$ = 4by gives 16ab/3; the area cut off by a parabola y = $ax^{2+bx+c}$ between its roots is |a|(beta-alpha)^3/6.

Parametric area uses A = |integral of y(t)*x'(t) dt|. Polar area uses A = $\frac{1}{2}$*integral of $r^2$ d(theta). These are less common but do appear in JEE.

The most common mistakes are: not taking absolute values for regions below the x-axis, missing intersection points, integrating with respect to the wrong variable, and forgetting to split at crossing points. Always sketch the region before computing.