Area Under Curves

The area under curves is one of the most direct and powerful applications of definite integration in JEE Main. The fundamental idea is that the definite integral of a function f(x) from a to b gives the signed area between the curve y = f(x), the x-axis, and the vertical lines x = a and x = b.

Basic Area Formula: If f(x) >= 0 on [a, b], then the area bounded by y = f(x), the x-axis, and the lines x = a, x = b is: A = integral from a to b of f(x) dx

When f(x) < 0 on some subinterval, the integral gives a negative contribution. To find the total geometric area, we must take the absolute value: A = integral from a to b of |f(x)| dx

Area Between Two Curves: If f(x) >= g(x) on [a, b], the area between y = f(x) and y = g(x) from x = a to x = b is: A = integral from a to b of [f(x) - g(x)] dx

Here, f(x) is the upper curve and g(x) is the lower curve. If the curves cross within [a, b], we must split the integral at the intersection points and take the absolute value of each piece.

Area Using Horizontal Strips: When curves are better expressed as x = f(y), integrate with respect to y: A = integral from c to d of [$f_{right}$(y) - $f_{left}$(y)] dy where $f_{right}$(y) is the rightmost curve and $f_{left}$(y) is the leftmost curve, with y ranging from c to d.

Standard Areas to Remember:

• Area of ellipse $\frac{x^{2}}{a^{2}}$ + $\frac{y^{2}}{b^{2}}$ = 1 is $\pi$ab
• Area under y = sin(x) from 0 to $\pi$ is 2
• Area bounded by $y^{2}$ = 4ax and x = a is ($8a^{2}$)/3
• Area bounded by y = $x^{2}$ and y = x is $\frac{1}{6}$
• Area bounded by $y^{2}$ = 4ax and y = mx is $8a^{2}$/($3m^{3}$)

Symmetry Shortcuts: If f(x) is even (symmetric about y-axis), integral from -a to a of f(x) dx = 2 * integral from 0 to a of f(x) dx. If f(x) is odd (antisymmetric about y-axis), integral from -a to a of f(x) dx = 0.

Shifting the Origin: Sometimes translating the coordinate system simplifies the integral. If the region is symmetric about a point (h, k), substitute u = x - h, v = y - k.

Area in Parametric Form: If x = x(t), y = y(t), then A = integral from t1 to t2 of y(t) * x'(t) dt (or the absolute value thereof).

Area in Polar Coordinates: A = ($\frac{1}{2}$) * integral from theta1 to theta2 of $r^{2}$ d($\theta$). This is occasionally tested but less common in JEE Main.

The key problem-solving concept is correctly identifying the bounded region by finding all intersection points, determining which curve is on top (or to the right), and splitting the integral at crossover points.