Area between curves

Two curves $y=f(x)$ and $y=g(x)$ typically enclose a region between two intersection points. Everything about the problem is decided at those two points, so finding them always comes first.

Step 1 — Intersect. Solve $f(x)=g(x)$ for the x-coordinates $x=a$ and $x=b$ (with $a<b$) where the curves cross. These become the limits of integration.

Step 2 — Order them. Between $a$ and $b$, one curve sits above the other. Test any point strictly inside $(a,b)$ — plug it into both functions and see which value is bigger. That one is "upper" on this stretch.

Step 3 — Integrate the gap. If $f(x)\ge g(x)$ on $[a,b]$, the enclosed area is

where: $f(x)$ is the upper curve, $g(x)$ is the lower curve, and $[a,b]$ is the interval between consecutive intersection points. The rule in words: area = integral of (top curve) minus integral of (bottom curve), taken only over the stretch where that ordering is fixed. If the curves swap places — cross a third time inside $[a,b]$ — a single integral is no longer valid; split at the crossing point $x=k$ and add $\int_a^k[f-g]\,dx+\int_k^b[g-f]\,dx$ instead, always subtracting lower from upper on each sub-piece. In CBSE problems the two curves are always drawn from a short, standard list — straight lines, circles $x^2+y^2=r^2$, parabolas in standard form, and ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ — precisely so that Step 1's intersection equation is always solvable by hand, usually reducing to a linear or quadratic equation once one curve's expression is substituted into the other. Recognising the shape of each curve from its equation, before attempting any algebra, is what makes Step 2's "which one is on top" judgment fast rather than guesswork. Sometimes it is easier to treat $x$ as a function of $y$ — for instance a circle written as $x=\sqrt{4-y^2}$ next to a vertical line $x=c$. The same logic runs sideways:

where: $f(y)$ is whichever curve lies further to the right and $g(y)$ further to the left, over the y-interval $[c,d]$ between their intersection points. Choosing $x$- or $y$-integration is purely a convenience call — pick whichever makes the algebra (usually a square root) simpler to handle.

Problem. Verify the same result by integrating with respect to $y$ instead, using the right-hand halves $x=\frac{y^2}{4}$ (from the parabola) and $x=y$ (from the line), between $y=0$ and $y=4$.

Step 1. For a fixed height $y$, the line's $x$-value is $y$ and the parabola's $x$-value is $\frac{y^2}{4}$. At $y=2$: line gives $x=2$, parabola gives $x=1$, so the line lies to the right of the parabola — line minus parabola is the correct order here.

Step 2. Set up $A=\displaystyle\int_0^4\left(y-\frac{y^2}{4}\right)dy$.

Step 3. Integrate: $\left[\frac{y^2}{2}-\frac{y^3}{12}\right]_0^4=\frac{16}{2}-\frac{64}{12}=8-\frac{16}{3}=\frac{24-16}{3}=\frac{8}{3}$. Both routes agree: $A=\dfrac{8}{3}$ square units, confirming the x-integration answer.

Competitive stretch

The same method scales to conics. For the circle $x^2+y^2=4$ cut by the line $x=1$, sketch first: the circle has radius $2$ centred at the origin, and the vertical line $x=1$ slices off a smaller segment on the right. Intersecting gives $y^2=4-1=3$, so $y=\pm\sqrt3$ — the chord runs from $(1,-\sqrt3)$ to $(1,\sqrt3)$. Doubling the upper half and integrating from $x=1$ to $x=2$, the smaller segment works out to $A=2\displaystyle\int_1^2\sqrt{4-x^2}\,dx=\frac{4\pi}{3}-\sqrt3$ square units, using the standard result $\int\sqrt{a^2-x^2}\,dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right)+C$ with $a=2$. The setup logic — intersect, order, integrate — is identical to the parabola-and-line case above; only the antiderivative changes because the boundary is curved rather than straight. This is exactly why the chapter insists on standard-form conics: their antiderivatives are all in the toolkit a Class 12 student already carries.

• Subtracting in the wrong order. Upper minus lower, always — swapping it silently flips the sign of the answer.
• Using one integral when the curves cross three times. If the "upper" curve changes partway through, split the integral at every crossing point instead of integrating straight across.
• Guessing the limits from a rough sketch. Always solve $f(x)=g(x)$ algebraically for exact intersection points; eyeballing a graph can shift the bounds and the answer.
• Picking the wrong segment in circle or ellipse problems. A chord cuts the conic into a "smaller" and a "larger" piece — reread the question to see which one is wanted before integrating.

Ask the tutor: Want to try a fresh pair of curves — say an ellipse and a line — and walk through the intersection-to-integral setup together?