JEE Corner — Area under Curves for JEE Main and Advanced

Where Areas Sit in JEE — and the One Formula NCERT Trimmed

JEE Main's integral calculus syllabus includes determining areas of the regions bounded by simple curves in standard form — expect one area question in most sessions. JEE Advanced pushes the same idea into multi-curve regions and parameter problems. Crucially, JEE freely asks the area between two curves, which the rationalized NCERT no longer covers — so it's the first add-on to learn here.

Area between two curves

If $f(x) \geq g(x)$ on $[a, b]$, the region between the two graphs has strip height $f(x) - g(x)$ (top minus bottom), so

$$A = \int_a^b \left[f(x) - g(x)\right]dx$$

The working routine:

1. Find the intersections: solve $f(x) = g(x)$ — these are the limits $a$ and $b$.
2. Decide who's on top on $(a, b)$ (test one point, or read the sketch).
3. Integrate the difference. If the curves cross inside the interval, split at the crossing and keep top-minus-bottom in each piece.
4. For regions leaning sideways, the mirror version is $A = \int_c^d\left[\text{right}(y) - \text{left}(y)\right]dy$.

Key Point: Because the strip runs curve-to-curve, the $x$-axis plays no role — no sign-splitting for axis crossings is needed. The only splits are where the two curves cross each other.

Results worth carrying into the exam

1. Line and parabola $y = x$, $y = x^2$: area $= \frac16$. More generally, the region between $y = x^2$ and $y = mx$ has area $\frac{m^3}{6}$.
2. Two parabolas $y^2 = 4ax$ and $x^2 = 4by$: area $= \frac{16ab}{3}$; the unit case $y^2 = x$, $x^2 = y$ gives $\frac13$.
3. Parabolic arch versus its chord: a parabola cuts off $\frac23$ of the bounding rectangle (Archimedes) — instant answers for latus-rectum and $y = c$ chord regions.

[JEE Tip] In the exam, compute intersections first and mark them on a ten-second sketch. Most wrong answers in area questions come from wrong limits or the wrong curve on top, not from wrong integration.

The Speed Toolkit

1. Symmetry first. Regions symmetric about an axis: compute one half and double. The lobes of $y = x^3$ versus $y = x$ are point-symmetric — twice the first-quadrant lobe.
2. Choose $dy$ strips deliberately. For regions bounded by $y^2 = 4ax$-type curves and horizontal lines, horizontal strips avoid square roots entirely: the region under $y^2 = 4x$ up to $y = 3$ against the $y$-axis is $\int_0^3\frac{y^2}{4}dy$ in one line.
3. Standard fractions of standard figures. Circle: $\pi r^2$ (full), $\frac{\pi r^2}{2}$, $\frac{\pi r^2}{4}$; ellipse: $\pi ab$ and its quarters; parabola-in-rectangle: $\frac23$ — assemble regions from these blocks before integrating anything.
4. Circle-segment template. The area cut from $x^2 + y^2 = r^2$ by the chord $x = c$ (right side) is $2\int_c^r\sqrt{r^2 - x^2}\,dx$ — root formula, two evaluations, done.
5. Inverse (parameter) problems. "Find $k$ so that the area is $S$": set up the integral with $k$ symbolic, integrate, solve the resulting equation. The area under $y = \sqrt x$ on $[0, k]$ is $\frac23 k^{3/2}$ — so area $18$ forces $k^{3/2} = 27$, i.e. $k = 9$.
6. Area splitting by a curve. "The line $y = mx$ divides the region into two equal parts" — write both sub-areas as integrals, set them equal, solve for $m$.

Traps that cost real marks

1. Wrong top curve. $f - g$ with the roles swapped flips the sign; if your area comes out negative, you subtracted in the wrong order — fix the order, don't just drop the sign blindly.
2. Curves crossing mid-interval. Between $\sin x$ and $\cos x$ on $[0, \pi]$, the top changes at $x = \frac{\pi}{4}$… and again the strip direction matters: split at every crossing.
3. Extraneous intersection roots. Squaring equations while solving $f = g$ can introduce fake intersections — check each root on the original curves.
4. Modulus boundaries. $y = |x|$-type boundaries mean the region's description changes branch at the corner — treat each branch as its own curve.

[JEE Tip] The +4/−1 economy rewards the memorised special results (items 1-3 above and $\frac16$, $\frac{16ab}{3}$): recognising the configuration is worth the full +4 in under a minute, and these five cover a large share of the area questions actually asked.

JEE-Pattern Worked Examples

Example 1: The classic sixth [JEE Main pattern]

Find the area of the region bounded by the parabola $y = x^2$ and the line $y = x$.

Solution:

1. Intersections: $x^2 = x$ gives $x = 0, 1$.
2. Top curve: on $(0, 1)$, $x > x^2$ — the line is on top.
3. Integrate: $A = \int_0^1\left(x - x^2\right)dx = \frac12 - \frac13 = \frac16$.

Final Answer: $\dfrac{1}{6}$ square units.

Example 2: Two parabolas [JEE Main pattern]

Find the area enclosed between the parabolas $y^2 = 4x$ and $x^2 = 4y$.

Solution:

1. Intersections: substituting $y = \frac{x^2}{4}$ into $y^2 = 4x$: $\frac{x^4}{16} = 4x$, so $x = 0$ or $x^3 = 64$, i.e. $x = 4$. Points: $(0,0)$ and $(4,4)$.
2. Top and bottom on $(0, 4)$: upper $y = 2\sqrt x$ (from $y^2 = 4x$), lower $y = \frac{x^2}{4}$.
3. Integrate: $A = \int_0^4\left(2\sqrt x - \frac{x^2}{4}\right)dx = \frac43\cdot8 - \frac{64}{12} = \frac{32}{3} - \frac{16}{3}$.

Final Answer: $\dfrac{16}{3}$ square units — matching $\frac{16ab}{3}$ with $a = b = 1$.

Example 3: Line through a parabola [JEE Main pattern]

Find the area bounded by $y = x^2$ and $y = 2x + 3$.

Solution:

1. Intersections: $x^2 = 2x + 3$ gives $x^2 - 2x - 3 = 0$, so $x = -1, 3$.
2. Top: the line is above the parabola between the roots.
3. Integrate: $A = \int_{-1}^{3}\left(2x + 3 - x^2\right)dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^{3} = 9 - \left(-\frac53\right)$.

Final Answer: $\dfrac{32}{3}$ square units.

Example 4: Trig curves crossing [JEE Main pattern]

Find the area bounded by $y = \cos x$, $y = \sin x$ and the $y$-axis, for $0 \leq x \leq \frac{\pi}{4}$.

Solution:

1. Who's on top: on $\left[0, \frac{\pi}{4}\right)$, $\cos x > \sin x$; they meet at $x = \frac{\pi}{4}$.
2. Integrate: $A = \int_0^{\pi/4}(\cos x - \sin x)dx = \left[\sin x + \cos x\right]_0^{\pi/4} = \sqrt2 - 1$.

Final Answer: $\sqrt 2 - 1$ square units.

Example 5: Parabola and an oblique line [JEE Main pattern]

Find the area of the region enclosed by $y = 2 - x^2$ and $y = -x$.

Solution:

1. Intersections: $2 - x^2 = -x$ gives $x^2 - x - 2 = 0$, so $x = -1, 2$.
2. Top: the parabola is above the line between the roots (test $x = 0$: $2 > 0$).
3. Integrate: $A = \int_{-1}^{2}\left(2 - x^2 + x\right)dx = \left[2x - \frac{x^3}{3} + \frac{x^2}{2}\right]_{-1}^{2} = \frac{10}{3} + \frac76$.

Final Answer: $\dfrac{9}{2}$ square units.

Example 6: An inverse problem [JEE Main pattern]

The area bounded by $y = \sqrt x$, the $x$-axis and $x = k$ is $18$ square units. Find $k$.

Solution:

1. Area as a function of $k$: $A(k) = \int_0^k\sqrt x\,dx = \frac23 k^{3/2}$.
2. Impose the condition: $\frac23 k^{3/2} = 18$, so $k^{3/2} = 27$.
3. Solve: $k = 27^{2/3} = 9$.

Final Answer: $k = 9$. (Check: $\frac23\cdot 9^{3/2} = \frac23\cdot27 = 18$. ✓)

Example 7: Cubic versus line — symmetric lobes [JEE Advanced pattern]

Find the total area enclosed between $y = x^3$ and $y = x$.

Solution:

1. Intersections: $x^3 = x$ gives $x = -1, 0, 1$ — two enclosed lobes.
2. Right lobe $(0,1)$: the line is on top: $\int_0^1(x - x^3)dx = \frac12 - \frac14 = \frac14$.
3. Left lobe $(-1,0)$: by the odd symmetry of both curves, congruent to the right lobe: another $\frac14$.
4. Add: $\frac14 + \frac14 = \frac12$.

Final Answer: $\dfrac{1}{2}$ square units.

Takeaway: Blindly integrating $x - x^3$ from $-1$ to $1$ gives $0$ — the lobes cancel in the signed integral exactly as axis-crossing regions do. Intersections inside the range are split points, always.