Area Under Curves (AOI)

Module 4 Objective

Area of bounded regions (AOI — Application of Integrals) is a guaranteed 4–8 marks in JEE Main every year. This module gives you every standard type, from simple curve-axis areas to intersecting curve regions, from absolute value problems to conic sections.

Chapter 1: Core Concept

Integration measures net signed area. For geometry (physical area), we need unsigned area:

Area between curve and x-axis:

\[A = \int_a^b |f(x)|\,dx\] Split at zeroes of \(f(x)\) within \([a,b]\) where the curve crosses the axis.

Area between curve and y-axis:

\[A = \int_c^d |g(y)|\,dy\] where \(x = g(y)\). Used when the bounding is horizontal.

Area between two curves:

\[A = \int_a^b [f_{\text{upper}}(x) - f_{\text{lower}}(x)]\,dx\] First find limits: set \(f(x)=g(x)\) to get intersection x-values.

Area using y as variable:

\[A = \int_c^d [g_{\text{right}}(y) - g_{\text{left}}(y)]\,dy\] Best for parabolas opening left/right likes \(x=y^2\).

Critical Warning Area is ALWAYS positive. The definite integral \(\displaystyle\int_a^b f(x)\,dx\) is negative when \(f(x)<0\) in the interval — but the area is its absolute value. Always sketch the curve mentally and split the integral at the x-intercepts.

Chapter 2: Standard Procedure (Step-by-Step)

The 4-Step Method

Sketch / Identify: Find what curves bound the region. Determine whether they're bounded above-below (use x-integration) or left-right (use y-integration).
Find Intersection Points: Set the two curves equal. These become your integration limits.
Determine Upper/Lower (or Right/Left): Verify which curve is on top by substituting a test point.
Integrate and Compute.

Type 1 — Curve vs x-axis

Q1. Area between \(y=x^2-4\) and the x-axis from \(x=-2\) to \(x=3\) Solution: \(y=x^2-4=0 \Rightarrow x=\pm2\). On \([-2,2]\): \(y\leq0\). On \([2,3]\): \(y\geq0\).
\(A = \left|\displaystyle\int_{-2}^2(x^2-4)\,dx\right| + \int_2^3(x^2-4)\,dx\)
\(\displaystyle\int_{-2}^2(x^2-4)\,dx = \left[\frac{x^3}{3}-4x\right]_{-2}^2 = \left(\frac{8}{3}-8\right)-\left(-\frac{8}{3}+8\right) = -\frac{32}{3}\). So \(\left|-\frac{32}{3}\right| = \frac{32}{3}\).
\(\displaystyle\int_2^3(x^2-4)\,dx = \left[\frac{x^3}{3}-4x\right]_2^3 = (9-12)-(\frac{8}{3}-8) = -3+\frac{16}{3} = \frac{7}{3}\)
Total \(A = \dfrac{32}{3}+\dfrac{7}{3} = \boxed{\dfrac{39}{3} = 13}\) sq. units.

Type 2 — Between Two Curves (Polynomial)

Q2. Area enclosed between \(y=x^2\) and \(y=x+2\) Solution: Intersections: \(x^2=x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0 \Rightarrow x=-1, 2\).
On \([-1,2]\): line \(y=x+2\) is above parabola \(y=x^2\).
\(A = \displaystyle\int_{-1}^2[(x+2)-x^2]\,dx = \left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]_{-1}^2 = \left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)\)
\(= \frac{10}{3}-\left(-\frac{7}{6}\right) = \frac{10}{3}+\frac{7}{6} = \frac{20+7}{6} = \boxed{\frac{27}{6} = \frac{9}{2}}\) sq. units.

Q3. Area bounded by parabolas \(y^2=4x\) and \(x^2=4y\) Solution: Intersection: sub \(y=\dfrac{x^2}{4}\) into \(y^2=4x\): \(\dfrac{x^4}{16}=4x \Rightarrow x^4=64x \Rightarrow x=0, 4\).
On \([0,4]\): \(y=2\sqrt{x}\) (from \(y^2=4x\)) is above \(y=\dfrac{x^2}{4}\) (from \(x^2=4y\)).
\(A = \displaystyle\int_0^4\left[2\sqrt{x}-\frac{x^2}{4}\right]dx = \left[\frac{4x^{3/2}}{3}-\frac{x^3}{12}\right]_0^4 = \frac{32}{3}-\frac{16}{3} = \boxed{\frac{16}{3}}\) sq. units.

Type 3 — Curve vs y-axis (Horizontal Integration)

Q4. Area between \(y^2=4x\) and \(x=3\) (left of the line) Solution — integrate w.r.t. y: \(y^2=4x \Rightarrow x=\dfrac{y^2}{4}\). Line is \(x=3\). Intersection: \(\dfrac{y^2}{4}=3 \Rightarrow y=\pm2\sqrt3\).
On \([-2\sqrt3,2\sqrt3]\): line \(x=3\) is to the right of parabola \(x=\dfrac{y^2}{4}\).
\(A = \displaystyle\int_{-2\sqrt3}^{2\sqrt3}\left[3-\frac{y^2}{4}\right]dy = 2\int_0^{2\sqrt3}\left(3-\frac{y^2}{4}\right)dy\) (by even symmetry)
\(= 2\left[3y-\dfrac{y^3}{12}\right]_0^{2\sqrt3} = 2\left[6\sqrt3-\dfrac{8\cdot3\sqrt3}{12}\right] = 2\left[6\sqrt3-2\sqrt3\right] = \boxed{8\sqrt3}\) sq. units.

Type 4 — Absolute Value / Modulus Functions

Q5. Area bounded by \(y=|x-1|+|x+1|\) and \(y=4\) Solution — Define piecewise: \(y = |x-1|+|x+1|\):
• \(x \leq -1\): \(y = (1-x)+(-1-x) = -2x\)
• \(-1 < x < 1\): \(y=(1-x)+(1+x)=2\)
• \(x \geq 1\): \(y = (x-1)+(x+1) = 2x\)
Intersection with \(y=4\): For \(x\leq-1\): \(-2x=4 \Rightarrow x=-2\). For \(x\geq1\): \(2x=4 \Rightarrow x=2\).
Region is between \(y=|x-1|+|x+1|\) (below) and \(y=4\) (above), from \(x=-2\) to \(x=2\).
\(A = \displaystyle\int_{-2}^{-1}[4-(-2x)]\,dx + \int_{-1}^1[4-2]\,dx + \int_1^2[4-2x]\,dx\)
\(= \displaystyle[4x+x^2]_{-2}^{-1} + [2x]_{-1}^1 + [4x-x^2]_1^2\)
\(= ((-4+1)-(-8+4)) + (2+2) + ((8-4)-(4-1)) = (-3+4)+4+(4-3) = 1+4+1 = \boxed{6}\) sq. units.

Type 5 — Conic Sections (JEE Advanced)

Q6. Area of ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) Solution — Use symmetry (4 × first quadrant area): \(y = \dfrac{b}{a}\sqrt{a^2-x^2}\)
\(A = 4\displaystyle\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{4b}{a}\cdot\frac{\pi a^2}{4} = \boxed{\pi ab}\) sq. units.
(The standard result \(\displaystyle\int_0^a\sqrt{a^2-x^2}\,dx = \frac{\pi a^2}{4}\) was used.)

Q7. Area of smaller region bounded by circle \(x^2+y^2=4\) and line \(x+y=2\) Solution: Circle radius 2. Line intersects it at \((2,0)\) and \((0,2)\).
\(A_{\text{segment}} = A_{\text{quarter circle}} - A_{\text{triangle}}\)
\(A_{\text{quarter circle}} = \dfrac{1}{4}\pi(4) = \pi\)
\(A_{\triangle} = \dfrac{1}{2}\cdot 2 \cdot 2 = 2\)
\(\boxed{A = \pi - 2}\) sq. units.

Q8. Area between parabola \(y=4x-x^2\) and the chord joining \((0,0)\) and \((3,3)\) Solution: The chord is the line through \((0,0)\) and \((3,3)\): \(y=x\).
Intersections of \(y=4x-x^2\) and \(y=x\): \(4x-x^2=x \Rightarrow x(3-x)=0 \Rightarrow x=0,3\).
Check: at \(x=1\), \(y_{\text{parabola}}=3\), \(y_{\text{line}}=1\) → parabola is on top.
\(A = \displaystyle\int_0^3[(4x-x^2)-x]\,dx = \int_0^3(3x-x^2)\,dx = \left[\frac{3x^2}{2}-\frac{x^3}{3}\right]_0^3 = \frac{27}{2}-9 = \boxed{\frac{9}{2}}\) sq. units.

Chapter 3: Standard Area Results — Quick Reference

Curve / Region	Area Formula	Derivation Trick
Circle \(x^2+y^2=a^2\)	\(\pi a^2\)	4 × first-quadrant integral of \(\sqrt{a^2-x^2}\)
Ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)	\(\pi ab\)	Scale circle formula
Parabola \(y^2=4ax\), from 0 to \(x=h\)	\(\frac{2}{3}\times h\times 2\sqrt{ah} = \frac{4}{3}\cdot h^{3/2}\sqrt{a}\)	2/3 of bounding rectangle
\(y=\sin x\) one arch, \([0,\pi]\)	2 sq. units	\(\int_0^\pi\sin x\,dx = 2\)
\(y=\|\sin x\|\) on \([0,n\pi]\)	\(2n\) sq. units	Periodicity
\(\|x\|+\|y\|=1\) (rhombus)	2 sq. units	4 triangles of area 1/2 each
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) and \(\frac{x^2}{c^2}+\frac{y^2}{d^2}=1\) (between ellipses)	\(\pi(ab-cd)\)	Subtract inner from outer

Chapter 4: Advanced Problems

Q9. Area of region \(\{(x,y): x^2\leq y\leq |x|\}\) Solution: \(y=x^2\) is below \(y=|x|\) when \(-1\leq x\leq 1\) (verify: at \(x=0.5\), \(x^2=0.25 < |x|=0.5\)).
Intersection: \(x^2=|x| \Rightarrow x=0,\pm1\).
By symmetry: \(A = 2\displaystyle\int_0^1(x-x^2)\,dx = 2\left[\frac{x^2}{2}-\frac{x^3}{3}\right]_0^1 = 2\times\frac{1}{6} = \boxed{\frac{1}{3}}\) sq. units.

Q10. Area between \(y=\sin x\) and \(y=\cos x\) from \(x=\pi/4\) to \(x=5\pi/4\) Solution: In \(\left[\frac{\pi}{4}, \frac{5\pi}{4}\right]\): \(\sin x \geq \cos x\) (verify at \(x=\pi/2\): \(1 > 0\) ✓).
\(A = \displaystyle\int_{\pi/4}^{5\pi/4}(\sin x-\cos x)\,dx = \Big[-\cos x-\sin x\Big]_{\pi/4}^{5\pi/4}\)
At \(5\pi/4\): \(-\cos\frac{5\pi}{4}-\sin\frac{5\pi}{4} = \frac{1}{\sqrt2}+\frac{1}{\sqrt2} = \sqrt2\).
At \(\pi/4\): \(-\cos\frac{\pi}{4}-\sin\frac{\pi}{4} = -\frac{1}{\sqrt2}-\frac{1}{\sqrt2} = -\sqrt2\).
\(A = \sqrt2-(-\sqrt2) = \boxed{2\sqrt2}\) sq. units.

Q11. Area between \(y=x^2-2x\) and \(y = x+4\) (includes sign analysis) Solution: \(x^2-2x = x+4 \Rightarrow x^2-3x-4=0 \Rightarrow (x-4)(x+1)=0 \Rightarrow x=-1, 4\).
On \([-1,4]\): at \(x=0\), \(y_{\text{line}}=4 > y_{\text{parabola}}=-0\). Line is on top.
\(A = \displaystyle\int_{-1}^4[(x+4)-(x^2-2x)]\,dx = \int_{-1}^4(-x^2+3x+4)\,dx\)
\(= \left[-\frac{x^3}{3}+\frac{3x^2}{2}+4x\right]_{-1}^4 = \left(-\frac{64}{3}+24+16\right)-\left(\frac{1}{3}+\frac{3}{2}-4\right)\)
\(= \frac{56}{3}-\left(-\frac{13}{6}\right) = \frac{112+13}{6} = \boxed{\frac{125}{6}}\) sq. units.

Q12. Find the area between \(y = e^x\), \(y = e^{-x}\), and \(x=1\) Solution: \(e^x=e^{-x} \Rightarrow e^{2x}=1 \Rightarrow x=0\). On \([0,1]\): \(e^x > e^{-x}\).
\(A = \displaystyle\int_0^1(e^x-e^{-x})\,dx = \Big[e^x+e^{-x}\Big]_0^1 = (e+e^{-1})-(1+1) = e+\frac{1}{e}-2\) sq. units.

Chapter 5: Error Traps for AOI

Error Trap 1 Forgetting to split at x-intercepts. If \(y=f(x)\) changes sign in \([a,b]\), you MUST split the integral. Otherwise positive and negative areas cancel, giving wrong (smaller) answers.

Error Trap 2 Wrong upper/lower identification. Always test a point inside the interval. Do not assume; verify which curve is on top.

Error Trap 3 Missing factor of 2 for symmetric regions. When a region is symmetric about the x-axis or y-axis, you can integrate half and double — just don't forget to actually double.

Error Trap 4 For curves like \(x=f(y)\), use y-integration. Integrating w.r.t. \(x\) for sideways parabolas forces you to split into two pieces, which is error-prone. Switch to y.

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