Area as Definite Integral

Q: What does area as a definite integral mean geometrically?

Geometrically, integral_a^b f(x), dx is the signed area between the curve y = f(x) and the x-axis on [a, b]. When f(x) ge 0, the integral equals the actual area. When f(x) lt 0, the integral is the negative of the area. To always get a positive area, use integral_a^b lvert f(x) rvert, dx.

Q: When should I integrate with respect to y instead of x?

Integrate with respect to y when the region is more naturally described as x = g(y) with horizontal boundary lines y = c and y = d. Typical cases: regions bounded by the y-axis and horizontal lines, or curves like x = sin y where inverting y = sin^{-1} x back would introduce awkward domain restrictions.

Q: How do I handle the case where the curve crosses the x-axis inside [a, b]?

Find all the zeros of f(x) in [a, b], split the integral at those points, and take the absolute value of each piece. For example, for y = x^3 on [-1, 1]: split at x = 0, giving integral_{-1}^0 (-x^3), dx + integral_0^1 x^3, dx = {1}{4} + {1}{4} = {1}{2}.

Q: What is the standard formula for the area of an ellipse, and how is it derived?

For the ellipse {x^2}{a^2} + {y^2}{b^2} = 1, the enclosed area is pi a b. Derivation: upper half is y = {b}{a}sqrt{a^2 - x^2}; integrate from -a to a and double. The integral integral_{-a}^{a}sqrt{a^2 - x^2}, dx equals {pi a^2}{2} (area of semicircle), so total area is 2 cdot {b}{a} cdot {pi a^2}{2} = pi a b.

Q: How do I find the area between two curves when I do not know which is above?

Find all intersection points by solving the two equations simultaneously. On each sub-interval between successive intersections, test which curve is higher by plugging in any convenient value. Then integrate (text{upper} - text{lower}) on each sub-interval and add. Equivalently, use integral_a^b lvert f(x) - g(x) rvert, dx.

Q: Why do we sketch the curve before computing area?

Sketching tells you: (a) whether the curve crosses the x-axis in [a, b] so you know where to split; (b) which curve is on top in a two-curve problem; (c) whether to integrate with respect to x or y; (d) the correct limits when they are not given explicitly. Curve-tracing checks: symmetry, intercepts, stationary points, monotonicity, behaviour at infinity, and asymptotes.

Q: What is the area enclosed by a parabola and its latus rectum?

For y^2 = 4ax, the latus rectum is x = a. The enclosed area is {8a^2}{3}. Derivation: by symmetry, twice the upper region: 2 integral_0^a 2sqrt{ax}, dx = 4sqrt{a}cdot {2 a^{3/2}}{3} = {8a^2}{3}. This is one of the most-asked standard results in JEE Main.

Q: How is the area between two parabolas y^2 = 4ax and x^2 = 4ay computed?

They intersect at (0, 0) and (4a, 4a). Between these, the upper curve is y = 2sqrt{ax} and the lower is y = {x^2}{4a}. The enclosed area is integral_0^{4a} left(2sqrt{ax} - {x^2}{4a}right), dx = {32 a^2}{3} - {16 a^2}{3} = {16 a^2}{3}.

MathsApplication Of IntegralsFor JEE aspirants

Area as a definite integral is the geometric interpretation of $\int_{a}^{b} f (x) d x$ as the signed area between the curve $y = f (x)$ and the x-axis. For a non-negative continuous function, the definite integral equals the actual area; when the curve dips below the x-axis or when we work with two curves, we take absolute values or differences so the area stays positive. This concept powers every JEE Main and JEE Advanced question on Application of Integrals, from a simple parabola bounded by ordinates to rotated ellipses, hyperbolic sectors, and area-ratio functional equations.

Key Formulas - Quick Reference

If $f (x) \geq 0$ on $[a, b]$ : area $= \int_{a}^{b} f (x) d x$
If $f (x) < 0$ on $[a, b]$ : area $= - \int_{a}^{b} f (x) d x$
Most general (curve and x-axis): area $= \int_{a}^{b} ∣ f (x)∣ d x$
Curve and y-axis: area $= \int_{c}^{d} ∣ g (y)∣ d y$ , where $x = g (y)$
Between two curves ( $f \geq g$ ): area $= \int_{a}^{b} [f (x) - g (x)] d x$
General (two curves): area $= \int_{a}^{b} ∣ f (x) - g (x)∣ d x$
Circle $x^{2} + y^{2} = a^{2}$ : area $= π a^{2}$
Ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ : area $= π ab$

1. Area Between a Curve, the x-axis and Two Ordinates

The area of the region enclosed by the curve $y = f (x)$ , the x-axis, and the two vertical lines $x = a$ and $x = b$ depends on the sign of $f (x)$ on the interval.

1.1 When $f (x) \geq 0$ on $[a, b]$

f (x) \geq 0

for all

x \in [a, b]

, then the area bounded by the curve

y = f (x)

, the x-axis, and the ordinates

x = a

x = b

Area = \int_{a}^{b} f (x) d x .

Figure 1: Area between

y = f (x)

, the x-axis, and the ordinates

x = a

x = b

when

f (x) \geq 0

Solved Example 1

Find the area enclosed between the curve

y = x^{2} + 2

, the x-axis, and the lines

x = 1

and

x = 2

Solution:

The parabola $y = x^{2} + 2$ lies entirely above the x-axis (minimum value is $2$ at $x = 0$ ), so on $[1, 2]$ we have $f (x) > 0$ and the graph is not strictly needed.

Example Figure: Area enclosed by

y = x^{2} + 2

, the x-axis, and lines

x = 1

x = 2

Required area $= \int_{1}^{2} (x^{2} + 2) d x = [\frac{x ^{3}}{3} + 2 x]_{1}^{2} = (\frac{8}{3} + 4) - (\frac{1}{3} + 2) = \frac{13}{3} sq. units.$

Note: If a function is known to be positive-valued on the interval, sketching the graph is not necessary; you can integrate directly.

Solved Example 2

Find the area bounded by the curve

y = ln x + tan^{- 1} x

and the x-axis between the ordinates

x = 1

and

x = 2

Solution:

Domain of $y$ : $x > 0$ . Then $\frac{d y}{d x} = \frac{1}{x} + \frac{1}{1 + x ^{2}} > 0$ , so $y$ is strictly increasing. At $x = 1$ , $y = 0 + \frac{π}{4} > 0$ ; hence $y > 0$ throughout $[1, 2]$ .

Required area $= \int_{1}^{2} (ln x + tan^{- 1} x) d x = [x ln x - x + x tan^{- 1} x - \frac{1}{2} ln (1 + x^{2})]_{1}^{2} .$

Evaluating: $= (2 ln 2 - 2 + 2 tan^{- 1} 2 - \frac{1}{2} ln 5) - (0 - 1 + tan^{- 1} 1 - \frac{1}{2} ln 2)$ $= \frac{5}{2} ln 2 - \frac{1}{2} ln 5 + 2 tan^{- 1} 2 - \frac{π}{4} - 1 sq. units.$

Solved Example 3

The area cut off from a parabola by any double ordinate is

k

times the corresponding rectangle contained by the double ordinate and its distance from the vertex. Find the value of

k

Solution:

Consider the parabola $y^{2} = 4 a x$ ( $a > 0$ ) and the double ordinate $x = c$ .

Example Figure: Parabola

y^{2} = 4 a x

cut off by the double ordinate at

x = c

, meeting the curve at

(c, 2 a c)

and

(c, - 2 a c)

Area cut off by the double ordinate $= 2 \int_{0}^{c} 2 a x d x = 4 a \cdot \frac{2}{3} c^{3/2} = \frac{8}{3} a c^{3/2} .$

The corresponding rectangle has width $c$ (distance from vertex) and height $4 a c$ (double ordinate), so its area is $c \cdot 4 a c = 4 a c^{3/2}$ .

Setting area $= k \cdot$ (rectangle): $\frac{8}{3} a c^{3/2} = k \cdot 4 a c^{3/2} \Rightarrow k = \frac{2}{3} .$

1.2 When $f (x) < 0$ on $[a, b]$

f (x) < 0

for all

x \in [a, b]

, then the area bounded by

y = f (x)

, the x-axis, and the ordinates

x = a

x = b

Area = - \int_{a}^{b} f (x) d x .

Figure 2: Region between

y = f (x)

and the x-axis when

f (x) < 0

[a, b]

; area equals

- \int_{a}^{b} f (x) d x

Solved Example 4

Find the area bounded by

y = lo g_{1/2} x

and the x-axis between

x = 1

and

x = 2

Solution:

Since the base $\frac{1}{2} < 1$ , the curve $y = lo g_{1/2} x$ is decreasing and lies below the x-axis for $x > 1$ . So $f (x) < 0$ on $[1, 2]$ .

Example Figure: Curve

y = lo g_{1/2} x

between

x = 1

and

x = 2

; the region lies below the x-axis.

Using the change-of-base identity $lo g_{1/2} x = lo g_{e} x \cdot lo g_{1/2} e$ : $Area = - \int_{1}^{2} lo g_{1/2} x d x = - lo g_{1/2} e \cdot \int_{1}^{2} ln x d x .$

Now $\int_{1}^{2} ln x d x = [x ln x - x]_{1}^{2} = (2 ln 2 - 2) - (0 - 1) = 2 ln 2 - 1$ . Hence $Area = - lo g_{1/2} e \cdot (2 ln 2 - 1) sq. units.$

Since $lo g_{1/2} e = - \frac{1}{ln 2} < 0$ , the answer simplifies to a positive value $\frac{2 ln 2 - 1}{ln 2} = 2 - \frac{1}{ln 2}$ .

Note: If

y = f (x)

does not change sign on

[a, b]

, the area bounded by

y = f (x)

, the x-axis, and

x = a

x = b

\int_{a}^{b} f (x) d x

1.3 When $f (x)$ changes sign on $[a, b]$

f (x) \geq 0

[a, c]

and

f (x) \leq 0

[c, b]

(with

a < c < b

), then the area bounded by

y = f (x)

and the x-axis is

Area = \int_{a}^{c} f (x) d x - \int_{c}^{b} f (x) d x .

Figure 3: When

f (x)

changes sign at

x = c

, total area equals

\int_{a}^{c} f (x) d x - \int_{c}^{b} f (x) d x

Solved Example 5

Find the area bounded by

y = x^{3}

and the x-axis between the ordinates

x = - 1

and

x = 1

Solution:

On $[- 1, 0]$ , $x^{3} \leq 0$ ; on $[0, 1]$ , $x^{3} \geq 0$ . Split the integral at $x = 0$ :

Example Figure: Area enclosed by

y = x^{3}

and the x-axis between

x = - 1

and

x = 1

; the curve changes sign at the origin.

Required area $= \int_{- 1}^{0} - x^{3} d x + \int_{0}^{1} x^{3} d x = [- \frac{x ^{4}}{4}]_{- 1}^{0} + [\frac{x ^{4}}{4}]_{0}^{1} = 0 - (- \frac{1}{4}) + \frac{1}{4} - 0 = \frac{1}{2} sq. units.$

General formula: The area bounded by

y = f (x)

and the x-axis between ordinates

x = a

and

x = b

\int_{a}^{b} ∣ f (x)∣ d x

. This single formula handles all three sign cases automatically.

2. Area Between a Curve, the y-axis and Two Abscissas

When the region is described more naturally by $x$ in terms of $y$ , we integrate with respect to $y$ instead. The setup mirrors Section 1, with $x = g (y)$ playing the role of $y = f (x)$ .

2.1 When $g (y) \geq 0$ on $[c, d]$

g (y) \geq 0

for all

y \in [c, d]

, then the area bounded by

x = g (y)

, the y-axis, and the horizontal lines

y = c

y = d

Area = \int_{c}^{d} g (y) d y .

Figure 4: Area bounded by

x = g (y)

, the y-axis and the abscissas

y = c

y = d

\int_{c}^{d} g (y) d y

Solved Example 6

Find the area bounded between

y = sin^{- 1} x

and the y-axis between

y = 0

and

y = \frac{π}{2}

Solution:

Rewrite $y = sin^{- 1} x$ as $x = sin y$ . On $[0, \frac{π}{2}]$ , $sin y \geq 0$ .

Example Figure: Area between

y = sin^{- 1} x

and the y-axis from

y = 0

y = π /2

, computed as

\int_{0}^{π /2} sin y d y

Required area $= \int_{0}^{π /2} sin y d y = [- cos y]_{0}^{π /2} = - (0 - 1) = 1 sq. unit.$

Alternative: The same area equals (rectangle formed by

x = 0

y = 0

x = 1

y = π /2

)

-

(area under

y = sin^{- 1} x

from

x = 0

x = 1

), i.e.

\frac{π}{2} \cdot 1 - \int_{0}^{1} sin^{- 1} x d x = \frac{π}{2} - [x sin^{- 1} x + 1 - x^{2}]_{0}^{1} = \frac{π}{2} - (\frac{π}{2} - 1) = 1

Solved Example 7

Find the area bounded by the parabola

x^{2} = y

, the y-axis, and the line

y = 1

Solution:

From $x^{2} = y$ , the right half is $x = y$ ( $x \geq 0$ ).

Example Figure: Area bounded by the parabola

x^{2} = y

, the y-axis, and the line

y = 1

; integrated with respect to

y

\int_{0}^{1} y d y

Area (region $O A E O$ , right side) $= \int_{0}^{1} x d y = \int_{0}^{1} y d y = [\frac{2 y ^{3/2}}{3}]_{0}^{1} = \frac{2}{3} sq. units.$

Solved Example 8

Find the area bounded by the parabola

x^{2} = y

and the line

y = 1

Solution:

The region is symmetric about the y-axis, so it is twice the region in Example 7.

Example Figure: Area bounded by the parabola

x^{2} = y

and the line

y = 1

; the region is symmetric about the y-axis.

Required area $= 2 \times \frac{2}{3} = \frac{4}{3}$ sq. units.

JEE Advanced

Solved Example 9

For any real

t

, the point

P = (\frac{e ^{t} + e ^{- t}}{2}, \frac{e ^{t} - e ^{- t}}{2})

lies on the hyperbola

x^{2} - y^{2} = 1

. Show that the area bounded by the hyperbola and the lines joining its centre

O

to the points corresponding to

t_{1}

and

- t_{1}

equals

t_{1}

Solution:

The two points are $P (\frac{e ^{t_{1}} + e ^{- t_{1}}}{2}, \frac{e ^{t_{1}} - e ^{- t_{1}}}{2})$ and $R (\frac{e ^{t_{1}} + e ^{- t_{1}}}{2}, - \frac{e ^{t_{1}} - e ^{- t_{1}}}{2})$ (obtained by replacing $t_{1}$ with $- t_{1}$ ).

Example Figure: Area of the region bounded by hyperbola

x^{2} - y^{2} = 1

and the lines from the centre

O

to the points at parameter

t_{1}

and

- t_{1}

Let $Q$ be the foot of the perpendicular from $P$ (and $R$ ) to the x-axis. Area under the hyperbola between $x = 1$ and $x = \frac{e ^{t_{1}} + e ^{- t_{1}}}{2}$ (region $P QR P$ ) is $2 \int_{1}^{(e^{t_{1}} + e^{- t_{1}}) /2} x^{2} - 1 d x = 2 [\frac{x}{2} x^{2} - 1 - \frac{1}{2} ln (x + x^{2} - 1)]_{1}^{(e^{t_{1}} + e^{- t_{1}}) /2} .$

Substituting the upper limit and simplifying gives area $(P QR P) = \frac{e ^{2 t_{1}} - e ^{- 2 t_{1}}}{4} - t_{1}$ .

Area of $△ O P Q$ (base $\times$ height, doubled for the symmetric triangle $O P R$ ) $= 2 \cdot \frac{1}{2} \cdot \frac{e ^{t_{1}} + e ^{- t_{1}}}{2} \cdot \frac{e ^{t_{1}} - e ^{- t_{1}}}{2} = \frac{e ^{2 t_{1}} - e ^{- 2 t_{1}}}{4} .$

Required (hyperbolic sector) area $=$ area $△ O P Q -$ area $(P QR P) = t_{1}$ .

2.2 When $g (y) \leq 0$ on $[c, d]$

g (y) \leq 0

for all

y \in [c, d]

, then the area bounded by

x = g (y)

, the y-axis, and

y = c

y = d

Area = - \int_{c}^{d} g (y) d y .

The general formula covering both signs is

\int_{c}^{d} ∣ g (y)∣ d y

3. Curve Tracing: A Toolkit for Sketching

To evaluate area problems accurately, you first need a rough sketch of the curve. The following checklist gives you the approximate shape without heavy computation.

3.1 Symmetry

About the x-axis: If every power of $y$ in the equation is even, the graph is symmetric about the x-axis. Example: $y^{2} = 4 a x$ .

Figure 5: The parabola $y^{2} = 4 a x$ (with $a > 0$ ) is symmetric about the x-axis.
About the y-axis: If every power of $x$ is even, the graph is symmetric about the y-axis. Example: $x^{2} = 4 a y$ .

Figure 6: The parabola $x^{2} = 4 a y$ (with $a > 0$ ) is symmetric about the y-axis.
About both axes: If every power of $x$ and every power of $y$ is even, the curve is symmetric about both axes. Example: $x^{2} + y^{2} = a^{2}$ .

Figure 7: The circle $x^{2} + y^{2} = a^{2}$ is symmetric about both axes.
About the line $y = x$ : If the equation is unchanged on interchanging $x$ and $y$ , the graph is symmetric about $y = x$ . Example: $x^{3} + y^{3} = 3 a x y$ (folium of Descartes).

Figure 8: The folium $x^{3} + y^{3} = 3 a x y$ (with $a > 0$ ) is symmetric about the line $y = x$ .
In opposite quadrants (symmetric about origin): If the equation is unchanged when $x, y$ are replaced by $- x, - y$ , then the graph has symmetry in opposite quadrants. Example: $x y = c^{2}$ .

Figure 9: The rectangular hyperbola $x y = c^{2}$ shows symmetry in opposite quadrants.

3.2 Intercepts, Stationary Points, Monotonicity, End Behaviour

Find the points where the curve crosses the x-axis (set $y = 0$ ) and the y-axis (set $x = 0$ ).
Compute $\frac{d y}{d x}$ and equate it to zero to locate horizontal tangents.
Determine intervals where $f (x)$ is increasing or decreasing.
Examine what happens to $y$ as $x \to \infty$ or $x \to - \infty$ .

3.3 Asymptotes

An asymptote is a line whose distance from the curve tends to zero as the point on the curve moves to infinity along a branch.

Vertical: If $x \to a lim f (x) = \pm \infty$ , then $x = a$ is a vertical asymptote of $y = f (x)$ .
Horizontal: If $x \to \infty lim f (x) = k$ (or the same limit as $x \to - \infty$ ), then $y = k$ is a horizontal asymptote.
Oblique (slant), to the right: If $x \to \infty lim \frac{f ( x )}{x} = m_{1}$ and $x \to \infty lim (f (x) - m_{1} x) = c_{1}$ , then $y = m_{1} x + c_{1}$ is a slant asymptote.
Oblique, to the left: Same test with $x \to - \infty$ giving $m_{2}$ and $c_{2}$ ; asymptote is $y = m_{2} x + c_{2}$ .

Solved Example 10

Find the asymptote of

y = e^{- x}

Solution:

$x \to \infty lim y = x \to \infty lim e^{- x} = 0$ , so $y = 0$ is a horizontal asymptote.

Example Figure: The line

y = 0

is a horizontal asymptote of

y = e^{- x}

Solved Example 11

Find the asymptotes of

x y = 1

and sketch the graph.

Solution:

Write $y = \frac{1}{x}$ . Then $x \to 0 lim y = x \to 0 lim \frac{1}{x} = \pm \infty$ , so $x = 0$ is a vertical asymptote. $x \to \infty lim y = x \to \infty lim \frac{1}{x} = 0$ , so $y = 0$ is a horizontal asymptote.

Example Figure: The curve

x y = 1

has

x = 0

and

y = 0

as its two asymptotes.

Solved Example 12

Find the asymptotes of

y = x + \frac{1}{x}

and sketch the curve.

Solution:

$x \to 0 lim y = x \to 0 lim (x + \frac{1}{x}) = \pm \infty$ , so $x = 0$ is a vertical asymptote. There is no horizontal asymptote of the form $y = k$ . For a slant asymptote: $x \to \infty lim \frac{y}{x} = x \to \infty lim (1 + \frac{1}{x ^{2}}) = 1,$ $x \to \infty lim (y - x) = x \to \infty lim \frac{1}{x} = 0.$ Hence $y = x$ is a slant asymptote (same conclusion as $x \to - \infty$ ).

Example Figure: The line

y = x

is a slant asymptote of

y = x + \frac{1}{x}

; the y-axis is a vertical asymptote.

4. Area Between Two Curves

f (x) \geq g (x)

for all

x \in [a, b]

, then the area bounded by the curves

y = f (x)

y = g (x)

and the ordinates

x = a

x = b

Area = \int_{a}^{b} [f (x) - g (x)] d x .

Figure 10: Area between two curves

y = f (x)

(upper) and

y = g (x)

(lower) from

x = a

x = b

\int_{a}^{b} [f (x) - g (x)] d x

Solved Example 13

Find the area enclosed by the curve

y = x^{2} + x + 1

and its tangent at

(1, 3)

between the ordinates

x = - 1

and

x = 1

Solution:

$\frac{d y}{d x} = 2 x + 1$ ; at $x = 1$ , slope $= 3$ . Tangent: $y - 3 = 3 (x - 1) \Rightarrow y = 3 x$ .

Example Figure: Region enclosed by

y = x^{2} + x + 1

and its tangent

y = 3 x

(1, 3)

between

x = - 1

and

x = 1

On $[- 1, 1]$ , the parabola lies above the tangent line (verify: $(x^{2} + x + 1) - 3 x = (x - 1)^{2} \geq 0$ ).

Required area $= \int_{- 1}^{1} [(x^{2} + x + 1) - 3 x] d x = \int_{- 1}^{1} (x^{2} - 2 x + 1) d x = [\frac{x ^{3}}{3} - x^{2} + x]_{- 1}^{1}$ $= (\frac{1}{3} - 1 + 1) - (- \frac{1}{3} - 1 - 1) = \frac{2}{3} + 2 = \frac{8}{3} sq. units.$

General formula: The area bounded by

y = f (x)

and

y = g (x)

between

x = a

and

x = b

(without knowing which is on top everywhere) is

\int_{a}^{b} ∣ f (x) - g (x)∣ d x

Solved Example 14

Find the area of the region bounded by

y = sin x

y = cos x

, and the ordinates

x = 0

x = \frac{π}{2}

Solution:

The two curves meet at $x = \frac{π}{4}$ . On $[0, \frac{π}{4}]$ , $cos x \geq sin x$ ; on $[\frac{π}{4}, \frac{π}{2}]$ , $sin x \geq cos x$ .

Example Figure: Region bounded by

y = sin x

y = cos x

, and the ordinates

x = 0

and

x = π /2

; the curves intersect at

x = π /4

Required area $= \int_{0}^{π /2} ∣ sin x - cos x ∣ d x = \int_{0}^{π /4} (cos x - sin x) d x + \int_{π /4}^{π /2} (sin x - cos x) d x .$

Each integral evaluates to $2 - 1$ , so total area $= 2 (2 - 1)$ sq. units.

JEE Advanced

Solved Example 15

Find the area contained by the ellipse

2 x^{2} + 6 x y + 5 y^{2} = 1

Solution:

Treat the equation as a quadratic in $y$ : $5 y^{2} + 6 x y + (2 x^{2} - 1) = 0$ . Solving, $y = \frac{- 6 x \pm 36 x ^{2} - 20 ( 2 x ^{2} - 1 )}{10} = \frac{- 3 x \pm 5 - x ^{2}}{5} .$

For $y$ to be real, $5 - x^{2} \geq 0$ , so $- 5 \leq x \leq 5$ .

Example Figure: Rotated ellipse

2 x^{2} + 6 x y + 5 y^{2} = 1

; the extreme

x

values are

\pm 5

and extreme

y

values are

\pm 2

At each $x$ , the vertical strip has height $\frac{- 3 x + 5 - x ^{2}}{5} - \frac{- 3 x - 5 - x ^{2}}{5} = \frac{2 5 - x ^{2}}{5} .$

So $Area = \int_{- 5}^{5} \frac{2 5 - x ^{2}}{5} d x = \frac{4}{5} \int_{0}^{5} 5 - x^{2} d x .$

Put $x = 5 sin θ$ , $d x = 5 cos θ d θ$ ; limits $θ : 0 \to \frac{π}{2}$ : $= \frac{4}{5} \int_{0}^{π /2} 5 - 5 sin^{2} θ \cdot 5 cos θ d θ = 4 \int_{0}^{π /2} cos^{2} θ d θ = 4 \cdot \frac{1}{2} \cdot \frac{π}{2} = π sq. units.$

5. Standard Curves in JEE: Circle, Ellipse and Parabolas

Certain standard-form area problems appear year after year in JEE Main. The five examples below are the classics, worth memorising as templates.

Solved Example 16

Find the area enclosed by the circle

x^{2} + y^{2} = a^{2}

Solution:

The circle is symmetric about both axes, so the total area is four times the area in the first quadrant.

Example Figure: Area enclosed by the circle

x^{2} + y^{2} = a^{2}

π a^{2}

, computed as

4 \int_{0}^{a} a^{2} - x^{2} d x

In the first quadrant, $y = a^{2} - x^{2}$ from $x = 0$ to $x = a$ . Thus $Area = 4 \int_{0}^{a} a^{2} - x^{2} d x = 4 [\frac{x}{2} a^{2} - x^{2} + \frac{a ^{2}}{2} sin^{- 1} \frac{x}{a}]_{0}^{a} .$

$= 4 (0 + \frac{a ^{2}}{2} \cdot \frac{π}{2}) = π a^{2} sq. units.$

Solved Example 17

Find the area enclosed by the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

Solution:

The ellipse is symmetric about both axes.

Example Figure: Area of the ellipse

\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1

π ab

In the upper half, $y = \frac{b}{a} a^{2} - x^{2}$ from $x = - a$ to $x = a$ . So $Area = 2 \cdot \frac{b}{a} \int_{- a}^{a} a^{2} - x^{2} d x = \frac{2 b}{a} \cdot \frac{π a ^{2}}{2} = π ab sq. units.$

(The integral $\int_{- a}^{a} a^{2} - x^{2} d x = \frac{π a ^{2}}{2}$ is the area of a semicircle of radius $a$ .)

Solved Example 18

Find the area of the region cut off from the parabola

y^{2} = 4 a x

by its latus rectum.

Solution:

The latus rectum is the chord $x = a$ through the focus, perpendicular to the axis.

Example Figure: Parabola

y^{2} = 4 a x

cut off by its latus rectum (

x = a

); the enclosed area equals

\frac{8 a ^{2}}{3}

The region is symmetric about the x-axis. In the upper half, $y = 2 a x$ from $x = 0$ to $x = a$ : $Area = 2 \int_{0}^{a} 2 a x d x = 4 a [\frac{2 x ^{3/2}}{3}]_{0}^{a} = 4 a \cdot \frac{2 a ^{3/2}}{3} = \frac{8 a ^{2}}{3} sq. units.$

Solved Example 19

Find the area of the region bounded by the parabola

y^{2} = 4 x

and the line

y = 2 x - 4

Solution:

Solving simultaneously: from the line, $x = \frac{y + 4}{2}$ . Substituting into $y^{2} = 4 x$ gives $y^{2} = 2 (y + 4)$ , so $y^{2} - 2 y - 8 = 0$ , giving $y = 4$ or $y = - 2$ . The intersection points are $(4, 4)$ and $(1, - 2)$ .

Example Figure: Area between the parabola

y^{2} = 4 x

and the line

y = 2 x - 4

; the two intersections are at

(1, - 2)

and

(4, 4)

Integrating with respect to $y$ (the line and the parabola are both single-valued as functions of $y$ ): $Area = \int_{- 2}^{4} [\frac{y + 4}{2} - \frac{y ^{2}}{4}] d y = [\frac{y ^{2}}{4} + 2 y - \frac{y ^{3}}{12}]_{- 2}^{4} .$

$= (4 + 8 - \frac{64}{12}) - (1 - 4 + \frac{8}{12}) = \frac{20}{3} - (- \frac{7}{3}) = 9 sq. units.$

Solved Example 20

Find the area of the region bounded by the parabolas

y^{2} = 4 a x

and

x^{2} = 4 a y

Solution:

From $x^{2} = 4 a y$ , $y = \frac{x ^{2}}{4 a}$ . Substituting into $y^{2} = 4 a x$ : $\frac{x ^{4}}{16 a ^{2}} = 4 a x \Rightarrow x^{4} = 64 a^{3} x \Rightarrow x (x^{3} - 64 a^{3}) = 0$ , so $x = 0$ or $x = 4 a$ . Intersection points: $(0, 0)$ and $(4 a, 4 a)$ .

Example Figure: Area between the parabolas

y^{2} = 4 a x

and

x^{2} = 4 a y

; they intersect at

(0, 0)

and

(4 a, 4 a)

; enclosed area equals

\frac{16 a ^{2}}{3}

Between $x = 0$ and $x = 4 a$ , the upper curve is $y = 2 a x$ (top branch of $y^{2} = 4 a x$ ) and the lower is $y = \frac{x ^{2}}{4 a}$ .

$Area = \int_{0}^{4 a} [2 a x - \frac{x ^{2}}{4 a}] d x = [\frac{4 a x ^{3/2}}{3} - \frac{x ^{3}}{12 a}]_{0}^{4 a} .$

$= \frac{4 a ( 4 a ) ^{3/2}}{3} - \frac{( 4 a ) ^{3}}{12 a} = \frac{32 a ^{2}}{3} - \frac{16 a ^{2}}{3} = \frac{16 a ^{2}}{3} sq. units.$

6. Advanced Problem Types

The problems below combine area computation with optimisation, calculus of variations of a parameter, and functional equations. They are typical JEE Advanced setups.

JEE Advanced

Solved Example 21

Find the area contained between the two arms of the curve

(y - x)^{2} = x^{3}

between

x = 0

and

x = 1

Solution:

Taking square roots: $y - x = \pm x^{3/2}$ , giving two arms $y = x + x^{3/2}$ (upper) and $y = x - x^{3/2}$ (lower), valid for $x \geq 0$ .

For the upper arm: $\frac{d y}{d x} = 1 + \frac{3}{2} x^{1/2} > 0$ , so it is strictly increasing.

For the lower arm: $\frac{d y}{d x} = 1 - \frac{3}{2} x^{1/2} = 0 \Rightarrow x = \frac{4}{9}$ ; and $\frac{d ^{2} y}{d x ^{2}} = - \frac{3}{4} x^{- 1/2} < 0$ , so a maximum occurs at $x = \frac{4}{9}$ .

Example Figure: Two arms of the curve

(y - x)^{2} = x^{3}

for

0 \leq x \leq 1

; the upper arm is

y = x + x^{3/2}

and the lower is

y = x - x^{3/2}

The upper arm lies above the lower arm throughout $[0, 1]$ : $Area = \int_{0}^{1} [(x + x^{3/2}) - (x - x^{3/2})] d x = 2 \int_{0}^{1} x^{3/2} d x = 2 \cdot \frac{2}{5} = \frac{4}{5} sq. units.$

JEE Advanced

Solved Example 22

Let

A (m)

be the area bounded by the parabola

y = x^{2} + 2 x - 3

and the line

y = m x + 1

. Find the least value of

A (m)

Solution:

Setting the two equal: $x^{2} + (2 - m) x - 4 = 0$ . Let $α, β$ be the roots; then $α + β = m - 2$ and $α β = - 4$ .

$A (m) = \int_{α}^{β} [(m x + 1) - (x^{2} + 2 x - 3)] d x = \int_{α}^{β} [- x^{2} + (m - 2) x + 4] d x .$

Standard result: for $\int_{α}^{β} [- (x - α) (x - β)] d x$ , the value is $\frac{( β - α ) ^{3}}{6}$ . Since $(β - α)^{2} = (α + β)^{2} - 4 α β = (m - 2)^{2} + 16$ , we get $A (m) = \frac{1}{6} [(m - 2)^{2} + 16]^{3/2} .$

The expression $(m - 2)^{2} + 16$ is minimised when $m = 2$ (giving $16$ ), so $A (m)_{m i n} = \frac{1}{6} (16)^{3/2} = \frac{64}{6} = \frac{32}{3} sq. units.$

JEE Advanced

Solved Example 23

A curve

y = f (x)

passes through the origin and lies entirely in the first quadrant. Through any point

P (x, y)

on the curve, lines are drawn parallel to the coordinate axes. If the curve divides the area formed by these lines and the coordinate axes in the ratio

m : n

, show that

f (x) = c x^{m / n}

f (x) = c x^{n / m}

(with

c

arbitrary).

Solution:

Example Figure: A curve

y = f (x)

divides the rectangle bounded by the axes and the lines through

P (x, y)

parallel to them into two regions in ratio

m : n

Area of rectangle $O A P B = x y$ . Area under the curve $O A P O = \int_{0}^{x} f (t) d t$ . Area above the curve $O P B O = x y - \int_{0}^{x} f (t) d t$ .

Given $\frac{area O A P O}{area O P B O} = \frac{m}{n}$ , we get $n \int_{0}^{x} f (t) d t = m (x y - \int_{0}^{x} f (t) d t) = m x f (x) - m \int_{0}^{x} f (t) d t .$

Rearranging: $(m + n) \int_{0}^{x} f (t) d t = m x f (x)$ .

Differentiating both sides with respect to $x$ : $(m + n) f (x) = m f (x) + m x f^{'} (x) \Rightarrow \frac{f ^{'} ( x )}{f ( x )} = \frac{n}{m} \cdot \frac{1}{x} .$

Integrating: $ln f (x) = \frac{n}{m} ln x + C$ , so $f (x) = c x^{n / m}$ . If the roles of $m$ and $n$ are interchanged in the ratio, we get $f (x) = c x^{m / n}$ .

Common Mistakes to Avoid

Watch out

Forgetting the modulus when the curve dips below the x-axis: Writing $\int_{a}^{b} f (x) d x$ for the area when $f$ is negative gives a negative number. Area is always positive; use $\int_{a}^{b} ∣ f (x)∣ d x$ , or split at zeros of $f$ and negate the negative parts.
Not splitting at points where the curve crosses the x-axis: For $y = x^{3}$ on $[- 1, 1]$ , integrating directly gives $0$ (positive and negative pieces cancel). The geometric area is $\frac{1}{2}$ .
Wrong choice of variable of integration: If the region is bounded by horizontal lines (say $y = c$ and $y = d$ ) and a curve $x = g (y)$ , integrate with respect to $y$ , not $x$ . Trying to invert the curve unnecessarily can lead to multi-valued branches.
Assuming one curve is always above the other: In $\int_{a}^{b} [f (x) - g (x)] d x$ , if $f$ and $g$ swap positions inside $[a, b]$ , you must split at the intersection points and use $∣ f - g ∣$ .
Confusing symmetry with area doubling: Symmetry lets you compute area in one region and multiply, but only when the region is fully bounded on both sides by the same functions. Half-symmetric regions (like Example 7) do not double automatically.
Forgetting units: Every area answer in JEE is in "square units" (dimensionally). Write "sq. units" at the end for full marks in Advanced-style subjective problems.
Mixing up limits of integration: For $\int_{c}^{d} g (y) d y$ , $c$ must be the lower and $d$ the upper $y$ -value. Swapping them flips the sign.

Frequently Asked Questions

Q1. What does area as a definite integral mean geometrically?

Geometrically, $\int_{a}^{b} f (x) d x$ is the signed area between the curve $y = f (x)$ and the x-axis on $[a, b]$ . When $f (x) \geq 0$ , the integral equals the actual area. When $f (x) < 0$ , the integral is the negative of the area. To always get a positive area, use $\int_{a}^{b} ∣ f (x)∣ d x$ .

Q2. When should I integrate with respect to $y$ instead of $x$ ?

Integrate with respect to $y$ when the region is more naturally described as $x = g (y)$ with horizontal boundary lines $y = c$ and $y = d$ . Typical cases: regions bounded by the y-axis and horizontal lines, or curves like $x = sin y$ where inverting $y = sin^{- 1} x$ back would introduce awkward domain restrictions.

Q3. How do I handle the case where the curve crosses the x-axis inside $[a, b]$ ?

Find all the zeros of $f (x)$ in $[a, b]$ , split the integral at those points, and take the absolute value of each piece. For example, for $y = x^{3}$ on $[- 1, 1]$ : split at $x = 0$ , giving $\int_{- 1}^{0} (- x^{3}) d x + \int_{0}^{1} x^{3} d x = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$ .

Q4. What is the standard formula for the area of an ellipse, and how is it derived?

For the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ , the enclosed area is $π ab$ . Derivation: upper half is $y = \frac{b}{a} a^{2} - x^{2}$ ; integrate from $- a$ to $a$ and double. The integral $\int_{- a}^{a} a^{2} - x^{2} d x$ equals $\frac{π a ^{2}}{2}$ (area of semicircle), so total area is $2 \cdot \frac{b}{a} \cdot \frac{π a ^{2}}{2} = π ab$ .

Q5. How do I find the area between two curves when I do not know which is above?

Find all intersection points by solving the two equations simultaneously. On each sub-interval between successive intersections, test which curve is higher by plugging in any convenient value. Then integrate $(upper - lower)$ on each sub-interval and add. Equivalently, use $\int_{a}^{b} ∣ f (x) - g (x)∣ d x$ .

Q6. Why do we sketch the curve before computing area?

Sketching tells you: (a) whether the curve crosses the x-axis in $[a, b]$ so you know where to split; (b) which curve is on top in a two-curve problem; (c) whether to integrate with respect to $x$ or $y$ ; (d) the correct limits when they are not given explicitly. Curve-tracing checks: symmetry, intercepts, stationary points, monotonicity, behaviour at infinity, and asymptotes.

Q7. What is an asymptote and how do I find one?

An asymptote is a line whose distance from the curve tends to zero as the point moves to infinity along the curve. Vertical: $x = a$ if $lim_{x \to a} f (x) = \pm \infty$ . Horizontal: $y = k$ if $lim_{x \to \pm \infty} f (x) = k$ . Slant: $y = m x + c$ if $lim_{x \to \infty} \frac{f ( x )}{x} = m$ and $lim_{x \to \infty} (f (x) - m x) = c$ .

Q8. What is the area enclosed by a parabola and its latus rectum?

For $y^{2} = 4 a x$ , the latus rectum is $x = a$ . The enclosed area is $\frac{8 a ^{2}}{3}$ . Derivation: by symmetry, twice the upper region: $2 \int_{0}^{a} 2 a x d x = 4 a \cdot \frac{2 a ^{3/2}}{3} = \frac{8 a ^{2}}{3}$ . This is one of the most-asked standard results in JEE Main.

Q9. How is the area between two parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ computed?

They intersect at $(0, 0)$ and $(4 a, 4 a)$ . Between these, the upper curve is $y = 2 a x$ and the lower is $y = \frac{x ^{2}}{4 a}$ . The enclosed area is $\int_{0}^{4 a} (2 a x - \frac{x ^{2}}{4 a}) d x = \frac{32 a ^{2}}{3} - \frac{16 a ^{2}}{3} = \frac{16 a ^{2}}{3}$ .

Previous year questions on Area as Definite Integral

48 questions from past papers, each with a step-by-step solution.

Show all 48 questions

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1. Area Between a Curve, the x-axis and Two Ordinates

1.1 When f(x)≥0 on [a,b]

1.2 When f(x)<0 on [a,b]

1.3 When f(x) changes sign on [a,b]

2. Area Between a Curve, the y-axis and Two Abscissas

2.1 When g(y)≥0 on [c,d]

2.2 When g(y)≤0 on [c,d]

3. Curve Tracing: A Toolkit for Sketching

3.1 Symmetry

3.2 Intercepts, Stationary Points, Monotonicity, End Behaviour

3.3 Asymptotes

4. Area Between Two Curves

5. Standard Curves in JEE: Circle, Ellipse and Parabolas

6. Advanced Problem Types

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What does area as a definite integral mean geometrically?

Q2. When should I integrate with respect to y instead of x?

Q3. How do I handle the case where the curve crosses the x-axis inside [a,b]?

Q4. What is the standard formula for the area of an ellipse, and how is it derived?

Q5. How do I find the area between two curves when I do not know which is above?

Q6. Why do we sketch the curve before computing area?

Q7. What is an asymptote and how do I find one?

Q8. What is the area enclosed by a parabola and its latus rectum?

Q9. How is the area between two parabolas y2=4ax and x2=4ay computed?

Previous year questions on Area as Definite Integral

Ready to master Application Of Integrals?

1.1 When $f (x) \geq 0$ on $[a, b]$

1.2 When $f (x) < 0$ on $[a, b]$

1.3 When $f (x)$ changes sign on $[a, b]$

2.1 When $g (y) \geq 0$ on $[c, d]$

2.2 When $g (y) \leq 0$ on $[c, d]$

Q2. When should I integrate with respect to $y$ instead of $x$ ?

Q3. How do I handle the case where the curve crosses the x-axis inside $[a, b]$ ?

Q9. How is the area between two parabolas $y^{2} = 4 a x$ and $x^{2} = 4 a y$ computed?