MHT-CET Maths · Applications of Definite Integral

Area Between Two Curves — Intersections First

The area between two curves is ∫(upper − lower) dx between their intersection points — so the intersections are found first, the top curve is decided over each stretch, and only then is anything integrated.

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Why this matters

21 PYQs at 38% HARD — the heart of the chapter, and the page where the difficulty is entirely in the setup. A parabola against a line is set every year; two parabolas through the origin, a parabola against |x|, and a region described by inequalities are the HARD variants, and each is wrong at the intersection step far more often than at the integral. Horizontal strips turn three of these from a two-piece integral into one piece, which is the single most useful habit on this page.

Concept 1 of 4

Solve for the Intersections, Then Integrate Upper Minus Lower

Intuition

Between two curves a vertical strip runs from the lower curve up to the upper one, so its height is

y_{\text{upper}} - y_{\text{lower}}

. The strips exist only between the points where the curves cross, and those are the limits.

Definition

Step 1: solve the curves simultaneously for the intersection x-values $x_1 < x_2$ .
Step 2: decide which curve is on top on $(x_1, x_2)$ — test one point, or sketch.
Step 3: $A = \int_{x_1}^{x_2}\left(y_{\text{upper}} - y_{\text{lower}}\right)dx$ .
Parabola and line: $y = x^2$ and $y = x + 2$ meet where $x^2 - x - 2 = 0$ , $x = -1, 2$ ; the line is on top: $\int_{-1}^{2}(x + 2 - x^2)\,dx = \dfrac92$ .
Two lines meeting at $x = 0$ , cut off by $x = 2$ : $\int_0^2\left[(4x + 1) - (3x + 1)\right]dx = 2$ .
When the stem supplies the vertical limits ( $x = 0$ and $x = 3$ ) and the curves do NOT cross inside them, there is no intersection step: $\int_0^3\left[(x^2 + 2) - (x + 1)\right]dx = \dfrac{15}{2}$ . Non-crossing pairs like $e^x$ and $\log x$ on $[1, 2]$ work the same way.

Area between two curves

A = \int_{x_1}^{x_2}\left(y_{\text{upper}} - y_{\text{lower}}\right)dx, \qquad x_1, x_2 \text{ the roots of } y_1(x) = y_2(x)

Worked example

Find the area between

y = x^2

and

y = 2x

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Applications of Definite IntegralMODERATE

The area (in sq. units) of the region bounded by

y-x=2

and

x^2=y

is equal to

[Q109 · 11th May Shift 2 · 2023]

Lower minus upper

Reversing the order gives the negative of the area, and the option list contains that negative or its magnitude with the wrong sign attached to another term. Test a single interior point to fix which curve is on top before writing the integrand.

Drill 7 more on solve for the intersections, then integrate upper minus lower

Concept 2 of 4

Two Parabolas Through the Origin, and Regions With Mirror Symmetry

Intuition

y^2 = 4ax

and

x^2 = 4ay

cross at

(0, 0)

and

(4a, 4a)

; between them the first is on top. For regions symmetric about an axis, compute one half and double — fewer integrals, fewer sign slips.

Definition

$y^2 = 4ax$ , $x^2 = 4ay$ : $A = \int_0^{4a}\left(2\sqrt{ax} - \dfrac{x^2}{4a}\right)dx = \dfrac{16a^2}{3}$ . The general $y = ax^2$ , $x = ay^2$ pair gives $\dfrac{1}{3a^2}$ ; setting that equal to $1$ gives $a = \dfrac{1}{\sqrt3}$ .
Parabola against $|x|$ : $y = x^2$ and $y = |x|$ meet at $x = 0, \pm1$ ; by symmetry $A = 2\int_0^1(x - x^2)\,dx = \dfrac13$ . $y^2 = 4x$ against $y = |x|$ lives only in $x \ge 0$ : $\int_0^4(2\sqrt x - x)\,dx = \dfrac83$ .
Two shifted parabolas and a horizontal line: $y = (x - 1)^2$ , $y = (x + 1)^2$ , $y = \dfrac14$ — symmetric about the y-axis; the enclosed region is $2\int_0^{1/2}\left[\dfrac14 - (x - 1)^2\right]dx$ … taken with the correct sign, $\dfrac13$ .
Always ask: is the region symmetric about the x-axis, the y-axis or $y = x$ ? If so, integrate half.

Two standard parabola results

\text{Area}\left(y^2 = 4ax,\ x^2 = 4ay\right) = \frac{16a^2}{3} \qquad \text{Area}\left(y = ax^2,\ x = ay^2\right) = \frac{1}{3a^2}

Worked example

Find the area between

y^2 = x

and

x^2 = y

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Applications of Definite IntegralMODERATE

The area (in sq. units) of the region bounded by the curves

y^2 = 4ax

and

x^2 = 4ay

a>0

[Q148 · May Shift 1 · 2021]

Counting a branch that is not there

y^2 = 4x

has no points with

x < 0

, so its region with

y = |x|

is NOT doubled — the answer is

\frac83

, not

\frac{16}{3}

. Symmetry doubles only when both curves exist on both sides.

Drill 5 more on two parabolas through the origin, and regions with mirror symmetry

Concept 3 of 4

Horizontal Strips: Integrate (Right − Left) in y

Intuition

For

y^2 = 2x

against

x = y + 4

, vertical strips need two pieces (the top boundary changes where the line takes over). Horizontal strips run from the parabola on the left to the line on the right for every

y

— one integral.

Definition

$A = \int_{y_1}^{y_2}\left(x_{\text{right}} - x_{\text{left}}\right)dy$ , limits from solving the curves in $y$ .
$\dfrac{y^2}{2} \le x \le y + 4$ : $y^2 - 2y - 8 = 0 \Rightarrow y = -2, 4$ ; $A = \int_{-2}^{4}\left(y + 4 - \dfrac{y^2}{2}\right)dy = 18$ .
$y^2 = 4x$ and $y = 2x - 4$ : in $y$ , $x = \dfrac{y^2}{4}$ and $x = \dfrac{y + 4}{2}$ , meeting at $y = -2, 4$ ; $A = \int_{-2}^{4}\left(\dfrac{y + 4}{2} - \dfrac{y^2}{4}\right)dy = 9$ .
Two parabolas $x^2 = \dfrac{y}{4}$ , $x^2 = 9y$ capped by $y = 2$ : at height $y$ the width is $2\left(3\sqrt y - \dfrac{\sqrt y}{2}\right) = 5\sqrt y$ , so $A = \int_0^2 5\sqrt y\,dy = \dfrac{20\sqrt2}{3}$ ; the same shape with $y = 9x^2$ , $y = \dfrac{x^2}{16}$ , $y = 1$ gives $\dfrac{44}{9}$ .
Rule of thumb: when a boundary is a sideways parabola or the region is capped by a horizontal line, slice horizontally.

Horizontal strips between two curves

A = \int_{y_1}^{y_2}\left(x_{\text{right}}(y) - x_{\text{left}}(y)\right)dy

Worked example

Find the area between

y^2 = x

and

x = y + 2

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Applications of Definite IntegralMODERATE

The area (in sq. units) of the region

A=\{(x,y): \frac{y^2}{2}\leq x\leq y+4\}

[Q141 · 9th May Shift 1 · 2024]

Vertical strips on a sideways parabola

For

y^2 = 2x

against a line, a vertical strip's top boundary switches from the parabola's upper arm to the line partway across — two integrals and a missed switch. One horizontal integral does the whole region.

Drill 4 more on horizontal strips: integrate (right − left) in y

Concept 4 of 4

Regions Described by Several Inequalities: Sketch, Then Split Where the Top Changes

Intuition

\{x \ge 0,\ x + y \le 3,\ x^2 \le 4y,\ y \le 1 + \sqrt x\}

is a region with one floor and two possible ceilings. The ceiling switches where

1 + \sqrt x = 3 - x

; each stretch gets its own integral with its own top curve.

Definition

Translate each inequality into a boundary: $x^2 \le 4y$ means above the parabola $y = \dfrac{x^2}{4}$ ; $y \le 1 + \sqrt x$ means below that curve; $x + y \le 3$ means below the line $y = 3 - x$ .
Find every pairwise intersection that matters: $1 + \sqrt x = 3 - x$ at $x = 1$ ; $3 - x = \dfrac{x^2}{4}$ at $x = 2$ .
Integrate stretch by stretch: $\int_0^1\left(1 + \sqrt x - \dfrac{x^2}{4}\right)dx + \int_1^2\left(3 - x - \dfrac{x^2}{4}\right)dx = \dfrac{19}{12} + \dfrac{11}{12} = \dfrac52$ .
'First quadrant, bounded by $y = \sqrt x$ , the line $2y - x + 3 = 0$ and the x-axis': the region under $\sqrt x$ up to the intersection $(9, 3)$ , minus the triangle under the line from $x = 3$ to $9$ : $18 - 9 = 9$ .
Subtracting a simple shape (triangle, rectangle) from an integral is often quicker than a second integral.

Piecewise ceiling

A = \int_{x_0}^{x_1}\left(\text{ceiling}_1 - \text{floor}\right)dx + \int_{x_1}^{x_2}\left(\text{ceiling}_2 - \text{floor}\right)dx

Worked example

Find the area of the region

\{(x, y):\ 0 \le y \le x^2,\ 0 \le x \le 1\} \cup \{(x, y):\ 0 \le y \le 2 - x,\ 1 \le x \le 2\}

Practice this conceptself-check

From the bank · past-year question

Example 4Applications of Definite IntegralHARD

The area (in sq. units) of the region

\{(x,y):\ x\geq 0,\; x+y\leq 3,\; x^2\leq 4y \text{ and } y\leq 1+\sqrt{x}\}

[Q148 · 10th May Shift 2 · 2023]

One ceiling for the whole interval

With

y \le 1 + \sqrt x

and

x + y \le 3

both in force, the ceiling is the LOWER of the two at each

x

, and it changes at

x = 1

. Using either curve alone over

[0, 2]

gives an answer that is not on the list — which is the paper's way of telling you to split.

Drill 1 more on regions described by several inequalities: sketch, then split where the top changes

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

Solve for the Intersections, Then Integrate Upper Minus Lower
Area between two curves
$A = \int_{x_1}^{x_2}\left(y_{\text{upper}} - y_{\text{lower}}\right)dx, \qquad x_1, x_2 \text{ the roots of } y_1(x) = y_2(x)$
Two Parabolas Through the Origin, and Regions With Mirror Symmetry
Two standard parabola results
$\text{Area}\left(y^2 = 4ax,\ x^2 = 4ay\right) = \frac{16a^2}{3} \qquad \text{Area}\left(y = ax^2,\ x = ay^2\right) = \frac{1}{3a^2}$
Horizontal Strips: Integrate (Right − Left) in y
Horizontal strips between two curves
$A = \int_{y_1}^{y_2}\left(x_{\text{right}}(y) - x_{\text{left}}(y)\right)dy$
Regions Described by Several Inequalities: Sketch, Then Split Where the Top Changes
Piecewise ceiling
$A = \int_{x_0}^{x_1}\left(\text{ceiling}_1 - \text{floor}\right)dx + \int_{x_1}^{x_2}\left(\text{ceiling}_2 - \text{floor}\right)dx$

Watch out for (4)

Lower minus upper
→ Solve for the Intersections, Then Integrate Upper Minus Lower
Counting a branch that is not there
→ Two Parabolas Through the Origin, and Regions With Mirror Symmetry
Vertical strips on a sideways parabola
→ Horizontal Strips: Integrate (Right − Left) in y
One ceiling for the whole interval
→ Regions Described by Several Inequalities: Sketch, Then Split Where the Top Changes

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