MHT-CET Maths · Applications of Definite Integral

Area Under a Curve — Between a Curve and an Axis

The area between y = f(x) and the x-axis from a to b is ∫|f(x)| dx — sketch, find where the curve meets the axis, and integrate each piece with a positive sign.

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Why this matters

14 PYQs at 14% HARD — the cheapest page in the whole MHT-CET Maths long tail, and the one where a sketch is the entire method. The recurring stems are a parabola cut off by the axis, a modulus or a cubic that crosses the axis (so the signed integral and the area differ), and a curve given as x in terms of y that wants a horizontal strip. Two stems hide the curve behind a derivative or two unknown coefficients, and one asks for the vertical line that halves an area — all of them are one integral once the setup is written.

Concept 1 of 5

Area Under a Curve Is an Integral of |y|

Intuition

Slice the region into thin vertical strips of width

dx

and height

y

; the area is the sum

\int y\,dx

. Where the curve dips below the axis the strip's height is

-y

, which is why the honest formula carries a modulus.

Definition

Area between $y = f(x)$ , the x-axis and $x = a$ , $x = b$ : $A = \int_a^b|f(x)|\,dx$ . When $f \ge 0$ on the interval this is just $\int_a^b f(x)\,dx$ .
Sketch first: find where $f = 0$ (the curve meets the axis) — those are the natural limits when the stem says 'bounded by the curve and the x-axis'.
$y = 4x - x^2$ meets the axis at $0$ and $4$ and is positive between: $A = \int_0^4(4x - x^2)\,dx = \dfrac{32}{3}$ .
A ratio of two areas under different curves over the same interval is just the ratio of the two integrals: $\int_0^{\pi/3}\cos x : \int_0^{\pi/3}\cos 2x = \dfrac{\sqrt3}{2} : \dfrac{\sqrt3}{4} = 2 : 1$ .
Units: 'sq. units' — the answer is a number, and it is never negative.

Area under a curve

A = \int_a^b\left|f(x)\right|dx \qquad (f \ge 0 \text{ on } [a, b] \Rightarrow A = \int_a^b f(x)\,dx)

Worked example

Find the area bounded by

y = 6x - x^2

and the x-axis.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Applications of Definite IntegralEASY

The area bounded by the curve

y= 4x-x^{2}

and X - axis in square units, is

\_\_\_\_

[Q123 · 25 April Shift II · 2025]

Integrating over the wrong interval

'Bounded by

y = 4x - x^2

and the x-axis' means from one axis-crossing to the other,

0

4

. Integrating to some other convenient number, or from

-4

, produces

16

32

— both offered.

Drill 1 more on area under a curve is an integral of |y|

Concept 2 of 5

When the Curve Crosses the Axis: Split and Take Each Piece Positive

Intuition

y = x(x - 2)(x + 1)

is above the axis on

(-1, 0)

and below on

(0, 2)

. The signed integral cancels part of one against the other; the AREA adds them, so integrate each piece and take absolute values.

Definition

Find the roots inside the interval, integrate between consecutive roots, take the absolute value of each piece, add.
$y = x(x - 2)(x + 1) = x^3 - x^2 - 2x$ : $\left|\int_{-1}^{0}\right| = \dfrac{5}{12}$ , $\left|\int_0^2\right| = \dfrac83$ , total $\dfrac{37}{12}$ .
$y = x|x|$ on $[-1, 1]$ : $-x^2$ on the left, $x^2$ on the right — each piece has area $\dfrac13$ , total $\dfrac23$ .
$y = |x - 2|$ between $x = 1$ and $x = 3$ is already non-negative: two triangles of area $\dfrac12$ each, total $1$ .
The signed integral $\int_{-1}^{2}(x^3 - x^2 - 2x)\,dx = \dfrac{5}{12} - \dfrac83 = -\dfrac94$ is what the setter offers as a distractor.

Area across a sign change

A = \left|\int_a^c f\,dx\right| + \left|\int_c^b f\,dx\right| \quad (f(c) = 0)

Worked example

Find the area bounded by

y = x^3 - x

and the x-axis.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Applications of Definite IntegralHARD

The area bounded by the X-axis and the curve

y=x(x-2)(x+1)

[Q109 · 10th May Shift 2 · 2024]

Reporting the signed integral as the area

\int_0^\pi\cos x\,dx = 0

, but the area between

\cos x

and the axis on

[0, \pi]

2

. Whenever the curve crosses the axis inside the interval, one integral is not the answer.

Drill 3 more on when the curve crosses the axis: split and take each piece positive

Concept 3 of 5

Curves Given as x = g(y): Integrate in y

Intuition

x = 2 - y - y^2

is a sideways parabola. Slicing it horizontally gives strips of width

x

and thickness

dy

, so the area is

\int x\,dy

between the y-values where it meets the y-axis.

Definition

Area between $x = g(y)$ and the y-axis from $y = c$ to $y = d$ : $A = \int_c^d|g(y)|\,dy$ .
$x = 2 - y - y^2$ meets the y-axis where $y^2 + y - 2 = 0$ : $y = -2, 1$ . $A = \int_{-2}^{1}(2 - y - y^2)\,dy = \dfrac92$ .
A region between two horizontal lines: $y = 4x^2$ , $x = 0$ , $y = 2$ , $y = 4$ — solve for $x = \dfrac{\sqrt y}{2}$ and integrate in $y$ : $\dfrac12\int_2^4\sqrt y\,dy = \dfrac13(8 - 2\sqrt2)$ .
A parabola symmetric about the x-axis between two vertical lines uses vertical strips of full height $2y$ : area inside $y^2 = 4ax$ from $x = a$ to $4a$ is $2\int_a^{4a}2\sqrt{ax}\,dx = \dfrac{56}{3}a^2$ .
Choose the strip direction that makes the region a single integral; a horizontal strip is the natural choice whenever the boundary is given as $x$ in terms of $y$ .

Horizontal strips

A = \int_c^d\left|g(y)\right|dy \qquad \text{(strip width } x = g(y)\text{, thickness } dy)

Worked example

Find the area bounded by

x = y^2

and the y-axis between

y = 0

and

y = 3

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Applications of Definite IntegralMODERATE

The area bounded by the curve

x= 2 -y-y^{2}

and the Y -axis is

[Q126 · 22 April Shift I · 2025]

Forgetting the lower half of a sideways parabola

y^2 = 4ax

has a branch below the axis. 'Area inside the parabola between

x = a

and

x = 4a

' counts both halves: the strip height is

2\sqrt{4ax}

, not

\sqrt{4ax}

. Half the correct answer is always in the options.

Drill 2 more on curves given as x = g(y): integrate in y

Concept 4 of 5

Recover the Curve First: Unknown Coefficients and a Given Derivative

Intuition

y = a\sqrt x + bx

passes through

(1, 2)

and the area under it to

x = 4

8

' is two equations for two unknowns — one from the point, one from the area integral. Set them up, solve, then answer whatever combination is asked.

Definition

Point condition: substitute the coordinates into the curve.
Area condition: integrate the curve with the unknowns as symbols; $\int_0^4(a\sqrt x + bx)\,dx = \dfrac{16a}{3} + 8b$ , set equal to $8$ , i.e. $2a + 3b = 3$ . With $a + b = 2$ : $a = 3$ , $b = -1$ .
Given a slope: $f'(x) = 2x + 1$ through $(1, 2)$ means $f(x) = x^2 + x + c$ with $c = 0$ ; then the area to $x = 1$ is $\int_0^1(x^2 + x)\,dx = \dfrac56$ .
Answer the asked quantity — $a - b = 4$ , the pair $(3, -1)$ , or the area — the same setup has been asked three ways.

Two conditions, two unknowns

\text{point: } y_0 = f(x_0;\,a, b) \qquad \text{area: } \int_0^{4}f(x;\,a, b)\,dx = 8

Worked example

The curve

y = ax^2 + b

passes through

(1, 4)

, and the area under it from

x = 0

x = 1

3

. Find

a

and

b

Practice this conceptself-check

From the bank · past-year question

Example 4Applications of Definite IntegralMODERATE

If a curve

y = a\sqrt{x}+ bx

passes through the point

(1,2)

and the area bounded by this curve, line

x = 4

and the X -axis is 8 sq. units, then the value of

a - b

[Q120 · 22 April Shift II · 2025]

Answering a when a − b was asked

With

a = 3

b = -1

a - b = 4

a + b = 2

ab = -3

. Every one of those is an option on some sitting of this question. Re-read the last line of the stem before choosing.

Drill 2 more on recover the curve first: unknown coefficients and a given derivative

Concept 5 of 5

Dividing an Area in Half, and the Integral as Accumulated Change

Intuition

x = \alpha

splits the area under

y = \dfrac{x^2}{4}

[0, 4]

into two equal parts, the area up to

\alpha

is half the total:

\dfrac{\alpha^3}{12} = \dfrac12\cdot\dfrac{64}{12}

. The same integral, read as a rate summed over an interval, gives the total change in a quantity.

Definition

Halving line: $\int_0^\alpha f\,dx = \dfrac12\int_0^4 f\,dx$ . For $f = \dfrac{x^2}{4}$ : $\alpha^3 = 32$ , $\alpha = 32^{1/3}$ .
Accumulated change: if $\dfrac{dP}{dx} = 100 - 12\sqrt x$ , the change in $P$ as $x$ goes $0 \to 9$ is $\int_0^9(100 - 12\sqrt x)\,dx = 900 - 216 = 684$ ; the new level is the old level plus $684$ .
Both are 'area under a curve' with a different question attached: an unknown limit, or a starting value to add.

Halving and accumulation

\int_0^\alpha f\,dx = \frac12\int_0^b f\,dx \qquad P(b) - P(a) = \int_a^b\frac{dP}{dx}\,dx

Worked example

The line

x = \alpha

divides the area under

y = x^2

[0, 2]

into two equal parts. Find

\alpha

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5Applications of Definite IntegralMODERATE

If the area bounded by the curve

x^{2}= 4y,X

-axis and the line

x= 4

is divided into equal areas by the line

x=\alpha

, then the value of

\alpha

is ...

[Q116 · 26 April Shift II · 2025]

Reporting the change instead of the new level

The integral gives the CHANGE (

684

items); the question asks for the new level (

1000 + 684 = 1684

). Both numbers are offered.

Drill 1 more on dividing an area in half, and the integral as accumulated change

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 (5)

Area Under a Curve Is an Integral of |y|
Area under a curve
$A = \int_a^b\left|f(x)\right|dx \qquad (f \ge 0 \text{ on } [a, b] \Rightarrow A = \int_a^b f(x)\,dx)$
When the Curve Crosses the Axis: Split and Take Each Piece Positive
Area across a sign change
$A = \left|\int_a^c f\,dx\right| + \left|\int_c^b f\,dx\right| \quad (f(c) = 0)$
Curves Given as x = g(y): Integrate in y
Horizontal strips
$A = \int_c^d\left|g(y)\right|dy \qquad \text{(strip width } x = g(y)\text{, thickness } dy)$
Recover the Curve First: Unknown Coefficients and a Given Derivative
Two conditions, two unknowns
$\text{point: } y_0 = f(x_0;\,a, b) \qquad \text{area: } \int_0^{4}f(x;\,a, b)\,dx = 8$
Dividing an Area in Half, and the Integral as Accumulated Change
Halving and accumulation
$\int_0^\alpha f\,dx = \frac12\int_0^b f\,dx \qquad P(b) - P(a) = \int_a^b\frac{dP}{dx}\,dx$

Watch out for (5)

Integrating over the wrong interval
→ Area Under a Curve Is an Integral of |y|
Reporting the signed integral as the area
→ When the Curve Crosses the Axis: Split and Take Each Piece Positive
Forgetting the lower half of a sideways parabola
→ Curves Given as x = g(y): Integrate in y
Answering a when a − b was asked
→ Recover the Curve First: Unknown Coefficients and a Given Derivative
Reporting the change instead of the new level
→ Dividing an Area in Half, and the Integral as Accumulated Change

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