Applications of Definite Integral — MHT-CET Maths Formulas and Traps

Area Under a Curve — Between a Curve and an Axis

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

Common traps

Integrating over the wrong interval

'Bounded by

y = 4x - x^2

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

0

to

4

. Integrating to some other convenient number, or from

-4

, produces

16

or

32

— both offered.

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]

is

2

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

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.

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.

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.

Area Between Two Curves — Intersections First

Learn this subtopic in the notes

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

Common traps

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.

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.

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.

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.

Areas of Circles, Ellipses and Hyperbolas — Sectors, Segments and Standard Integrals

Learn this subtopic in the notes

The Circle Integral ∫√(a² − x²) dx, Quarter Discs and Sectors

Circle integral and sector

\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} \qquad \text{sector} = \frac12 r^2\theta

The Smaller Segment of a Circle Cut by a Vertical Line

Minor segment

A = 2\int_c^a\sqrt{a^2 - x^2}\,dx = a^2\theta - \frac{a^2}{2}\sin 2\theta, \quad \cos\theta = \frac{c}{a}

Ellipse Arc Minus Chord: Quarter-Ellipse Minus Triangle

Arc minus chord

A = \frac{\pi ab}{4} - \frac{ab}{2} = \frac{ab}{4}(\pi - 2)

Hyperbola Segments and Circle-Parabola Regions

Hyperbola integral and the circle-parabola region

\int\sqrt{x^2 - a^2}\,dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log\left|x + \sqrt{x^2 - a^2}\right| \qquad \{x^2 + y^2 \le 1,\ y^2 \le 1 - x\}:\ \frac{\pi}{2} + \frac43

Common traps

Reading x = y√3 as a 60° line

x = y\sqrt3

is

y = \dfrac{x}{\sqrt3}

, slope

\dfrac{1}{\sqrt3}

, angle

30^\circ

. The

60^\circ

reading gives

\dfrac{2\pi}{3}

, which is offered.

Forgetting the factor 2 for the lower half

\int_c^a\sqrt{a^2 - x^2}\,dx

is only the part above the x-axis. The segment is symmetric, so double it; the un-doubled value is always an option.

Using πab/2 for the quadrant

A quadrant is a QUARTER of the ellipse,

\dfrac{\pi ab}{4}

. Halving instead of quartering doubles the first term and lands on

\dfrac{15}{2}(\pi - 2)

-type distractors.

Setting up the circle-parabola region as one integral

The binding boundary is the circle for

x \le 0

and the parabola for

x \ge 0

. One integral from

-1

to

1

with either curve gives the wrong number; split at

x = 0

and add a half-disc to a parabolic cap.

Applications of Definite Integral formulas

Area Under a Curve — Between a Curve and an Axis

Area Under a Curve Is an Integral of |y|

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

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

Recover the Curve First: Unknown Coefficients and a Given Derivative

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

Common traps

Area Between Two Curves — Intersections First

Solve for the Intersections, Then Integrate Upper Minus Lower

Two Parabolas Through the Origin, and Regions With Mirror Symmetry

Horizontal Strips: Integrate (Right − Left) in y

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

Common traps

Areas of Circles, Ellipses and Hyperbolas — Sectors, Segments and Standard Integrals

The Circle Integral ∫√(a² − x²) dx, Quarter Discs and Sectors

The Smaller Segment of a Circle Cut by a Vertical Line

Ellipse Arc Minus Chord: Quarter-Ellipse Minus Triangle

Hyperbola Segments and Circle-Parabola Regions

Common traps

More MHT-CET Maths formula sheets