2. Calculus — Integration

Area under simple curves

The computation of the area under simple curves forms the cornerstone of definite integration applications in calculus, translating accumulation problems into precise geometric measurements.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Fundamental Principle of Area under a Curve

Let $f: [a, b] \to \mathbb{R}$ be a continuous and non-negative function. The area bounded by the curve $y = f(x)$, the $x$-axis, and the ordinates $x = a$ and $x = b$ is defined by the Riemann integral \text{Area} = \int_{a}^{b} f(x)\,dx = \lim_{n \to \infty} \sum_{i=1}^{n} f(x_i^*)\, \Delta x_i.

📐 Geometric Interpretation

Geometrically, this represents the net measure of the two-dimensional region trapped between the curve $y=f(x)$ and the horizontal $x$-axis from $x=a$ to $x=b$. Portions of the curve lying below the $x$-axis contribute negative signed area, necessitating absolute value integration or splitting the interval when total physical area is required.

⚙️ Algebraic & Structural Meaning

Algebrically, it bridges discrete summation limits with continuous antiderivatives via the Fundamental Theorem of Calculus, transforming the evaluation of infinite rectangular strips into evaluating the difference of antiderivative values at boundary endpoints: $F(b) - F(a)$ where $F'(x) = f(x)$.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Continuity of the bounding function	`[a, b] (Closed interval)`	Strictly Mandatory / Necessary	If $f(x)$ possesses jump discontinuities or unbounded singularities on $[a, b]$, standard Riemann integrability fails, and improper integrals or measure-theoretic Lebesgue integration must be invoked.
Sign Consistency (or controlled sign changes)	`[a, b] (Closed interval)`	Strictly Mandatory for physical area	Failing to account for sign changes across the $x$-axis results in algebraic cancellation of positive and negative regions, yielding the net displacement rather than the true geometrical area.
Smoothness or piecewise smoothness of bounding curves	`(a, b) (Open interval)`	Essential for explicit integration	Highly erratic or non-rectifiable boundaries can prevent explicit parametrization or lead to undefined integral bounds.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Area of Standard Parabola and Line/Parabola Intersection

$$ \text{Area} = \frac{8}{3} \frac{a^2}{b} \quad \text{or} \quad \text{Area bounded by } y^2 = 4ax \text{ and } y = mx \text{ is } \frac{8a^2}{3m^3} $$

Rule: Directly memorize standard parabolic segment formulas bounded by chords or lines to bypass lengthy intersection point and integration steps.

Application: undefined

Area of Ellipse and Circle via Affine Scaling

$$ \text{Area of } \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \text{ is } \pi ab; \quad \text{Area of circle } x^2 + y^2 = r^2 \text{ is } \pi r^2 $$

Rule: Recognize standard conic scaling forms instantly without computing $\int_{-a}^{a} \frac{b}{a}\sqrt{a^2-x^2}\,dx$.

Astroid and Parametric Loop Symmetry

$$ \text{Area of astroid } x^{2/3} + y^{2/3} = a^{2/3} \text{ is } \frac{3}{8}\pi a^2 $$

Rule: Use four-quadrant symmetry: evaluate the integral in the first quadrant and multiply by 4.

Application: Direct result tested in UP PGT Mathematics.

Area Bounded by Loops of Folium of Descartes / Lemniscate

$$ \text{Area of loop of } x^3 + y^3 = 3axy \text{ is } \frac{3}{2} a^2 $$

Rule: Remember standard symmetric loop formulas or convert to polar coordinates where $r^2 = a^2 \cos 2\theta$ yields $\frac{1}{2} \int r^2 d\theta = a^2$.

Application: High-level PGT questions targeting curve tracing.

Area Between Two Parabolas $y^2 = 4ax$ and $x^2 = 4by$

$$ \text{Area} = \frac{16}{3} ab $$

Rule: For intersecting parabolas of the form $y^2 = k_1 x$ and $x^2 = k_2 y$, the enclosed area is always $\frac{16}{3} \times \text{(coefficient product terms adjustments)}$. Specifically, between $y^2=4ax$ and $x^2=4by$, area is $\frac{16ab}{3}$.

Rapid Elimination via Dimensional Homogeneity

$$ [\text{Area}] = [\text{Length}]^2 $$

Rule: Check the degree of parameters in options. Area terms must always be quadratic in the given linear dimension parameter $a$ or $r$ (e.g., $\pi a^2$, $\frac{3}{8}\pi a^2$). Discard options with incorrect dimensions instantly.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Ignoring Sign Changes in Definite Integrals

Common Pitfall: Integrating a function that dips below the $x$-axis across the interval $[a, b]$ directly as $\int_a^b f(x)\,dx$ without taking absolute values.

Counterexample: $f(x) = \sin x$ on $[0, 2\pi]$

Why it fails: Direct integration yields $\int_0^{2\pi} \sin x\,dx = 0$, completely missing the actual geometric area of $4$ square units.

🛡️ Exam Defense: Always sketch the curve or check roots inside the interval; split into subintervals where $f(x) \ge 0$ and $f(x) \le 0$, taking $\int |f(x)|\,dx$.

⚠️ COMMON ERROR #02

Mistaking Horizontal Integration for Vertical Integration

Common Pitfall: Integrating with respect to $x$ ($dx$) when curves are far easier to express as functions of $y$ ($dy$), leading to complicated radical splits.

Counterexample: Area between $y^2 = x$ and $y = x - 2$

Why it fails: Using $dx$ requires splitting the region into two distinct parts due to the lower boundary changing from the lower half-parabola to the upper half-parabola. Using $dy$ ($x_{right} - x_{left}$) solves it in a single integral.

🛡️ Exam Defense: Inspect whether the region is more naturally 'vertically simple' or 'horizontally simple'.

Comparison

Concept / Curve	Standard Equation	Key Formula / Area	Symmetry Property	Exam Identification Clue
Circle	$x^2 + y^2 = a^2$	$\pi a^2$	Symmetric about both axes and origin	Radial parameter $a$
Ellipse	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$\pi ab$	Symmetric about both axes	Semi-axes $a$ and $b$ present
Parabola	$y^2 = 4ax$	$\frac{8}{3} a^2$ (with latus rectum)	Symmetric about $x$-axis	Single squared variable
Astroid	$x^{2/3} + y^{2/3} = a^{2/3}$	$\frac{3}{8} \pi a^2$	Symmetric in all four quadrants	Fractional powers $2/3$

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the area of the region bounded by the parabola $y^2 = 4x$ and the line $y = 2x$.

(A) \frac{1}{3}(B) \frac{2}{3}(C) \frac{4}{3}(D) \frac{8}{3}

📚 Method A: Formal Textbook Derivation

Step 1: Find intersection points by substituting $y = 2x$ into $y^2 = 4x \implies (2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies x = 0, 1$. Intersection points are $(0,0)$ and $(1,2)$. Step 2: Set up the integral with respect to $y$ or $x$. Using $x$: $\int_{0}^{1} (2\sqrt{x} - 2x)\,dx = \left[ 2 \frac{x^{3/2}}{3/2} - x^2 \right]_0^1 = \frac{4}{3} - 1 = \frac{1}{3}$.

⚡ Method B: 25-Second Exam Shortcut

Apply the direct shortcut for area between $y^2 = 4ax$ and $y = mx$: Area = $\frac{8a^2}{3m^3}$. Here $4a = 4 \implies a = 1$ and $m = 2$. Thus Area = $\frac{8(1)^2}{3(2)^3} = \frac{8}{3(8)} = \frac{1}{3}$.

💡 Tip: Recognize standard line-parabola intersections to deploy formula coefficients instantly.

CANONICAL PROBLEM 02 Dual-Method Breakdown

Find the area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$.

(A) $6\pi$(B) $9\pi$(C) $12\pi$(D) $36\pi$

📚 Method A: Formal Textbook Derivation

Step 1: Identify $a^2 = 9 \implies a = 3$ and $b^2 = 4 \implies b = 2$. Step 2: Use integral Area = $4 \int_{0}^{3} \frac{2}{3}\sqrt{9-x^2}\,dx$. Step 3: Evaluate standard integral to get $4 \times \frac{2}{3} \left[ \frac{x}{2}\sqrt{9-x^2} + \frac{9}{2}\arcsin\frac{x}{3} \right]_0^3 = 4 \times \frac{2}{3} \left( \frac{9}{2} \frac{\pi}{2} \right) = 6\pi$.

⚡ Method B: 25-Second Exam Shortcut

Instantly apply standard ellipse area formula $\text{Area} = \pi a b = \pi (3)(2) = 6\pi$.

💡 Tip: Never integrate ellipses from scratch in competitive exams; use the product of semi-axes times $\pi$.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The area bounded by the curve $y = x^3$, the $x$-axis, and the ordinates $x = -2$ and $x = 2$ is:

A $0$

B $4$

C $8$

D $16$

Derivation & Analysis:

Since $y = x^3$ is an odd function lying below the $x$-axis for $x \in [-2, 0]$ and above for $x \in [0, 2]$, the total physical area is given by $\int_{-2}^{2} |x^3|\,dx = 2 \int_{0}^{2} x^3\,dx = 2 \left[ \frac{x^4}{4} \right]_0^2 = 2 \left( \frac{16}{4} \right) = 8$.

MCQ 2: The area of the circle $x^2 + y^2 = 16$ is:

A $4\pi$

B $8\pi$

C $16\pi$

D $32\pi$

Derivation & Analysis:

For a circle $x^2 + y^2 = r^2$, the radius $r = 4$. Area = $\pi r^2 = \pi (4)^2 = 16\pi$.

MCQ 3: The area enclosed by the parabola $y^2 = 4ax$ and its latus rectum is:

A $\frac{4}{3} a^2$

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

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

D $4a^2$

Derivation & Analysis:

The latus rectum is the line $x = a$. The region is bounded by $y^2 = 4ax$ and $x = a$. Area = $2 \int_{0}^{a} 2\sqrt{a}\sqrt{x}\,dx = 4\sqrt{a} \left[ \frac{x^{3/2}}{3/2} \right]_0^a = \frac{8}{3}a^2$.

MCQ 4: What is the area of the region bounded by $y = |x|$ and $y = 1$?

A $1$

B $2$

C $\frac{1}{2}$

D $4$

Derivation & Analysis:

The curves intersect at $(-1, 1)$ and $(1, 1)$. The region forms a triangle with base $2$ (from $x = -1$ to $1$) and height $1$ (from $y = 0$ to $1$). Area = $\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} (2)(1) = 1$ square unit? Wait, let's check bounds: $y$ goes from $|x|$ to $1$, so $x$ ranges from $-1$ to $1$. $\int_{-1}^{1} (1 - |x|)\,dx = 2 \int_{0}^{1} (1 - x)\,dx = 2 [x - x^2/2]_0^1 = 2(1/2) = 1$. Option A is correct.

MCQ 5: The area of the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ is given by:

A $\frac{3}{8}\pi a^2$

B $\frac{3}{4}\pi a^2$

C $\frac{3}{2}\pi a^2$

D $\pi a^2$

Derivation & Analysis:

Using standard formula for astroids, the total area is $\frac{3}{8}\pi a^2$.

MCQ 6: The area bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$ is:

A $0$

B $2$

C $4$

D $8$

Derivation & Analysis:

The curve $\cos x$ is positive on $[0, \pi/2)$ and $(3\pi/2, 2\pi]$, and negative on $(\pi/2, 3\pi/2)$. Total area = $\int_{0}^{2\pi} |\cos x|\,dx = 4 \int_{0}^{\pi/2} \cos x\,dx = 4 [\sin x]_0^{\pi/2} = 4$.

MCQ 7: The area of the region bounded by $y^2 = x$ and $x^2 = y$ is:

A $\frac{1}{3}$

B $\frac{2}{3}$

C $1$

D $\frac{4}{3}$

Derivation & Analysis:

Intersection points are $(0,0)$ and $(1,1)$. Area = $\int_{0}^{1} (\sqrt{x} - x^2)\,dx = \left[ \frac{2}{3}x^{3/2} - \frac{x^3}{3} \right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$.

MCQ 8: Assertion (A): The area bounded by an even function $f(x)$ symmetric about the $y$-axis on $[-a, a]$ is $2 \int_{0}^{a} f(x)\,dx$. Reason (R): Definite integrals over symmetric intervals obey parity properties.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation of (A)

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are valid properties of definite integrals of even functions.

MCQ 9: The area bounded by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ in the first quadrant is:

A $\pi ab$

B $\frac{1}{2}\pi ab$

C $\frac{1}{4}\pi ab$

D $\frac{1}{8}\pi ab$

Derivation & Analysis:

The total area of the ellipse is $\pi ab$. Since the ellipse is symmetric across all four quadrants, the area in the first quadrant is exactly one-fourth of the total, i.e., $\frac{1}{4}\pi ab$.

Q10

MCQ 10: Area enclosed by the curve $y = e^x$, $x$-axis, and lines $x = 0$, $x = 1$ is:

A $e$

B $e - 1$

C $e + 1$

D $1$

Derivation & Analysis:

Area = $\int_{0}^{1} e^x\,dx = [e^x]_0^1 = e^1 - e^0 = e - 1$.

Common questions

Definite integration computes net signed area where regions below the $x$-axis contribute negative values. When total physical area is required, the absolute value of the function must be integrated, or the domain must be split at roots where the curve crosses the axis.

Choose $dx$ (vertical strips) when the upper and lower bounding functions are easily expressed explicitly as $y = f(x)$. Choose $dy$ (horizontal strips) when the region is bounded on the left and right by functions of $y$ ($x = g(y)$), avoiding unnecessary splitting of radicals.

Yes, standard curves like astroids, cycloids, and ellipses in parametric form appear regularly in PGT exams. Memorizing standard quadrant multipliers and symmetry rules is essential.

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