Calculus

Area between curves

The computation of the area between curves relies on definite integration by partitioning regions into infinitesimal vertical or horizontal strips. This topic covers intersection points, curve sketching, and proper choice of integration variables (with respect to $x$ or $y$).

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 Between Curves

Let $f(x)$ and $g(x)$ be continuous functions on the closed interval $[a, b]$ such that $f(x) \ge g(x)$ for all $x \in [a, b]$. The net area $A$ of the region bounded by the curves $y = f(x)$, $y = g(x)$, and the lines $x = a$ and $x = b$ is given by $$A = \int_{a}^{b} [f(x) - g(x)]\,dx$$.

📐 Geometric Interpretation

Geometrically, the integral accumulates the areas of infinitesimal vertical rectangles of height $f(x) - g(x)$ and width $dx$ from $x = a$ to $x = b$. If the curves intersect within $[a, b]$, the interval must be partitioned at the intersection roots where $f(x) = g(x)$, integrating $|f(x) - g(x)|$ across sub-intervals.

⚙️ Algebraic & Structural Meaning

Algebraically, finding the exact boundaries $a$ and $b$ requires solving the equation $f(x) = g(x)$ to determine the abscissae of the points of intersection. The absolute difference ensures non-negativity of the calculated geometric measure.

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 bounding functions	`[a, b] (Closed interval)`	Strictly Mandatory / Necessary	If either function possesses a jump or infinite discontinuity, the Riemann integral fails to exist in the standard sense, rendering the net area undefined or requiring improper integration techniques.
Non-negativity of integrand magnitude	`[a, b]`	Strictly Mandatory	Ignoring intersection points where the upper and lower curves switch roles results in negative cancellation, yielding signed area instead of total geometric area.
Finiteness of integration limits	`Finite bounds $a$ and $b$`	Essential for conclusion	Unbounded domains lead to improper integrals that may diverge, yielding infinite areas unless evaluated via limits.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Area Bounded by a Parabola and a Line

$$A = \frac{(beta - \alpha)^3}{6a}$$

Rule: For the area bounded by a parabola $y^2 = 4ax$ (or $y = ax^2 + bx + c$) and a line intersecting it at two points with abscissae $\alpha$ and $\beta$, the area is given directly by the cubic difference formula divided by $6a$ (where $a$ is the leading coefficient).

Archimedes' Parabola Theorem

$$A = \frac{2}{3} \times (base \times height)$$

Rule: The area enclosed by a parabola and any chord is exactly two-thirds of the area of the circumscribed parallelogram formed by the chord and the tangent parallel to the chord.

Application: Use this result in the stated type of problem.

Area of Standard Ellipse

$$A = \pi ab$$

Rule: The total area enclosed by the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is $\pi ab$. For shifted ellipses, the area remains invariant under translation.

Application: Standard baseline question appearing as a direct formula or preliminary step in composite curve areas.

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

$$A = \frac{16ab}{3}$$

Rule: The region enclosed by symmetric standard parabolas $y^2 = 4ax$ and $x^2 = 4by$ has a fixed area ratio calculated directly from parameters $a$ and $b$.

Application: Directly featured in DSSSB PGT Mathematics multiple-choice questions.

Symmetry Exploitation (Even/Odd Functions)

$$ \int_{-a}^{a} f(x)\,dx = 2 \int_{0}^{a} f(x)\,dx \text{ (if even)} $$

Rule: When calculating areas bounded symmetrically about the $y$-axis, compute the integral from $0$ to $a$ and multiply by 2.

Application: Saves up to 45 seconds on trigonometric and polynomial curve bounding problems.

Integration with respect to $y$ ($dy$ strips)

$$A = \int_{c}^{d} [x_{right}(y) - x_{left}(y)]\,dy$$

Rule: If curves are expressed more cleanly as functions of $y$ (e.g., parabolas opening left/right), integrate horizontally with respect to $y$ to avoid splitting into multiple $x$-integrals.

Application: Essential strategy to eliminate algebraic complexity in composite regions.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Intersection Blindness (Crossing Curves)

Common Pitfall: Integrating $\int_a^b (f(x) - g(x))\,dx$ without checking if $f(x)$ and $g(x)$ cross each other inside $[a, b]$.

Counterexample: $y = \sin x$ and $y = \cos x$ on $[0, \pi/2]$

Why it fails: At $x = \pi/4$, $\sin x = \cos x$. Integrating without splitting gives a net signed area of zero instead of the true geometric area.

🛡️ Exam Defense: Always solve $f(x) = g(x)$ across the domain to locate interior intersection nodes and split integrals accordingly.

⚠️ COMMON ERROR #02

Confusing Vertical and Horizontal Strips

Common Pitfall: Forcing $dx$ integration on curves defined naturally as $x = f(y)$, leading to difficult square roots and multiple piecewise integrals.

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

Why it fails: Using $dx$ requires splitting the region into two parts because the lower boundary of the region changes from the lower branch of the parabola to the upper branch. Using $dy$ unifies it into a single integral.

🛡️ Exam Defense: Examine whether $x$ is easily expressed as a function of $y$. If yes, integrate with respect to $y$ ($dy$).

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Vertical Strip Area	Continuity of $f, g$ on $[a, b]$	$$ \int_{a}^{b} [f(x) - g(x)]\,dx $$	Sum of vertical rectangles between upper and lower curves	Explicit functions $y = f(x)$ with clear top and bottom bounds
Horizontal Strip Area	Continuity of $u, v$ on $[c, d]$	$$ \int_{c}^{d} [x_{R}(y) - x_{L}(y)]\,dy $$	Sum of horizontal rectangles between right and left curves	Explicit relations $x = f(y)$ or parabolas opening sideways
Parametric Area	Smooth curves $x = x(t), y = y(t)$	$$ \int_{t_1}^{t_2} y(t) x'(t)\,dt $$	Area swept under parametric trajectory	Equations given in terms of parameter $t$ (e.g., circle, astroid)
Polar Area	Continuous r($\theta$) on $[\alpha, \beta]$	$$ \frac{1}{2} \int_{\alpha}^{\beta} r^2\,d\theta $$	Sum of infinitesimal circular sectors	Equations in polar coordinates ($r, \theta$) like cardioids, roses

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) $\frac{1}{3}$ sq. units(B) $\frac{2}{3}$ sq. units(C) $\frac{4}{3}$ sq. units(D) $\frac{8}{3}$ sq. units

📚 Method A: Formal Textbook Derivation

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

⚡ Method B: 25-Second Exam Shortcut

Use Shortcut 1 for $y^2 = 4ax$ and $y = mx$: Here $4a = 4 \implies a = 1$. The line is $y = 2x \implies m = 2$. The area formula for $y^2 = 4ax$ and $y = mx$ is $A = \frac{8a^2}{3m^3} = \frac{8(1)^2}{3(2)^3} = \frac{8}{24} = \frac{1}{3}$.

💡 Tip: Memorize standard parameter formulas for parabola-line intersections to solve.

CANONICAL PROBLEM 02 Dual-Method Breakdown

The area of the region bounded by the curve $y = x^3$ and the lines $y = x$ for $x \ge 0$ is:

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

📚 Method A: Formal Textbook Derivation

Step 1: Find intersection points: $x^3 = x \implies x(x^2 - 1) = 0 \implies x = -1, 0, 1$. Given $x \ge 0$, integration limits are $0$ to $1$. Step 2: Check upper curve on $[0,1]$: For $x = 0.5$, $x = 0.5$ and $x^3 = 0.125$, so line $y = x$ is above $y = x^3$. Step 3: Compute integral: $A = \int_{0}^{1} (x - x^3)\,dx = \left[ \frac{x^2}{2} - \frac{x^4}{4} \right]_0^1 = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.

⚡ Method B: 25-Second Exam Shortcut

Recognize polynomial intersection bounds. For $y = x^n$ and $y = x$, area between $0$ and $1$ is $\frac{1}{2} - \frac{1}{n+1} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$.

💡 Tip: Pay close attention to domain restrictions like $x \ge 0$ to avoid integrating over negative regions unnecessarily.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: The area bounded by the curve $y = x^2$ and the line $y = 4$ is:

A $\frac{32}{3}$

B $\frac{16}{3}$

C $\frac{8}{3}$

D $\frac{64}{3}$

Derivation & Analysis:

The parabola $y = x^2$ intersects $y = 4$ at $x = -2$ and $x = 2$. By symmetry, $A = 2 \int_{0}^{2} (4 - x^2)\,dx = 2 \left[ 4x - \frac{x^3}{3} \right]_0^2 = 2 \left( 8 - \frac{8}{3} \right) = 2 \left( \frac{16}{3} \right) = \frac{32}{3}$.

MCQ 2: What is the area of the circle $x^2 + y^2 = a^2$?

A $2\pi a^2$

B $\pi a^2$

C $\frac{1}{2}\pi a^2$

D $\pi a$

Derivation & Analysis:

Using symmetry across four quadrants, $A = 4 \int_{0}^{a} \sqrt{a^2 - x^2}\,dx = 4 \left[ \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) \right]_0^a = 4 \left( \frac{a^2}{2} \cdot \frac{\pi}{2} \right) = \pi a^2$.

MCQ 3: The area enclosed by the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ is:

A $6\pi$

B $12\pi$

C $3\pi$

D $36\pi$

Derivation & Analysis:

Using the standard ellipse area formula $A = \pi ab$, where $a = 3$ and $b = 2$, we get $A = \pi(3)(2) = 6\pi$.

MCQ 4: The area of the region bounded by $y = \sin x$, $x = 0$, and $x = 2\pi$ is:

A $0$

B $2$

C $4$

D $2\pi$

Derivation & Analysis:

Sine wave area over one period $[0, \pi]$ is $2$, and over $[\pi, 2\pi]$ is $2$ (taking absolute value). Thus total geometric area is $\int_{0}^{\pi} \sin x\,dx + \int_{\pi}^{2\pi} -\sin x\,dx = 2 + 2 = 4$.

MCQ 5: The area of the region bounded by $y^2 = 9x$ and $y = 3x$ is:

A $\frac{1}{2}$

B $\frac{1}{3}$

C $\frac{1}{4}$

D $\frac{1}{6}$

Derivation & Analysis:

Here $4a = 9 \implies a = 9/4$, and line slope $m = 3$. Using shortcut $A = \frac{8a^2}{3m^3} = \frac{8(81/16)}{3(27)} = \frac{8 \times 81}{16 \times 3 \times 27} = \frac{648}{1296}$ wait, let's recompute: $y^2 = 9x, y = 3x \implies (3x)^2 = 9x \implies 9x^2 - 9x = 0 \implies x = 0, 1$. Limits $0$ to $1$. $A = \int_0^1 (3\sqrt{x} - 3x)\,dx = 3 [\frac{2}{3}x^{3/2} - \frac{x^2}{2}]_0^1 = 3(\frac{2}{3} - \frac{1}{2}) = 3(\frac{1}{6}) = \frac{1}{2}$. Let's verify options: option (A) is $1/2$.

MCQ 6: The area bounded by the curves $y = \ln x$, $x = 1$, $x = e$, and the $x$-axis is:

A $1$

B $e - 1$

C $e$

D $1 - e$

Derivation & Analysis:

Area $A = \int_{1}^{e} \ln x\,dx = [x\ln x - x]_1^e = (e\ln e - e) - (1\ln 1 - 1) = (e - e) - (0 - 1) = 1$.

MCQ 7: The area of the region bounded by $|x| + |y| = 1$ is:

A $1$

B $2$

C $\sqrt{2}$

D $4$

Derivation & Analysis:

The equation $|x| + |y| = 1$ represents a square rotated by $45^\circ$ with vertices $(1,0), (0,1), (-1,0), (0,-1)$. The diagonals have lengths $d_1 = 2$ and $d_2 = 2$. Area of square = $\frac{1}{2} d_1 d_2 = \frac{1}{2}(2)(2) = 2$. Alternatively, $4 \times \text{Area in 1st quadrant} = 4 \int_0^1 (1-x)\,dx = 4(1 - 1/2) = 2$.

MCQ 8: The area bounded by the parabola $x^2 = 4y$ and the line $x = 4y - 2$ is:

A $\frac{9}{8}$

B $\frac{9}{4}$

C $\frac{3}{4}$

D $\frac{9}{2}$

Derivation & Analysis:

Rewrite line as $4y = x + 2$. Substitute $4y$ into parabola: $x^2 = x + 2 \implies x^2 - x - 2 = 0 \implies (x-2)(x+1) = 0 \implies x = -1, 2$. Using integration with respect to $x$: $A = \int_{-1}^{2} \left( \frac{x+2}{4} - \frac{x^2}{4} \right)\,dx = \frac{1}{4} \int_{-1}^{2} (2 + x - x^2)\,dx = \frac{1}{4} \left[ 2x + \frac{x^2}{2} - \frac{x^3}{3} \right]_{-1}^2 = \frac{1}{4} \left( \frac{9}{2} \right) = \frac{9}{8}$.

MCQ 9: The area of the region bounded by $y = |x|-1$ and $y = 1-|x|$ is:

A $1$

B $2$

C $4$

D $8$

Derivation & Analysis:

The curves form a rhombus with vertices $(0,1), (1,0), (0,-1), (-1,0)$. The diagonals are along the coordinate axes with lengths $2$ and $2$. Area = $\frac{1}{2} \times 2 \times 2 = 2$.

Q10

MCQ 10: The area enclosed by the astroid $x^{2/3} + y^{2/3} = a^{2/3}$ is:

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

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

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

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

Derivation & Analysis:

By parametric integration with $x = a\cos^3\theta, y = a\sin^3\theta$, the total area is $4 \int_0^{\pi/2} y \frac{dx}{d\theta}\,d\theta = 4 \int_0^{\pi/2} (a\sin^3\theta)(3a\cos^2\theta(-\sin\theta))\,d\theta = \frac{3}{8}\pi a^2$.

Common questions

Choose $dx$ integration if the upper and lower boundaries are functions of $x$ ($y = f(x)$) and the region has vertical slice continuity. Choose $dy$ integration if the region is bounded on the left and right by functions of $y$ ($x = g(y)$), or if $dx$ requires splitting the region into multiple unnecessary sub-integrals.

Definite integrals compute net signed area where regions below the $x$-axis contribute negatively. To find true geometric area between two curves that cross, you must compute $\int |f(x) - g(x)|\,dx$ by splitting the integral at intersection points.

Yes! Always check for symmetry (even/odd functions or symmetric geometric shapes like circles, ellipses, and astroids) to reduce integration limits from $[-a, a]$ to $[0, a]$ and multiply by 2 or 4.

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