When should I integrate with respect to $y$ instead of $x$?

Integrate with respect to $y$ using $A = \int_{c}^{d} (x_{right} - x_{left}) \, dy$ whenever expressing $y$ as a function of $x$ results in multiple separate non-smooth sub-regions, or when bounding limits are explicitly horizontal lines $y = c$ and $y = d$.

How do I determine the region for systems of inequalities like $x^2 + y^2 \le 4$ and $y \ge |x|$?

First convert inequalities to equality curves $x^2 + y^2 = 4$ and $y = |x|$ to plot boundary lines. Select a test point such as $(0, 1)$ inside a bounded region and plug it into both inequalities ($0^2 + 1^2 \le 4$ and $1 \ge |0|$). Since both hold true, shade the region containing $(0, 1)$.

What is the key rule for area bounded by parametric equations $x = f(t), y = g(t)$?

When integrating parametric curves, replace $dx$ with $f'(t) dt$ and transform $x$-limits $[a, b]$ into parameter limits $[t_1, t_2]$. Compute area as $A = \int_{t_1}^{t_2} g(t) f'(t) \, dt$, ensuring $t_1$ to $t_2$ traces the boundary in a consistent direction.

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Solved High-Yield Sample Questions

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Question #1JEE Main 2024

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

Solution Locked

Step-by-Step Solution & Concept

Step 1

Find points of intersection by setting $(2x)^2 = 4x \implies 4x^2 - 4x = 0 \implies x = 0$ or $x = 1$ . The corresponding $y$ -values are $0$ and $2$ .

Step 2

Identify the upper boundary curve $y_{upper} = 2\sqrt{x}$ and lower boundary line $y_{lower} = 2x$ for $x \in [0, 1]$ .

Step 3

Integrate the difference: $A = \int_{0}^{1} (2\sqrt{x} - 2x) \, dx = \left[ 2 \cdot \frac{2}{3} x^{3/2} - x^2 \right]_0^1 = \frac{4}{3} - 1 = \frac{1}{3}$ .

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Question #2JEE Main 2023

The area of the region enclosed by the curves $y = |x - 1|$ and $y = 3 - |x|$ is equal to:

Solution Locked

Step-by-Step Solution & Concept

Step 1

Sketch both V-shaped curves. $y = |x - 1|$ has a vertex at $(1, 0)$ pointing upwards, while $y = 3 - |x|$ has a vertex at $(0, 3)$ pointing downwards.

Step 2

Calculate intersection points: Solve $|x - 1| = 3 - |x|$ . For $x < 0$ , $1 - x = 3 + x \implies 2x = -2 \implies x = -1$ . For $x \ge 1$ , $x - 1 = 3 - x \implies 2x = 4 \implies x = 2$ . Intersection bounds are $x = -1$ and $x = 2$ .

Step 3

Integrate upper minus lower curve: $A = \int_{-1}^{2} ((3 - |x|) - |x - 1|) \, dx$ . Splitting at critical points $x = 0$ and $x = 1$ yields total area $A = 4$ square units.

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Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Area Bounded by $y = f(x)$ and $x$-axis

Area Between Two Curves $y = f(x)$ and $y = g(x)$

Area Bounded along $y$-axis

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

Area Enclosed by Standard Ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

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