What is the Areas topic in JEE Mathematics?

The Areas topic is part of Applications of Integrals. It deals with calculating the area enclosed by curves, straight lines, coordinate axes, and other boundaries using definite integration.

Which concepts are important for JEE Areas questions?

Important concepts include the area under a curve, area between two curves, intersection points, definite integrals, symmetry, piecewise integration, and choosing whether to integrate with respect to x or y.

How many questions are asked from Areas in JEE Main?

JEE Main generally includes zero to one direct question from Areas or Applications of Integrals. The topic may contribute up to four marks when tested directly.

Are Areas questions important for JEE Advanced?

Yes. JEE Advanced may combine Areas with definite integration, coordinate geometry, functions, and inequalities. These questions often require students to identify boundaries and divide the region into multiple parts.

How do I find the area between two curves?

First, calculate the intersection points of the curves. When integrating with respect to x, subtract the lower curve from the upper curve. When integrating with respect to y, subtract the left boundary from the right boundary.

Why should I draw a graph before solving an Areas question?

A rough graph helps identify the required region, intersection points, integration limits, and relative positions of the curves. It also shows whether the integral must be divided into multiple intervals.

What are the common mistakes in JEE Areas questions?

Common mistakes include using incorrect limits, subtracting the curves in the wrong order, overlooking a change in sign, missing symmetry, and treating a negative definite integral as a negative geometric area.

How can I improve my speed in solving Areas questions?

Practise sketching standard curves, memorise important definite integration properties, calculate intersection points carefully, and use symmetry whenever possible. Solving previous-year questions also improves boundary identification and calculation speed.

JEE Areas Questions

Areas is an important application-based topic in JEE Mathematics that connects definite integration with the graphical interpretation of curves. It focuses on calculating the area enclosed by curves, coordinate axes, straight lines, and other boundaries. Questions from this topic require students to identify the correct region, determine intersection points, choose suitable limits, and form the appropriate definite integral. The chapter covers the area under a curve, area between two curves, regions divided by the coordinate axes, integration with respect to x or y, symmetry-based calculations, and piecewise integration. JEE Main generally tests direct area calculations involving standard curves, while JEE Advanced may combine multiple boundaries, inequalities, symmetry, and non-standard regions. Regularly solving topic-wise JEE Questions helps students improve their graphical understanding and translate bounded regions into definite integrals accurately.

Question 1 JEE Mains

What is the area bounded by $$y=x^3-7x^2+14x-8$$ and the x-axis in the interval $$[0,2]$$?

$$2$$

$$5/2$$

$$3$$

$$7/2$$

Solution

Factor the polynomial to find where the curve meets the x axis:

$$y = x^3 - 7x^2 + 14x - 8$$

$$y = (x - 1)(x - 2)(x - 4)$$

The required interval goes from zero to two.

The roots of the polynomial inside this range are at $$x = 1$$ and $$x = 2$$.

The curve changes sign at $$x = 1$$:

From zero to one the function is negative so we take the negative of the integral
From one to two the function is positive so we take the standard integral

Set up the area calculation:

$$\text{Area} = -\int_0^1 (x^3 - 7x^2 + 14x - 8) \, dx + \int_1^2 (x^3 - 7x^2 + 14x - 8) \, dx$$

Find the antiderivative function:

$$F(x) = \frac{x^4}{4} - \frac{7x^3}{3} + 7x^2 - 8x$$

Evaluate the first integral from zero to one:

$$-\left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = -\left( \frac{3 - 28 - 12}{12} \right) = \frac{37}{12}$$

Evaluate the second integral from one to two:

$$\left( 4 - \frac{56}{3} + 28 - 16 \right) - \left( \frac{1}{4} - \frac{7}{3} + 7 - 8 \right) = \left( 16 - \frac{56}{3} \right) - \left( -\frac{37}{12} \right)$$

$$= -\frac{8}{3} + \frac{37}{12} = \frac{-32 + 37}{12} = \frac{5}{12}$$

Sum both portions to get the total area:

$$\text{Total Area} = \frac{37}{12} + \frac{5}{12} = \frac{42}{12} = \frac{7}{2}$$

The correct option is D.

Question 2 JEE Mains

Let $$A_{1}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,x+y=8$$ and y-axis that lies in the first quadrant. Let $$A_{2}$$ be the bounded area enclosed by the curves $$y=x^{2}+2,y^{2}=x,x=2$$ and y-axis that lies in the first quadrant. Then $$A_{1}-A_{2}$$ is equal to

$$\frac{2}{3}(2\sqrt{2}+1)$$

$$\frac{2}{3}(\sqrt{2}+1)$$

$$\frac{2}{3}(4\sqrt{2}+1)$$

$$\frac{2}{3}(3\sqrt{2}+1)$$

Solution

Given curves for $$A_1$$:

$$y = x^2 + 2, \quad x + y = 8 \implies y = 8 - x, \quad x = 0 \quad (\text{y-axis})$$

Finding the point of intersection of $$y = x^2 + 2$$ and $$y = 8 - x$$:

$$x^2 + 2 = 8 - x \implies x^2 + x - 6 = 0 \implies (x+3)(x-2) = 0 \implies x = 2 \quad (\text{since } x \ge 0)$$

$$A_1 = \int_{0}^{2} \left[ (8 - x) - (x^2 + 2) \right] dx = \int_{0}^{2} (6 - x - x^2) dx$$

$$A_1 = \left[ 6x - \frac{x^2}{2} - \frac{x^3}{3} \right]_{0}^{2} = 12 - 2 - \frac{8}{3} = 10 - \frac{8}{3} = \frac{22}{3}$$

Given curves for $$A_2$$:

$$y = x^2 + 2, \quad y^2 = x \implies y = \sqrt{x}, \quad x = 2, \quad x = 0$$

$$A_2 = \int_{0}^{2} \left[ (x^2 + 2) - \sqrt{x} \right] dx$$

$$A_2 = \left[ \frac{x^3}{3} + 2x - \frac{2}{3}x^{3/2} \right]_{0}^{2} = \frac{8}{3} + 4 - \frac{2}{3}(2\sqrt{2}) = \frac{20}{3} - \frac{4\sqrt{2}}{3}$$

$$A_1 - A_2 = \frac{22}{3} - \left( \frac{20}{3} - \frac{4\sqrt{2}}{3} \right) = \frac{2}{3} + \frac{4\sqrt{2}}{3} = \frac{2}{3}(2\sqrt{2} + 1)$$

Question 1 JEE Mains

Let the area of the region $$\{(x, y) : x - 2y + 4 \geq 0, \; x + 2y^2 \geq 0, \; x + 4y^2 \leq 8, \; y \geq 0\}$$ be $$\frac{m}{n}$$, where $$m$$ and $$n$$ are coprime numbers. Then $$m + n$$ is equal to _______.

Solution

We need the area of the region defined by:

1) $$x - 2y + 4 \geq 0$$ (i.e., $$x \geq 2y - 4$$)

2) $$x + 2y^2 \geq 0$$ (i.e., $$x \geq -2y^2$$)

3) $$x + 4y^2 \leq 8$$ (i.e., $$x \leq 8 - 4y^2$$)

4) $$y \geq 0$$

Let us integrate with respect to $$y$$. From condition 3: $$x \leq 8 - 4y^2$$, so $$y \leq \sqrt{2}$$ (for $$x \geq 0$$, max).

The right boundary is $$x = 8 - 4y^2$$.

The left boundary is $$x = \max(2y - 4, -2y^2)$$.

For $$y \geq 0$$: When is $$2y - 4 \geq -2y^2$$?

$$2y^2 + 2y - 4 \geq 0$$, i.e., $$y^2 + y - 2 \geq 0$$, i.e., $$(y+2)(y-1) \geq 0$$. Since $$y \geq 0$$: $$y \geq 1$$.

So left boundary = $$-2y^2$$ for $$0 \leq y \leq 1$$ and $$2y - 4$$ for $$y \geq 1$$.

Upper limit from right boundary: $$8 - 4y^2 \geq 2y - 4$$ gives $$4y^2 + 2y - 12 \leq 0$$, $$2y^2 + y - 6 \leq 0$$, $$(2y - 3)(y + 2) \leq 0$$, so $$y \leq 3/2$$.

Also $$8 - 4y^2 \geq -2y^2$$ gives $$2y^2 \leq 8$$, $$y \leq 2$$.

Area = $$\int_0^1 (8 - 4y^2 - (-2y^2)) dy + \int_1^{3/2} (8 - 4y^2 - (2y - 4)) dy$$

$$= \int_0^1 (8 - 2y^2) dy + \int_1^{3/2} (12 - 4y^2 - 2y) dy$$

First integral: $$\left[8y - \frac{2y^3}{3}\right]_0^1 = 8 - \frac{2}{3} = \frac{22}{3}$$

Second integral: $$\left[12y - \frac{4y^3}{3} - y^2\right]_1^{3/2}$$

At $$y = 3/2$$: $$18 - \frac{4 \cdot 27/8}{3} - \frac{9}{4} = 18 - \frac{9}{2} - \frac{9}{4} = 18 - \frac{27}{4} = \frac{45}{4}$$

At $$y = 1$$: $$12 - \frac{4}{3} - 1 = 11 - \frac{4}{3} = \frac{29}{3}$$

Second integral = $$\frac{45}{4} - \frac{29}{3} = \frac{135 - 116}{12} = \frac{19}{12}$$

Total area = $$\frac{22}{3} + \frac{19}{12} = \frac{88 + 19}{12} = \frac{107}{12}$$

$$m + n = 107 + 12 = 119$$.

The answer is $$\boxed{119}$$.

Question 1 JEE Mains

If the tangent at a point P on the parabola $$y^2 = 3x$$ is parallel to the line $$x + 2y = 1$$ and the tangents at the points Q and R on the ellipse $$\frac{x^2}{4} + \frac{y^2}{1} = 1$$ are perpendicular to the line $$x - y = 2$$, then the area of the triangle $$PQR$$ is:

$$\frac{9}{\sqrt{5}}$$

$$5\sqrt{3}$$

$$\frac{3}{2}\sqrt{5}$$

$$3\sqrt{5}$$

Solution

We need to find the area of triangle PQR. To determine point P on the parabola $$y^2 = 3x$$, note that the tangent at point $$(at^2, 2at)$$ on $$y^2 = 4ax$$ has slope $$\frac{1}{t}$$. Here $$4a = 3$$, so $$a = \frac{3}{4}$$. The line $$x + 2y = 1$$ has slope $$-\frac{1}{2}$$. Thus $$\frac{1}{t} = -\frac{1}{2}$$, giving $$t = -2$$. It follows that $$P = \left(\frac{3}{4} \times 4, 2 \times \frac{3}{4} \times (-2)\right) = (3, -3)$$.

Next, we find points Q and R on the ellipse $$\frac{x^2}{4} + y^2 = 1$$. Lines perpendicular to $$x - y = 2$$ (slope 1) have slope $$-1$$. The general tangent to the ellipse is given by $$y = mx \pm \sqrt{a^2m^2 + b^2}$$ where $$a^2 = 4, b^2 = 1$$. Hence the tangents with slope -1 are $$y = -x \pm \sqrt{4 + 1} = -x \pm \sqrt{5}$$.

For the tangent $$y = -x + \sqrt{5}$$, substituting into the ellipse equation yields

$$\frac{x^2}{4} + (-x + \sqrt{5})^2 = 1$$

$$\frac{x^2}{4} + x^2 - 2\sqrt{5}x + 5 = 1$$

$$\frac{5x^2}{4} - 2\sqrt{5}x + 4 = 0$$

$$5x^2 - 8\sqrt{5}x + 16 = 0$$

$$x = \frac{8\sqrt{5} \pm \sqrt{320 - 320}}{10} = \frac{8\sqrt{5}}{10} = \frac{4\sqrt{5}}{5}$$.

Then

$$y = -\frac{4\sqrt{5}}{5} + \sqrt{5} = \frac{\sqrt{5}}{5} = \frac{1}{\sqrt{5}}$$, so $$Q = \left(\frac{4\sqrt{5}}{5}, \frac{\sqrt{5}}{5}\right) = \left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$. Similarly, for the tangent $$y = -x - \sqrt{5}$$ one finds $$R = \left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$.

Finally, the area of triangle PQR with vertices $$P(3,-3)$$, $$Q\left(\frac{4}{\sqrt{5}}, \frac{1}{\sqrt{5}}\right)$$, and $$R\left(-\frac{4}{\sqrt{5}}, -\frac{1}{\sqrt{5}}\right)$$ is given by

Area $$= \frac{1}{2}|x_P(y_Q - y_R) + x_Q(y_R - y_P) + x_R(y_P - y_Q)|$$

$$= \frac{1}{2}\left|3 \cdot \frac{2}{\sqrt{5}} + \frac{4}{\sqrt{5}}\left(-\frac{1}{\sqrt{5}} - (-3)\right) + \left(-\frac{4}{\sqrt{5}}\right)\left(-3 - \frac{1}{\sqrt{5}}\right)\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4}{\sqrt{5}} \cdot \frac{-1 + 3\sqrt{5}}{\sqrt{5}} - \frac{4}{\sqrt{5}} \cdot \frac{-3\sqrt{5} - 1}{\sqrt{5}}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{4(-1+3\sqrt{5})}{5} + \frac{4(3\sqrt{5}+1)}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{-4+12\sqrt{5}+12\sqrt{5}+4}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24\sqrt{5}}{5}\right|$$

$$= \frac{1}{2}\left|\frac{6}{\sqrt{5}} + \frac{24}{\sqrt{5}}\right| = \frac{1}{2} \cdot \frac{30}{\sqrt{5}} = \frac{15}{\sqrt{5}} = 3\sqrt{5}$$.

Therefore, the area is $$\boxed{3\sqrt{5}}$$, which corresponds to Option D.

Areas Topic Overview

Parameter	Details
Topic Name	Areas
Subject	Mathematics
Chapter	Applications of Integrals
JEE Main Weightage	~3-4% (0-1 direct question on average)
JEE Advanced Weightage	~3-5% (often combined with Calculus or Coordinate Geometry)
Difficulty Level	Moderate
Important Concepts	Area Under Curves, Area Between Curves, Intersection Points, Symmetry, Piecewise Integration
Recommended Practice Level	High – attempt 50+ mixed problems

Why Practice JEE Areas Questions?

Direct scoring potential: Questions based on standard curves and definite integrals can be solved quickly when the graph is understood correctly.
Graphical application: The topic strengthens the connection between algebraic equations, graphs, and integration.
Calculus integration: Areas questions test concepts from functions, limits, definite integration, and coordinate geometry together.
Boundary identification: Regular practice helps students recognise which curve lies above, below, to the left, or to the right.
Symmetry advantage: Symmetric regions often reduce lengthy calculations and make solutions faster.
Advanced problem-solving: JEE Advanced may describe a region through inequalities instead of displaying a direct graph.
Pattern familiarity: Solving JEE Mains previous papers helps students understand frequently tested curve combinations and integration patterns.

Important Concepts and Subtopics

Concept	Importance	Difficulty Level	Frequently Asked In
Area Under a Curve	Very High	Easy-Moderate	JEE Main
Area Between Two Curves	Very High	Moderate	JEE Main and Advanced
Intersection Points and Limits	Very High	Moderate	JEE Main and Advanced
Area Bounded by Curves and Axes	High	Moderate	JEE Main
Integration with Respect to x or y	High	Moderate	JEE Main and Advanced
Symmetry of Bounded Regions	High	Moderate	JEE Main and Advanced
Piecewise Area Calculation	Very High	Moderate-High	JEE Main and Advanced
Regions Defined by Inequalities	High	High	JEE Advanced

Preparation Strategy for JEE Areas

Concept learning: Begin by understanding the geometric meaning of a definite integral. Learn how the sign of a function affects the integral and why geometric area must always be taken as positive. Then study the area between a curve and an axis, followed by the area enclosed between two curves. Practise sketching lines, parabolas, circles, and other standard curves before setting up integrals.

Graphical preparation: Always sketch a rough graph before integrating. Mark the intersection points and identify the upper and lower curves when integrating with respect to x. When integrating with respect to y, determine the right and left boundaries. A structured JEE Maths Course can help students develop this graphical approach through guided examples and progressively challenging problems.

Formula revision: Revise the standard equations and shapes of straight lines, parabolas, circles, ellipses, and basic trigonometric curves. Keep definite integration properties, standard integrals, and symmetry rules together for quick reference. A concise JEE Mains formula sheet is useful for revising these results before practice sessions and examinations.

Practice method: Start with questions involving a single curve and an axis. Progress to areas between two curves, regions that require splitting into intervals, and problems defined through inequalities. Reliable JEE Study Material should include diagrams, solved examples, previous-year questions, and mixed exercises involving both x-based and y-based integration.

Common mistakes: Students often use incorrect limits, subtract the lower curve from the upper curve in the wrong order, overlook a change in the sign of the function, or calculate signed area instead of geometric area. Another common mistake is using a single integral when the boundaries change within the required interval.

Exam strategy: Draw the region before starting the calculation. If the graph and boundaries are immediately clear, attempt the question early. If the region involves several inequalities or changing boundaries, mark it for a later review and return after completing more direct questions.

JEE Main and Advanced Weightage Analysis

Exam	Average Questions	Expected Marks
JEE Main	0-1 direct question	Up to 4 marks
JEE Advanced	0-1 question, often integrated with other topics	Varies with the paper pattern

Areas is generally tested as part of Applications of Integrals in JEE Main. Questions commonly involve polynomial curves, straight lines, parabolas, coordinate axes, and definite integrals. In JEE Advanced, the region may be described through multiple curves or inequalities, requiring students to identify boundaries before forming the integral.

Tips to Solve Areas Questions Faster

Sketch the curves and shade the required region before setting up the integral.
Calculate all relevant intersection points to determine the correct limits.
Use upper curve minus lower curve when integrating with respect to x.
Use right boundary minus left boundary when integrating with respect to y.
Split the integral whenever the curves exchange positions or the function changes sign.
Use symmetry to calculate one part of the region and multiply it appropriately.
Check whether integrating with respect to x or y produces the simpler expression.
Remember that geometric area is positive even when the definite integral is negative.

After learning the standard methods, solve timed sets containing both direct and graph-based questions. This develops the visual interpretation, boundary selection, and calculation speed required to handle Areas questions confidently in JEE Main and JEE Advanced.

JEE Areas Questions

Areas Topic Overview

Why Practice JEE Areas Questions?

Important Concepts and Subtopics

Preparation Strategy for JEE Areas

JEE Main and Advanced Weightage Analysis

Tips to Solve Areas Questions Faster

Frequently Asked Questions

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