IIT JEE Main Maths -Unit 8- Area under curves- Study Notes-New Syllabus

IIT JEE Main Maths -Unit 8- Area under curves – Study Notes – New syllabus

IIT JEE Main Maths -Unit 8- Area under curves – Study Notes -IIT JEE Main Maths – per latest Syllabus.

Key Concepts:

Area under curves

IIT JEE Main Maths -Study Notes – All Topics

Area Under a Curve – Geometrical Meaning and Evaluation using Definite Integrals

One of the most important applications of definite integrals is to find the area under a curve. If a function $ y = f(x) $ is continuous and non-negative on an interval $[a, b]$, then the area bounded by the curve, the x-axis, and the vertical lines $ x = a $ and $ x = b $ is given by the definite integral:

$ A = \int_a^b f(x)\,dx $

Geometrical Meaning of Definite Integral

The definite integral $ \displaystyle \int_a^b f(x)\,dx $ represents the net signed area under the curve $ y = f(x) $ from $ x = a $ to $ x = b $.

If $ f(x) \ge 0 $ → area is positive (above x-axis).
If $ f(x) \le 0 $ → area is negative (below x-axis).
If the curve crosses the x-axis, areas above and below cancel out algebraically.

Formula for Area Under a Curve

$ A = \int_a^b |f(x)|\,dx $

Here, we take the absolute value of $ f(x) $ when the curve lies below the x-axis to ensure total area is always positive.

Common Situations for Area Evaluation

Situation	Formula for Area
Area between curve and x-axis	$ A = \displaystyle \int_a^b \|f(x)\|\,dx $
Area between two curves $ y_1 = f(x) $ and $ y_2 = g(x) $	$ A = \displaystyle \int_a^b \|f(x) – g(x)\|\,dx $
Area bounded by a curve and y-axis	$ A = \displaystyle \int_{y_1}^{y_2} x\,dy $

Steps to Find Area Under a Curve

Sketch the curve to identify the region clearly.
Determine the points of intersection or limits of integration.
Decide whether the function lies above or below the x-axis.
Set up the appropriate definite integral.
Integrate and simplify to get the required area.

Example

Find the area under the curve $ y = x^2 $ between $ x = 0 $ and $ x = 2 $.

▶️ Answer / Explanation

Step 1: Since $ y = x^2 \ge 0 $ in [0, 2], area = $ \displaystyle \int_0^2 x^2\,dx. $

Step 2: Integrate: $ \displaystyle \int x^2\,dx = \dfrac{x^3}{3}. $

Step 3: Apply limits: $ \left[\dfrac{x^3}{3}\right]_0^2 = \dfrac{8}{3}. $

Answer: $ \dfrac{8}{3} \text{ square units.} $

Example

Find the area enclosed by $ y = \sin x $ and the x-axis between $ x = 0 $ and $ x = \pi $.

▶️ Answer / Explanation

Step 1: $ y = \sin x \ge 0 $ in [0, π].

Step 2: $ A = \displaystyle \int_0^{\pi} \sin x\,dx = [-\cos x]_0^{\pi} = -(-1) + (1) = 2. $

Answer: $ 2 \text{ square units.} $

Example

Find the area enclosed between $ y = x^2 $ and $ y = 4x – x^2 $.

▶️ Answer / Explanation

Step 1: Find intersection points by equating:

$ x^2 = 4x – x^2 \Rightarrow 2x^2 – 4x = 0 \Rightarrow x(2x – 4) = 0 \Rightarrow x = 0, 2. $

Step 2: Since $ 4x – x^2 \ge x^2 $ between 0 and 2, $ A = \displaystyle \int_0^2 [(4x – x^2) – x^2]\,dx = \int_0^2 (4x – 2x^2)\,dx. $

Step 3: Integrate:

$ \int (4x – 2x^2)\,dx = 2x^2 – \dfrac{2x^3}{3}. $

Step 4: Apply limits:

$ [2x^2 – \dfrac{2x^3}{3}]_0^2 = (8 – \dfrac{16}{3}) = \dfrac{8}{3}. $

Answer: $ \dfrac{8}{3} \text{ square units.} $

Area under Curve with Negative Region

If the curve lies below the x-axis for some interval, the total area is obtained by splitting and taking the absolute value:

$ A = \int_a^c f(x)\,dx – \int_c^b f(x)\,dx, \quad \text{if } f(x) \text{ changes sign at } x = c. $

Special Case — Symmetric Curves

If $ f(x) $ is even → $ A = 2\int_0^a f(x)\,dx $
If $ f(x) $ is odd → $ A = 0 $ (area cancels out above and below x-axis)

Area Under the Curve

The area under a curve $ y = f(x) $ between $ x = a $ and $ x = b $ is:

$ \text{Area} = \int_a^b f(x)\,dx. $

If the curve is above x-axis → area positive. If below x-axis → integral gives negative value, but area is taken as positive (modulus).

Case 1 — Area Between Curve and x–Axis

If $ f(x) \ge 0 $ in $ [a,b] $, then:

$ A = \int_a^b f(x)\,dx. $

Example

Find area under $ y = x^2 $ from $ x=0 $ to $ x=2 $.

▶️ Answer / Explanation

$ A = \int_0^2 x^2 dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}. $

Area = $ \dfrac{8}{3} $

Case 2 — Area Between Curve and y–Axis

Rewrite curve as $ x = g(y) $, then:

$ A = \int_{y_1}^{y_2} g(y)\,dy. $

Example

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

▶️ Answer / Explanation

Points: $ y = -2 $ to $ y = 2 $.

$ A = \int_{-2}^{2} (4 – y^2)\,dy = \left[4y – \frac{y^3}{3}\right]_{-2}^{2} = \frac{32}{3}. $

Area = $ \dfrac{32}{3} $

Case 3 — Area Between Two Curves

If curves intersect at $ x = a $ and $ x = b $:

$ A = \int_a^b (f(x) – g(x))\,dx $

where $ f(x) $ is the upper curve.

Example

Find area between $ y = x $ and $ y = x^2 $.

▶️ Answer / Explanation

Intersections: $ x=0,1 $.

Upper curve: $ y = x $.

$ A = \int_0^1 (x – x^2)\,dx = \left[\frac{x^2}{2} – \frac{x^3}{3}\right]_0^1 = \frac{1}{2} – \frac{1}{3} = \frac{1}{6}. $

Area = $ \dfrac{1}{6} $

Case 4 — Area Using Symmetry

Even function: $ f(-x) = f(x) $

$ \int_{-a}^{a} f(x)\,dx = 2\int_0^{a} f(x)\,dx. $

Odd function: $ f(-x) = -f(x) $

$ \int_{-a}^{a} f(x)\,dx = 0. $

Example

Evaluate area under $ y = |x| $ from -3 to 3.

▶️ Answer / Explanation

Use symmetry: $ A = 2\int_0^3 x\,dx = 2\left[\frac{x^2}{2}\right]_0^3 = 9. $

Area = 9

Case 5 — Area Involving Modulus Functions

Whenever $ y = |f(x)| $, split the interval where $ f(x) \ge 0 $ and where $ f(x) < 0 $.

Important JEE note: Area is always taken as positive.

Example

Find area under $ y = |x – 2| $ from $ x = 0 $ to $ x = 4 $.

▶️ Answer / Explanation

Critical point x = 2.

$ A = \int_0^2 (2 – x)\,dx + \int_2^4 (x – 2)\,dx $

Both integrals give 2.

Total area = 4.

Case 6 — Area Between Curve & Vertical/Horizontal Lines

Typical JEE types:

between $ y = f(x) $ and x-axis
between $ y = f(x) $ and a line $ y = k $
between $ y = f(x) $ and a vertical line $ x = k $

General idea: integrate the difference.

Case 7 — Area of a Loop (Important for JEE)

Some curves form loops: e.g., $ y^2 = x(4-x) $, cycloids, lemniscates.

Use symmetry / find intersection points / integrate carefully.

Example

Find the area bounded by curves $ y = \sin x $ and $ y = \cos x $ between their points of intersection.

▶️ Answer / Explanation

Intersection: $ \sin x = \cos x \Rightarrow x = \frac{\pi}{4}. $ Next intersection: $ \sin x = \cos x \Rightarrow x = \frac{5\pi}{4}. $

Upper curve in this interval is $ \cos x $.

$ A = \int_{\pi/4}^{5\pi/4} (\cos x – \sin x)\,dx. $

$ A = [\sin x + \cos x]_{\pi/4}^{5\pi/4} $

Compute both values → Area = 2√2.

Final Answer = $ 2\sqrt{2} $

Notes and Study Materials

Examples and Exercise

IIT JEE (Main) Mathematics ,”Area Under Curve” Notes ,Test Papers, Sample Papers, Past Years Papers , NCERT , S. L. Loney and Hall & Knight Solutions and Help from Ex- IITian

About this unit

Integral as an anti – derivative. Fundamental integrals involving algebraic, trigonometric, exponential and logarithmic functions.Integration by substitution, by parts, and by partial fractions. Integration using trigonometric identities Integral as limit of a sum. Evaluation of simple integrals: Fundamental Theorem of Calculus. Properties of Area Under Curve, evaluation of Area Under Curve, determining areas of the regions bounded by simple curves in standard form.

IITian Academy Notes for IIT JEE (Main) Mathematics – Area Under Curve

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S. L. Loney IIT JEE (Main) Mathematics

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IIT JEE Main Maths -Unit 8- Area under curves- Study Notes-New Syllabus