What is the weightage of Chapter 8 Applications of Integrals in CBSE Class 12?

It contributes 4 to 6 marks in Unit III (Calculus), typically tested as a 3-mark or 5-mark subjective question requiring a labeled sketch.

How do you find the area bounded between two curves y = f(x) and y = g(x)?

Find intersection points x = a and x = b. The enclosed area is given by Integral from a to b of [f(x) - g(x)] dx, taking upper curve minus lower curve.

Is finding the area of triangles bounded by lines included in the CBSE syllabus?

Yes, finding the area of triangular regions with given vertices or line equations using integration is an essential board exam question.

Do you need to draw a rough sketch for Area under Curves questions?

Yes, CBSE stepwise marking assigns 1 mark explicitly for drawing and labeling a rough sketch with region boundaries and intersection coordinates.

Class 12 Maths Chapter 8: Applications of Integrals Notes & Formulas

2026?27 Syllabus Focus

The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.

Finding areas under simple curves.
Areas involving lines, circles, parabolas and ellipses in standard form.

CBSE Mathematics (Code 041) curriculum, 2026?27

Last 24-Hour Checklist

Area under curve: $A = \int _a^b y \, dx$
Area with y-axis: $A = \int _c^d x \, dy$
Circle Area: $π a^2$ (Use symmetry $4 ×$)
Ellipse Area: $π ab$ (Use symmetry $4 ×$)

Negative Areas: Take modulus $|\int f(x) dx|$
Between Curves: $\int _a^b [f(x) - g(x)] dx$
Intersection: Find points where curves meet
Standard Integral: $\int \sqrt{a^2-x^2} dx$

Area Under Simple Curves

Definition: The area bounded by the curve $y = f(x)$, x-axis and ordinates $x = a$ and $x = b$ is given by: $$A = \int _a^b y \, dx = \int _a^b f(x) \, dx$$

Vertical Strips ($dx$)

Use when $y = f(x)$ from $x=a$ to $x=b$.

$Area = \int _a^b y \, dx$

Horizontal Strips ($dy$)

Use when $x = g(y)$ from $y=c$ to $y=d$.

$Area = \int _c^d x \, dy$

Note: If the curve lies below the x-axis, the integral value will be negative. Take the absolute value (modulus) to get the area.

Standard Areas (Circle & Ellipse)

Area of Circle

Equation: $x^2 + y^2 = a^2$

$Area = 4 × \int _0^a \sqrt{a^2-x^2} dx = \mathbf{π a^2}$

Shortcut: Full area is $π a^2$. Area in 1st quadrant is $\frac{π a^2}{4}$.

Area of Ellipse

Equation: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

$Area = 4 × \int _0^a \frac{b}{a}\sqrt{a^2-x^2} dx = \mathbf{π ab}$

Shortcut: Full area bounded by ellipse is always $π a b$.

⚡ High-Yield AOI Conceptual Rules & Verification Formulas (CBSE 3 & 5 Marks):

Standard Area Formula Bank (Instant MCQ Verification):
• Circle ($x^2 + y^2 = a^2$): Total Area $= \pi a^2$. First quadrant area $= \frac{1}{4}\pi a^2$.
• Ellipse ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$): Total Area $= \pi ab$. First quadrant area $= \frac{1}{4}\pi ab$.
• Parabola & Latus Rectum ($y^2 = 4ax$ & $x = a$): Area $= \frac{8}{3}a^2$.
• Parabola & Line ($y^2 = 4ax$ & $y = mx$): Area $= \frac{8a^2}{3m^3}$.
Symmetry Multiplication Strategy:
• For curves symmetric about both axes (Circle $x^2+y^2=a^2$, Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$): $$\text{Total Area} = 4 \times \int_0^a y \, dx$$ • For curves symmetric about the x-axis (e.g. $y^2 = 4ax$): $$\text{Total Area} = 2 \times \int_0^a \sqrt{4ax} \, dx$$
Integration with respect to $y$ ($dy$ strips):
When integrating curves bounded against the y-axis or when solving for $x$ in terms of $y$ is simpler (e.g. horizontal parabolas $x = \frac{y^2}{4a}$ bounded by lines $y=c, y=d$): $$\text{Area} = \int_c^d x \, dy = \int_c^d g(y) \, dy$$
Net Geometric Area vs. Definite Integral:
If the curve drops below the x-axis ($y < 0$) in part of $[a, b]$, you MUST split at the x-intercept $c$ and take the absolute value: $\text{Area} = \int_a^c y \, dx + \left|\int_c^b y \, dx\right|$.
📌 Syllabus Clarification: Finding area between two intersecting non-linear curves (e.g. intersection of circle and parabola, NCERT Ex 8.2) has been removed from the current CBSE syllabus. CBSE board questions strictly test area under simple standard curves (lines, circles, parabolas, and ellipses in standard form).

Key Integration Formula

$$\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$$ This is the most used formula in this chapter. Memorize it!

Solved Examples (Board Marking Scheme)

Q1. Find the area of the region bounded by the line $y = 3x + 2$, the x-axis and the ordinates $x = -1$ and $x = 1$. (3 Marks)

Step 1: Find where line meets x-axis 1 Mark

When $y = 0$: $3x + 2 = 0 \Rightarrow x = -2/3$.

For $-1 ≤ x < -2/3$, curve is below x-axis. For $-2/3 < x ≤ 1$, above x-axis.

Step 2: Split the integral 1 Mark

Area $= |\int _{-1}^{-2/3} (3x + 2) dx| + \int _{-2/3}^{1} (3x + 2) dx$.

Step 3: Calculate 1 Mark

$= |[\frac{3x^2}{2} + 2x]_{-1}^{-2/3}| + [\frac{3x^2}{2} + 2x]_{-2/3}^{1}$

$= \frac{1}{6} + \frac{25}{6} = \frac{26}{6} = \frac{13}{3}$ sq units.

Q2. Find the area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$. (3 Marks)

Step 1: Identify parameters 0.5 Mark

$a^2 = 16 \Rightarrow a=4$, $b^2=9 \Rightarrow b=3$.

Step 2: Use Symmetry 1 Mark

Area $= 4 × \int _0^4 \frac{3}{4}\sqrt{16-x^2} dx = 3 \int _0^4 \sqrt{4^2-x^2} dx$.

Step 3: Evaluate 1.5 Marks

$= 3 [\frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}\frac{x}{4}]_0^4$

$= 3 [0 + 8(\frac{π}{2}) - 0] = 3(4π) = 12π$ sq units.

Q3. Find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$. (4 Marks)

Step 1: Sketch & Symmetry 1 Mark

Parabola $y^2=4x$ is symmetric about x-axis. Line $x=3$ is vertical.

Required Area $= 2 × (Area in 1st quadrant)$.

Step 2: Integrate 2 Marks

$A = 2 \int _0^3 y dx = 2 \int _0^3 2\sqrt{x} dx = 4 \int _0^3 x^{1/2} dx$.

Step 3: Final Answer 1 Mark

$= 4 [\frac{2}{3}x^{3/2}]_0^3 = \frac{8}{3} (3\sqrt{3}) = 8\sqrt{3}$ sq units.

Previous Year Questions (PYQs)

2023 (1 Mark MCQ): Area lying in the first quadrant and bounded by the circle $x^2 + y^2 = 4$ and the lines $x = 0$ and $x = 2$ is:
(A) $π$ (B) $\frac{π}{2}$ (C) $\frac{π}{3}$ (D) $\frac{π}{4}$
Ans: (A). Radius $r=2$. Area of quadrant $= \frac{1}{4}π r^2 = \frac{1}{4}π(4) = π$.

2021 (4 Marks): Find the area enclosed between the parabola $y^2 = 4ax$ and the line $y = mx$.
Hint: Find intersection points: $m^2x^2 = 4ax \Rightarrow x = 0, x = 4a/m^2$.
Area $= \int _0^{4a/m^2} (2\sqrt{ax} - mx) dx = \frac{8a^2}{3m^3}$ sq units.

2019 (1 Mark MCQ): Area of the region bounded by the curve $y^2 = 4x$, y-axis and the line $y = 3$ is:
(A) 2 (B) $\frac{9}{4}$ (C) $\frac{9}{3}$ (D) $\frac{9}{2}$
Ans: (D). Use horizontal strips: $∫_0^3 x dy = ∫_0^3 \frac{y^2}{4} dy = [\frac{y^3}{12}]_0^3 = \frac{27}{12} = \frac{9}{4}$. Wait, option B is 9/4. Let's recheck. $y^2=4x \Rightarrow x=y^2/4$. Integral is correct. $27/12 = 9/4$. So Ans is (B).

Exam Strategy & Mistake Bank

Common Mistakes

Mistake 1: Forgetting to take absolute value when area is below x-axis. Area cannot be negative!

Mistake 2: Not multiplying by 4 for Circle/Ellipse area when limits are $0$ to $a$.

Mistake 3: Incorrect limits of integration. Always draw a rough sketch to identify intersection points.

Scoring Tips

Tip 1: Always draw a rough sketch of the curve. It carries marks and prevents errors in limits.

Tip 2: Memorize the standard integral $\int \sqrt{a^2-x^2} dx$. It appears in almost every 4-mark question.

Tip 3: For area between curves, identify which curve is "upper" ($y_1$) and which is "lower" ($y_2$). Area $= \int (y_1 - y_2) dx$.

Practice Problems (Self-Assessment)

Level 1: Basic (2 Marks Each)

Q1. Find the area of the region bounded by the curve $y = x^2$, x-axis and the lines $x = 1$ and $x = 2$.

Answer: $∫_1^2 x^2 dx = [\frac{x^3}{3}]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ sq units.

Q2. Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = π$.

Answer: $\int _0^π \sin x dx = [-\cos x]_0^π = -(-1 - 1) = 2$ sq units.

Level 2: Intermediate (4 Marks Each)

Q3. Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Answer: Here $a=2, b=3$. Area $= π ab = 6π$ sq units.

Q4. Find the area of the region bounded by the parabola $y^2 = 4x$ and the line $y = 2x$.

Hint: Intersection points: $(2x)^2 = 4x \Rightarrow 4x^2-4x=0 \Rightarrow x=0, 1$.
Area $= ∫_0^1 (2\sqrt{x} - 2x) dx = [\frac{4}{3}x^3/2 - x^2]_0^1 = \frac{4}{3} - 1 = \frac{1}{3}$ sq units.

Chapter 8 Board Mastery Test

Practice test • 10 marks • 25 min

Q1 (1 Mark): Write the integral expression for the area bounded by the curve $y = f(x)$, the $x$-axis and the ordinates $x = a$ and $x = b$ ($b > a$).

Q2 (2 Marks): Find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$ in the first quadrant.

Q3 (3 Marks): Find the area of the region bounded by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Q4 (4 Marks): Find the area of the region in the first quadrant enclosed by the $x$-axis, the line $y = x$, and the circle $x^2 + y^2 = 32$.

Marking Scheme & Solutions

Solution 1:

$Area A = \mathbf{\int _a^b y\,dx = \int _a^b f(x)\,dx}$. [1 Mark]

Solution 2:

In first quadrant, $y = \sqrt{4x} = 2\sqrt{x}$.

$Area = \int _0^3 2\sqrt{x}\,dx = 2 [ \frac{2}{3} x^{3/2} ]_0^3 = \frac{4}{3} (3^{3/2}) = \frac{4}{3} (3\sqrt{3}) = \mathbf{4\sqrt{3} sq units}$. [2 Marks]

Solution 3:

By symmetry about both axes, Total Area $= 4 × (Area in 1st quadrant)$.

In 1st quadrant: $y = 3\sqrt{1 - \frac{x^2}{16}} = \frac{3}{4}\sqrt{16 - x^2}$. [1 Mark]

$Area = 4 \int _0^4 \frac{3}{4}\sqrt{16 - x^2}\,dx = 3 [ \frac{x}{2}\sqrt{16-x^2} + \frac{16}{2}\sin^{-1}(\frac{x}{4}) ]_0^4$ [1 Mark]

$= 3 [ 0 + 8\sin^{-1}(1) - 0 ] = 3 × 8 × \frac{π}{2} = \mathbf{12π sq units}$. [1 Mark]

Solution 4:

Point of intersection of line $y = x$ and circle $x^2 + y^2 = 32$ in 1st quadrant:

$x^2 + x^2 = 32 \implies 2x^2 = 32 \implies x^2 = 16 \implies x = 4, y = 4$. Intersection point is $(4, 4)$. [1 Mark]

Circle crosses $x$-axis at $x = \sqrt{32} = 4\sqrt{2}$.

Required Area $= \int _0^4 y_{line}\,dx + \int _4^{4\sqrt{2}} y_{circle}\,dx = \int _0^4 x\,dx + \int _4^{4\sqrt{2}} \sqrt{32 - x^2}\,dx$. [1 Mark]

$\int _0^4 x\,dx = [\frac{x^2}{2}]_0^4 = \frac{16}{2} = 8$.

$\int _4^{4\sqrt{2}} \sqrt{32 - x^2}\,dx = [ \frac{x}{2}\sqrt{32-x^2} + \frac{32}{2}\sin^{-1}(\frac{x}{\sqrt{32}}) ]_4^{4\sqrt{2}}$

$= [ 0 + 16\sin^{-1}(1) ] - [ \frac{4}{2}\sqrt{16} + 16\sin^{-1}(\frac{1}{\sqrt{2}}) ] = 16(\frac{π}{2}) - [ 8 + 16(\frac{π}{4}) ] = 8π - 8 - 4π = 4π - 8$. [1 Mark]

$Total Area = 8 + (4π - 8) = \mathbf{4π sq units}$. [1 Mark]

Chapter 8: Application of Integrals

Overview