Home Class 12 Application of Integrals

Chapter 8: Application of Integrals

Complete Board Exam Focused Notes with Area Calculations, Curves & PYQs

Exam Weightage & Blueprint

Total: ~8-10 Marks

Application of Integrals is crucial for finding areas under curves and between curves. This chapter directly applies integration concepts to practical geometry problems.

Question Type Marks Frequency Focus Topic
MCQ 1 High Basic area calculations (Circle, Parabola)
Short Answer (2M) 2 Medium Area under simple curves
Long Answer (4M/5M) 4/5 Very High Area between curves, Ellipse, Parabola

Last 24-Hour Checklist

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 $y$ is easily expressed as $f(x)$.

Area $= \int_a^b y \, dx$

Horizontal Strips ($dy$)

Use when $x$ is easily expressed as $g(y)$.

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 $= 4 \int_0^a \sqrt{a^2-x^2} dx = \pi a^2$

Area of Ellipse

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

Area $= 4 \int_0^a \frac{b}{a}\sqrt{a^2-x^2} dx = \pi ab$

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 $-1 \le x < -2/3$, curve is below x-axis. For $-2/3 < x \le 1$, 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 $= 4 \times \int_0^4 \frac{3}{4}\sqrt{16-x^2} dx = 3 \int_0^4 \sqrt{4^2-x^2} dx$.

Step 3: Evaluate 1.5 Marks

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

$= 3 [0 + 8(\frac{\pi}{2}) - 0] = 3(4\pi) = 12\pi$ 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 $= 2 \times (\text{Area in 1st quadrant})$.

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) $\pi$   (B) $\frac{\pi}{2}$   (C) $\frac{\pi}{3}$   (D) $\frac{\pi}{4}$
Ans: (A). Radius $r=2$. Area of quadrant $= \frac{1}{4}\pi r^2 = \frac{1}{4}\pi(4) = \pi$.
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: $\int_0^3 x dy = \int_0^3 \frac{y^2}{4} dy = [\frac{y^3}{12}]_0^3 = \frac{27}{12} = \frac{9}{4}$. 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: $\int_1^2 x^2 dx = [\frac{x^3}{3}]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$ sq units.

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

Answer: $\int_0^\pi \sin x dx = [-\cos x]_0^\pi = -(-1 - 1) = 2$ 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 $= \pi ab = 6\pi$ 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 $= \int_0^1 (2\sqrt{x} - 2x) dx = [\frac{4}{3}x^{3/2} - x^2]_0^1 = \frac{4}{3} - 1 = \frac{1}{3}$ sq units.
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