Chapter 8: Application of Integrals - Board Exam Notes

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
Short Answer (2M)	2	Medium	Area under simple curves
Long Answer (4M)	4	Very High	Area between curves, ellipse, parabola
Long Answer (6M)	6	Medium	Complex area problems with multiple regions

⏰ Last 24-Hour Checklist

Area under curve: $A = \int_a^b y \, dx = \int_a^b f(x) \, dx$
Area with y-axis: $A = \int_c^d x \, dy = \int_c^d g(y) \, dy$
Area of circle: $x^2 + y^2 = a^2$ gives $A = \pi a^2$
Area of ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ gives $A = \pi ab$

Symmetry: Use 4× first quadrant for symmetric curves
Negative areas: Take absolute value $\left|\int_a^b f(x) \, dx\right|$
Between curves: $A = \int_a^b |f(x) - g(x)| \, dx$
Strips method: Vertical strips (dx) or horizontal strips (dy)

Area Under Simple Curves ★★★★★

Fundamental Concept

Definition: The area bounded by the curve $y = f(x)$, x-axis and ordinates $x = a$ and $x = b$ is given by: $$A = \int_a^b y \, dx = \int_a^b f(x) \, dx$$ where $y = f(x)$ represents the height of the elementary strip and $dx$ is its width.

Method of Elementary Strips:

Vertical Strips: When integrating with respect to $x$, use $dA = y \, dx$
Horizontal Strips: When integrating with respect to $y$, use $dA = x \, dy$
Choose the strip orientation that makes integration easier
For curves symmetric about axes, calculate area in one quadrant and multiply

Area with Respect to Y-axis

The area bounded by the curve $x = g(y)$, y-axis and lines $y = c$ and $y = d$ is: $$A = \int_c^d x \, dy = \int_c^d g(y) \, dy$$

Important Note: If the curve lies below the x-axis (i.e., $f(x) < 0$), the integral gives negative value. Always take the absolute value: $A = \left|\int_a^b f(x) \, dx\right|$

Handling Curves Above and Below X-axis

When a curve has portions both above and below x-axis between $x = a$ and $x = b$:
• Find the points where curve crosses x-axis
• Calculate area of each portion separately
• Total area = $|A_1| + |A_2| + |A_3| + ...$
• Take absolute values to ensure all areas are positive

Area of Circle ⭕ ★★★★☆

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

For the circle centered at origin with radius $a$:
• Equation: $x^2 + y^2 = a^2$
• This gives: $y = \pm\sqrt{a^2 - x^2}$
• Circle is symmetric about both x-axis and y-axis

Total Area of Circle: $$A = 4 \times \text{(Area in first quadrant)}$$ $$A = 4\int_0^a \sqrt{a^2 - x^2} \, dx$$ $$= 4 \left[\frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a}\right]_0^a$$ $$= 4 \times \frac{a^2\pi}{4} = \pi a^2$$

Alternative Using Horizontal Strips:
Using $x = \sqrt{a^2 - y^2}$ and horizontal strips: $$A = 4\int_0^a \sqrt{a^2 - y^2} \, dy = \pi a^2$$ Both methods give the same result!

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 formula is essential - memorize it!

Area of Ellipse 🔵 ★★★★★

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

For the ellipse centered at origin:
• Standard form: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
• This gives: $y = \pm\frac{b}{a}\sqrt{a^2 - x^2}$
• Ellipse is symmetric about both axes
• When $a = b$, it becomes a circle

Total Area of Ellipse: $$A = 4 \times \text{(Area in first quadrant)}$$ $$A = 4\int_0^a \frac{b}{a}\sqrt{a^2 - x^2} \, dx$$ $$= \frac{4b}{a}\int_0^a \sqrt{a^2 - x^2} \, dx$$ $$= \frac{4b}{a} \times \frac{\pi a^2}{4} = \pi ab$$

Alternative Using Horizontal Strips:
From $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get $x = \frac{a}{b}\sqrt{b^2 - y^2}$ $$A = 4\int_0^b \frac{a}{b}\sqrt{b^2 - y^2} \, dy = \pi ab$$

Special Cases:

When $a = b$: Ellipse becomes circle with area $\pi a^2$
When $a > b$: Major axis along x-axis
When $b > a$: Major axis along y-axis

Solved Examples (Board Marking Scheme)

Example 1: Area bounded by line, x-axis and ordinates

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

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

When $y = 0$: $3x + 2 = 0 \Rightarrow x = -\frac{2}{3}$

For $-1 \leq x < -\frac{2}{3}$: curve is below x-axis

For $-\frac{2}{3} < x \leq 1$: curve is above x-axis

Step 2: Split the integral 2 Marks

Area = $\left|\int_{-1}^{-2/3} (3x + 2) \, dx\right| + \int_{-2/3}^{1} (3x + 2) \, dx$

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

Step 3: Calculate 1 Mark

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

$= \left|-\frac{2}{3} + \frac{1}{2}\right| + \frac{7}{2} + \frac{2}{3}$

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

Example 2: Area bounded by $y = \cos x$

Q. Find the area bounded by the curve $y = \cos x$ between $x = 0$ and $x = 2\pi$.

Step 1: Identify regions 1 Mark

$\cos x > 0$ for $0 \leq x \leq \frac{\pi}{2}$ and $\frac{3\pi}{2} \leq x \leq 2\pi$

$\cos x < 0$ for $\frac{\pi}{2} \leq x \leq \frac{3\pi}{2}$

Step 2: Set up integral 1.5 Marks

Area = $\int_0^{\pi/2} \cos x \, dx + \left|\int_{\pi/2}^{3\pi/2} \cos x \, dx\right| + \int_{3\pi/2}^{2\pi} \cos x \, dx$

Step 3: Evaluate 1.5 Marks

$= [\sin x]_0^{\pi/2} + |[\sin x]_{\pi/2}^{3\pi/2}| + [\sin x]_{3\pi/2}^{2\pi}$

$= (1 - 0) + |(-1 - 1)| + (0 - (-1))$

$= 1 + 2 + 1 = 4$ square units

Example 3: Area of Ellipse

Q. Find the area enclosed by the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$.

Step 1: Identify parameters 0.5 Mark

Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we get $a^2 = 16$ and $b^2 = 9$

Therefore, $a = 4$ and $b = 3$

Step 2: Use symmetry 1 Mark

From symmetry: Total area = $4 \times$ (Area in first quadrant)

$y = \frac{3}{4}\sqrt{16 - x^2}$ in first quadrant

Area = $4\int_0^4 \frac{3}{4}\sqrt{16 - x^2} \, dx$

Step 3: Evaluate integral 1.5 Marks

$= 3\int_0^4 \sqrt{16 - x^2} \, dx$

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

$= 3\left[0 + 8 \times \frac{\pi}{2} - 0\right] = 3 \times 4\pi = 12\pi$ square units

Direct Formula: Area of ellipse = $\pi ab = \pi \times 4 \times 3 = 12\pi$ square 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) $\frac{\pi}{4}$
Answer: (A) $\pi$. This is exactly one quarter of the circle with radius 2, so area = $\frac{1}{4}\pi(2)^2 = \pi$ square units.

2022 (4 Marks): Using integration, find the area of the region bounded by the curve $y^2 = 4x$ and the line $x = 3$.
Solution:
• The parabola $y^2 = 4x$ is symmetric about x-axis
• Area = $2\int_0^3 y \, dx = 2\int_0^3 2\sqrt{x} \, dx$
• $= 4\left[\frac{2x^{3/2}}{3}\right]_0^3 = \frac{8}{3}[3\sqrt{3}] = 8\sqrt{3}$ square units

2021 (4 Marks): Find the area enclosed between the parabola $y^2 = 4ax$ and the line $y = mx$.
Hint: Find intersection points, then integrate: $\int_0^{4a/m^2} (2\sqrt{ax} - mx) \, dx$
Answer: $\frac{8a^2}{3m^3}$ square units

2020 (6 Marks): Find the area of the region: $\{(x,y): 0 \leq y \leq x^2 + 1, 0 \leq y \leq x + 1, 0 \leq x \leq 2\}$
Solution: Need to find intersection of $y = x^2 + 1$ and $y = x + 1$, then split integral appropriately.
Intersection at $x = 0$ (both give $y = 1$), so integrate the difference from 0 to 2.

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) $\frac{9}{2}$
Answer: (D) $\frac{9}{2}$. Using $x = \frac{y^2}{4}$, Area = $\int_0^3 \frac{y^2}{4} \, dy = \frac{1}{4} \times \frac{27}{3} = \frac{9}{2}$ square units.

Exam Strategy & Mistake Bank

Common Mistakes 🚨

Mistake 1: Forgetting to take absolute value when area is below x-axis. Always use $|A|$ for negative integrals.

Mistake 2: Not splitting integrals when curve crosses axis. Check where $f(x) = 0$ and split accordingly.

Mistake 3: Wrong limits of integration. Always sketch the curve first to identify correct limits.

Mistake 4: Forgetting the factor 4 when using symmetry for circles/ellipses.

Mistake 5: Incorrect formula: Using $\int \sqrt{a^2 - x^2} \, dx$ without proper substitution or standard formula.

Scoring Tips 🏆

Tip 1: Always draw a rough sketch of the curve - it helps identify regions and limits correctly.

Tip 2: For symmetric curves (circle, ellipse), use $4 \times$ first quadrant area to save time.

Tip 3: Check if using horizontal or vertical strips is easier - choose wisely!

Tip 4: Memorize standard formulas: area of circle = $\pi a^2$, ellipse = $\pi ab$.

Tip 5: Show all steps clearly including substitution of limits - partial marks are given!

Important Formulas & Results

Standard Area Formulas:

1. Circle: $x^2 + y^2 = a^2 \Rightarrow$ Area $= \pi a^2$

2. Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow$ Area $= \pi ab$

3. Parabola: $y^2 = 4ax$ from $x = 0$ to $x = h \Rightarrow$ Area $= \frac{4}{3}\sqrt{ah^3}$

4. Under curve: $A = \int_a^b f(x) \, dx$ (take absolute value if negative)

5. Between curves: $A = \int_a^b |f(x) - g(x)| \, dx$

Essential Integration Formulas:

1. $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$

2. $\int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x + \sqrt{x^2+a^2}| + C$

3. $\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2}\ln|x + \sqrt{x^2-a^2}| + C$

Quick Reference Table

Curve Type	Standard Form	Area Formula
Circle	$x^2 + y^2 = a^2$	$\pi a^2$
Ellipse	$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$	$\pi ab$
Parabola (y² form)	$y^2 = 4ax$	Use integration
Parabola (x² form)	$x^2 = 4ay$	Use integration

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 $x = 2$.

Hint: $\int_1^2 x^2 \, dx$

Q2. Find the area bounded by the curve $y = x^4$, x-axis and the lines $x = 1$ and $x = 5$.

Q3. Sketch the graph of $y = |x + 3|$ and evaluate $\int_0^6 |x + 3| \, dx$.

Level 2: Intermediate (4 Marks Each)

Q4. Find the area of the region bounded by the ellipse $\frac{x^2}{4} + \frac{y^2}{9} = 1$.

Hint: Use $a = 2, b = 3$ and formula $\pi ab$

Q5. Find the area bounded by the curve $y = \sin x$ between $x = 0$ and $x = 2\pi$.

Q6. Find the area of the region bounded by the parabola $y^2 = 4x$ and the line $y = 2x$.

Level 3: Advanced (6 Marks Each)

Q7. Using integration, find the area of the triangular region whose sides have equations $y = 2x + 1$, $y = 3x + 1$ and $x = 4$.

Q8. Find the area of the region: $\{(x,y): x^2 + y^2 \leq 1 \leq x + y\}$

Hint: Area between circle and line - requires careful setup

Formula Sheet (Must Remember!) 📝

Core Formulas

1. Area under curve (x-axis): $A = \int_a^b y \, dx = \int_a^b f(x) \, dx$

2. Area with y-axis: $A = \int_c^d x \, dy = \int_c^d g(y) \, dy$

3. Area between curves: $A = \int_a^b |f(x) - g(x)| \, dx$

4. Circle area: $x^2 + y^2 = a^2 \Rightarrow A = \pi a^2$

5. Ellipse area: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \Rightarrow A = \pi ab$

6. Symmetry: For symmetric curves, $A = 4 \times A_{\text{first quadrant}}$

Integration Formulas

7. $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C$

8. $\int_0^a \sqrt{a^2 - x^2} \, dx = \frac{\pi a^2}{4}$

9. For negative areas: Always take $\left|\int_a^b f(x) \, dx\right|$

Exam Day Checklist:

✓ Sketch the curve before solving
✓ Check for symmetry to simplify calculations
✓ Identify where curve crosses axes
✓ Split integrals at crossing points
✓ Take absolute values for negative areas
✓ Write limits clearly and substitute carefully