Antiderivatives(IV) | NEB Class 11 Mathematics Notes

Chapter Overview

How do you find the exact area of a shape bounded by a curve? You can't use simple rectangle or triangle formulas. This chapter answers that question using Definite Integrals — the most powerful application of calculus. The core idea is beautiful: slice the region into infinitely many infinitely thin vertical (or horizontal) rectangles, add up their areas, and take the limit. That limit IS the integral. This chapter bridges theoretical integration with real, measurable quantities like area, distance, and volume cross-sections.

What this chapter covers

Area Bounded by a Curve and the x-axisArea Bounded by a Curve and the y-axisArea Between Two CurvesProof of the Area Under a Curve TheoremArea Below the x-axis (Sign Convention)Finding Intersection Points as LimitsArea of a Circle Using IntegrationArea of an Ellipse Using IntegrationParabola-Line and Parabola-Parabola Region Areas

Learning Objectives

1

State and prove the theorem that the area under y = f(x) between x = a and x = b is ∫ₐᵇ y dx

2

Calculate areas bounded by curves and the x-axis, handling regions above and below the axis separately

3

Calculate areas bounded by curves and the y-axis using ∫ x dy

4

Find the area of regions enclosed between two curves by first locating their intersection points

5

Derive the formula for the area of a circle (πr²) and ellipse (πab) using definite integrals

6

Handle area problems where the curve crosses the x-axis by splitting the integral at the crossing point

Integration = Infinite Thin Rectangles Summed Up

Key Insight

The area under a curve is found by imagining infinitely many vertical rectangles, each of width dx and height y = f(x), stacked side by side.

One rectangle has area = f(x) · dx. Summing all of them from x = a to x = b gives the total area. The symbol $\int_a^b f(x)\,dx$ literally means 'sum (∫) of f(x) times dx (tiny width) from a to b'.

Critical sign rule: Area is always POSITIVE. But the integral can be negative if the curve is below the x-axis. Always check which side of the x-axis your curve is on before computing. If below, negate the integral.

Key Definitions

Definite Integral

An integral with an upper limit $b$ and a lower limit $a$, giving a specific numerical value representing the net signed area between the curve and the x-axis.

$$\int_a^b f(x)\,dx = [F(x)]_a^b = F(b) - F(a)$$

Ordinates

The vertical boundary lines $x = a$ and $x = b$ that mark the left and right boundaries of a region when integrating with respect to x.

Abscissae

The horizontal boundary lines $y = c$ and $y = d$ that mark the bottom and top boundaries of a region when integrating with respect to y.

Latus Rectum

For a parabola $y^2 = 4ax$, the latus rectum is the chord through the focus parallel to the directrix. Its equation is $x = a$. It appears frequently in NEB area questions.

$$\text{Latus rectum of } y^2 = 4ax \text{ is the line } x = a$$

Net Signed Area vs True Area

The definite integral $\int_a^b f(x)\,dx$ gives the net signed area — positive above the x-axis, negative below. The true geometric area is always positive and requires taking the absolute value of each piece.

x-axis Integration vs y-axis Integration

The direction you integrate depends on which axis forms the boundary and how the curve equation is expressed.

Integrating along x (dx)

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

Curve is expressed as y = f(x)

Boundaries are vertical lines: x = a and x = b

The region is bounded by the x-axis (y = 0)

Rectangles are vertical — width dx, height y

Use when: 'area bounded by x-axis and ordinates x=a, x=b'

VS

Integrating along y (dy)

$$A = \int_c^d x\,dy = \int_c^d f(y)\,dy$$

Curve is expressed as x = f(y)

Boundaries are horizontal lines: y = c and y = d

The region is bounded by the y-axis (x = 0)

Rectangles are horizontal — height dy, width x

Use when: 'area bounded by y-axis' or x is given as a function of y

Memory: x-axis boundary → integrate dx (vertical strips). y-axis boundary → integrate dy (horizontal strips). Between two curves → Top minus Bottom (dx) or Right minus Left (dy).

All Area Formulas with Derivations

Area bounded by curve and x-axis (above)

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

Valid when f(x) ≥ 0 throughout [a, b]. If the curve dips below, split the integral.

1Let $A(x)$ be the area from the starting ordinate up to the vertical line at position $x$. Consider two nearby points $P(x, y)$ and $Q(x+\Delta x, y+\Delta y)$ on the curve.

2The strip of area $\Delta A$ between $x$ and $x+\Delta x$ is sandwiched between two rectangles: $y \cdot \Delta x < \Delta A < (y+\Delta y) \cdot \Delta x$.

3Divide by $\Delta x$: $y < \frac{\Delta A}{\Delta x} < y + \Delta y$. As $\Delta x \to 0$, also $\Delta y \to 0$, squeezing $\frac{dA}{dx} = y$.

4Integrate both sides from $a$ to $b$: $\int_a^b \frac{dA}{dx}\,dx = \int_a^b y\,dx$. So $A(b) - A(a) = \int_a^b y\,dx$. Since $A(a) = 0$ (no area at starting point): $$A = \int_a^b y\,dx \quad \blacksquare$$

Area bounded by curve and x-axis (below)

A = -\int_a^b y\,dx = \left|\int_a^b y\,dx\right|

When f(x) ≤ 0 in [a, b], the integral is negative but area is positive. Always negate.

1When the curve lies below the x-axis, $y = f(x) < 0$ throughout the interval. The definite integral $\int_a^b f(x)\,dx$ comes out negative.

2Geometrically, area is always a positive quantity — it cannot be negative. So we negate the negative result.

3$$A = -\int_a^b y\,dx \qquad \text{(when y < 0)}$$ If the curve crosses the x-axis, find the crossing point $c$, then: $A = \int_a^c y\,dx - \int_c^b y\,dx$ (negate the negative part).

Area bounded by curve and y-axis

A = \int_c^d x\,dy = \int_c^d f(y)\,dy

Use when the curve is to the right of the y-axis and expressed as x = f(y).

1This is the horizontal-strip version of the x-axis theorem. Now $B(y)$ is the area from y = c up to y = y.

2By the same squeeze argument: $\frac{dB}{dy} = x$.

3Integrating: $$B(d) - B(c) = \int_c^d x\,dy \implies A = \int_c^d f(y)\,dy$$

Area Between Two Curves

A = \int_a^b (y_{\text{upper}} - y_{\text{lower}})\,dx

Always subtract lower from upper. First find intersection points a and b by solving y₁ = y₂.

1Two curves $y_1 = f_1(x)$ (upper) and $y_2 = f_2(x)$ (lower) enclose a region from $x = a$ to $x = b$.

2Each vertical strip has height = $y_1 - y_2$ and width = $dx$.

3Sum all strips: $$A = \int_a^b (y_1 - y_2)\,dx$$ This is valid as long as $y_1 \ge y_2$ throughout $[a,b]$. If they cross, split the integral at the crossing point.

Area of a Circle

A = \pi a^2 \quad (\text{for circle } x^2 + y^2 = a^2)

Derived by integrating the first-quadrant area and multiplying by 4.

1From $x^2 + y^2 = a^2$, solve for y: $y = \sqrt{a^2 - x^2}$ (upper semicircle).

2Area of first quadrant = $\int_0^a \sqrt{a^2-x^2}\,dx$. Substitute $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$: integral becomes $\int_0^{\pi/2} a^2\cos^2\theta\,d\theta = \frac{a^2\pi}{4}$.

3Total circle area = $4 \times \frac{\pi a^2}{4} = \pi a^2$. ✓ Matches familiar formula.

Area of an Ellipse

A = \pi ab \quad \left(\text{for ellipse } \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\right)

Generalisation of circle formula. When a = b, gives πa² (circle). Use the quarter-rule: integrate first quadrant and multiply by 4.

1From the ellipse equation: $y = \frac{b}{a}\sqrt{a^2 - x^2}$.

2First-quadrant area = $\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx = \frac{b}{a} \cdot \frac{\pi a^2}{4} = \frac{\pi ab}{4}$.

3Total area = $4 \times \frac{\pi ab}{4} = \pi ab$.

Which Formula to Use? — Decision Tree

Read the question carefully and follow these steps. The wording tells you everything.

Are the limits given? (x = a, x = b or y = c, y = d)

Yes

Use those as integration limits directly → continue to next step

No

Find limits by solving the curve equations simultaneously to get intersection points

Is the boundary the x-axis (y = 0) or two vertical lines?

Yes

Use ∫ y dx (vertical strips). Check: is curve above x-axis? If yes → ∫y dx. If below → −∫y dx

No

Continue ↓

Is the boundary the y-axis (x = 0) or two horizontal lines?

Yes

Express curve as x = f(y), use ∫ x dy with horizontal limits y = c to y = d

No

Continue ↓

Is the region enclosed between TWO curves?

Yes

∫(y_upper − y_lower) dx. Identify which curve is on top in the region first.

No

Check if it's a circle or ellipse → use ¼ of area × 4 (Quarter Rule)

Memory Tricks — Use These Every Exam

Trick: Four simple rules that cover 95% of area questions in NEB.

Steps to Remember:

Top Minus Bottom: Area between two curves along the x-direction = ∫(upper curve − lower curve) dx. Never subtract the wrong way or you get a negative area.
Right Minus Left: When working along the y-axis, Area = ∫(right curve − left curve) dy. Think of horizontal strips where the right boundary is farther from the y-axis.
The Quarter Rule: For symmetric shapes like circles ($x^2+y^2=a^2$) or ellipses ($\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$), integrate only the first quadrant ($x,y \ge 0$) and multiply the result by 4. Saves time and avoids sign errors.
No Limits Given = Find Intersections First: If the problem says 'area bounded by y = x² and y = x' without giving x = a and x = b, set x² = x → x(x−1) = 0 → x = 0 and x = 1. These become your limits.
Always Write sq. units: NEB examiners deduct marks if you omit the unit 'sq. units' at the end of every area answer.

Mnemonic: Top−Bottom (dx) | Right−Left (dy) | Quarter Rule for circles/ellipses | No limits → find intersections first

Common Mistakes

These mistakes account for the most lost marks in NEB area questions.

Forgetting negative area: If a curve dips below the x-axis, $\int y\,dx$ is negative. Area is always positive — negate the integral for that region or take the absolute value. Many students just write the negative value and lose all marks.
Wrong limits: If limits aren't given, you MUST find the intersection points by solving the curve equations. Students often guess x = 0 and x = 1 when the actual intersections are different.
Wrong direction of subtraction: For area between two curves, always do Upper − Lower (or Right − Left for y-integration). Doing Lower − Upper gives a negative result — wrong.
Forgetting dx or dy: Always write $\int y\,dx$ or $\int x\,dy$ — never just $\int y$ or $\int x$. Missing the differential is a notation error that loses marks.
Not using the Quarter Rule for full circles/ellipses: Integrating the full circle by going from x = −a to x = +a with $y = \sqrt{a^2-x^2}$ only gives the upper semicircle. Multiply by 2, not 4.
Forgetting 'sq. units': Area answers must always end with 'sq. units'. NEB board marks specifically check for this.

NEB Exam Traps — These Appear Every Year

NEB Exam Warning

Board examiners test these specific edge cases. Know them to protect your marks.

Proof of 'Area under a curve = ∫y dx' is a frequently asked long question. Learn the squeeze argument: $y\,\Delta x < \Delta A < (y+\Delta y)\,\Delta x$, divide by $\Delta x$, take limit → dA/dx = y.
Area between $y^2 = 4ax$ and $x^2 = 4ay$ is $\frac{16a^2}{3}$ — this exact problem appears almost every year. Know it.
For $y = x^3$ from $x = -1$ to $x = 2$: the curve is negative for $x \in [-1, 0]$ and positive for $x \in [0, 2]$. Must split into two integrals. Answer is $\frac{1}{4} + 4 = \frac{17}{4}$, NOT $\frac{15}{4}$.
For area of ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$: the formula is $\pi ab$ — but show the integration steps using $x = a\sin\theta$ substitution. Don't just write the formula without proof.
When a parabola and line are given (like $y = x^2$ and $y = 2x+3$), students forget to check which curve is on top. Sketch or test a point between the intersections.

Question Recognition Guide

Keyword / Phrase in Question	Use This
'Bounded by x-axis, x = a, x = b' (curve above x-axis)	∫ₐᵇ y dx directly
Curve is below x-axis in [a, b]	−∫ₐᵇ y dx (negate to get positive area)
Curve crosses x-axis at x = c within [a, b]	∫ₐᶜ y dx − ∫꜀ᵇ y dx (split and negate negative part)
'Bounded by y-axis' or curve given as x = f(y)	∫꜀ᵈ x dy with horizontal limits
Area between two curves, limits given	∫ₐᵇ (y_upper − y_lower) dx
Area between two curves, NO limits given	First solve y₁ = y₂ for intersection points, THEN ∫(y_upper − y_lower) dx
Circle x² + y² = a²	4 × ∫₀ᵃ √(a²−x²) dx = πa² (Quarter Rule)
Ellipse x²/a² + y²/b² = 1	4 × ∫₀ᵃ (b/a)√(a²−x²) dx = πab (Quarter Rule)
Parabola y² = 4ax and its latus rectum x = a	2∫₀ᵃ 2√(ax) dx (use symmetry above/below x-axis)
Two parabolas y² = 4ax and x² = 4ay	Intersections at (0,0) and (4a,4a); ∫₀^(4a) (2√(ax) − x²/4a) dx

Solved Examples (Easy → Hard)

1

Find the area bounded by $y = 2x$, the x-axis, $x = 1$ and $x = 3$.

Easy

Solution

The curve $y = 2x$ is a line through the origin. For $x \in [1, 3]$, y is always positive (above x-axis).

$$A = \int_1^3 y\,dx = \int_1^3 2x\,dx = 2\left[\frac{x^2}{2}\right]_1^3 = [x^2]_1^3 = 9 - 1 = 8$$

∴ 8 sq. units

2

Find the area bounded by $y = x^2$ and the x-axis from $x = 0$ to $x = 2$.

Easy

Solution

The parabola $y = x^2$ is always above the x-axis for $x \in [0, 2]$.

$$A = \int_0^2 x^2\,dx = \left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3} - 0 = \frac{8}{3}$$

∴ 8/3 sq. units

3

Find the area bounded by $x = y^2$, the y-axis, $y = 1$ and $y = 2$.

Easy

Solution

The curve is expressed as $x = f(y)$ and the boundary is the y-axis → integrate with respect to y.

$$A = \int_1^2 x\,dy = \int_1^2 y^2\,dy = \left[\frac{y^3}{3}\right]_1^2 = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}$$

∴ 7/3 sq. units

4

Find the area bounded by $y = x^2 - 4$ and the x-axis.

Medium

Solution

Step 1 — Find x-intercepts: $x^2 - 4 = 0 \Rightarrow x = \pm 2$.

Step 2 — Check sign: At $x = 0$: $y = -4 < 0$. So the curve is below the x-axis between $-2$ and $2$.

Step 3 — Negate the integral:
$$A = -\int_{-2}^{2}(x^2-4)\,dx = -\left[\frac{x^3}{3}-4x\right]_{-2}^{2}$$

$$= -\left[\left(\frac{8}{3}-8\right)-\left(-\frac{8}{3}+8\right)\right] = -\left[-\frac{16}{3}-\frac{16}{3}\right] \cdot (-1)$$

Actually: $= -\left[\frac{-16}{3} + \frac{-16}{3}\right]$... let me compute carefully:

At $x=2$: $\frac{8}{3}-8 = -\frac{16}{3}$. At $x=-2$: $-\frac{8}{3}+8 = +\frac{16}{3}$.

$-\left(-\frac{16}{3} - \frac{16}{3}\right) = -\left(-\frac{32}{3}\right) = \frac{32}{3}$

∴ 32/3 sq. units

5

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

Medium

Solution

Step 1 — Find intersections: From $x^2 = y$, substitute into $y^2 = x$: $(x^2)^2 = x \Rightarrow x^4 = x \Rightarrow x^4 - x = 0 \Rightarrow x(x^3-1) = 0$.
So $x = 0$ and $x = 1$. Points: $(0,0)$ and $(1,1)$.

Step 2 — Identify upper/lower: At $x = 0.5$: $\sqrt{0.5} \approx 0.707$ and $(0.5)^2 = 0.25$. So $y_1 = \sqrt{x}$ is upper, $y_2 = x^2$ is lower.

Step 3 — Integrate:
$$A = \int_0^1(\sqrt{x} - x^2)\,dx = \left[\frac{2}{3}x^{3/2} - \frac{x^3}{3}\right]_0^1 = \frac{2}{3} - \frac{1}{3} = \frac{1}{3}$$

∴ 1/3 sq. units

6

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

Hard

Solution

Here $a = 4$, $b = 3$. From the ellipse: $y = \frac{3}{4}\sqrt{16-x^2}$.

Quarter Rule: Find first-quadrant area and multiply by 4:
$$A = 4\int_0^4 \frac{3}{4}\sqrt{16-x^2}\,dx = 3\int_0^4\sqrt{16-x^2}\,dx$$

Substitute $x = 4\sin\theta$, $dx = 4\cos\theta\,d\theta$. When $x=0$: $\theta=0$; when $x=4$: $\theta=\pi/2$:
$$= 3\int_0^{\pi/2}4\cos\theta \cdot 4\cos\theta\,d\theta = 48\int_0^{\pi/2}\cos^2\theta\,d\theta$$

$$= 48\int_0^{\pi/2}\frac{1+\cos 2\theta}{2}\,d\theta = 24\left[\theta+\frac{\sin 2\theta}{2}\right]_0^{\pi/2} = 24 \cdot \frac{\pi}{2} = 12\pi$$

Check: $\pi ab = \pi(4)(3) = 12\pi$ ✓

∴ 12π sq. units

7

Find the area bounded by $y^2 = 8x$ and the line $y = 2x$.

Hard

Solution

Step 1 — Find intersections: Substitute $y = 2x$ into $y^2 = 8x$: $(2x)^2 = 8x \Rightarrow 4x^2 = 8x \Rightarrow 4x(x-2) = 0 \Rightarrow x = 0, 2$.

Step 2 — Identify curves: $y_1 = \sqrt{8x} = 2\sqrt{2}\sqrt{x}$ (upper parabola branch), $y_2 = 2x$ (line).
At $x=1$: $y_1 = 2\sqrt{2} \approx 2.83 > y_2 = 2$. ✓ Parabola is upper.

Step 3 — Integrate:
$$A = \int_0^2(2\sqrt{2}\,x^{1/2} - 2x)\,dx = \left[\frac{4\sqrt{2}}{3}x^{3/2} - x^2\right]_0^2$$

$$= \frac{4\sqrt{2}}{3}(2\sqrt{2}) - 4 = \frac{4\sqrt{2} \cdot 2\sqrt{2}}{3} - 4 = \frac{16}{3} - 4 = \frac{4}{3}$$

∴ 4/3 sq. units

MCQ Practice

Mixed Level — MCQ

1

The area under $y = x$ from $x = 0$ to $x = 2$ is:

Correct!

Incorrect. Correct: 2

Solution

$$A = \int_0^2 x\,dx = \left[\frac{x^2}{2}\right]_0^2 = \frac{4}{2} - 0 = 2 \text{ sq. units}$$

∴ 2

2

The area of the circle $x^2 + y^2 = 9$ is:

Correct!

Incorrect. Correct: $9\pi$

Solution

Circle $x^2+y^2=9$ has radius $r = 3$.

$$A = \pi r^2 = \pi(3)^2 = 9\pi \text{ sq. units}$$

∴ $9\pi$

3

The integral $\int_a^b y\,dx$ represents:

Correct!

Incorrect. Correct: Area bounded by x-axis

Solution

By definition, $\int_a^b y\,dx = \int_a^b f(x)\,dx$ gives the area bounded by the curve $y=f(x)$, the x-axis, and the ordinates $x=a$ and $x=b$.

∴ Area bounded by x-axis

4

If a curve is below the x-axis, its area is calculated as:

Correct!

Incorrect. Correct: $-\int y\,dx$

Solution

When $f(x) < 0$ (curve below x-axis), $\int_a^b f(x)\,dx < 0$.

Since area is always positive: $$A = -\int_a^b y\,dx$$

∴ $-\int y\,dx$

5

The area of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is:

Correct!

Incorrect. Correct: $\pi ab$

Solution

Using the Quarter Rule: $A = 4\int_0^a \frac{b}{a}\sqrt{a^2-x^2}\,dx = 4 \cdot \frac{b}{a} \cdot \frac{\pi a^2}{4} = \pi ab$

∴ $\pi ab$

6

The area bounded by $y = \cos x$ from $0$ to $\pi/2$ is:

Correct!

Incorrect. Correct: 1

Solution

$$A = \int_0^{\pi/2}\cos x\,dx = [\sin x]_0^{\pi/2} = \sin\frac{\pi}{2} - \sin 0 = 1 - 0 = 1 \text{ sq. unit}$$

∴ 1

7

The intersection points of $y = x^2$ and $y = x$ are:

Correct!

Incorrect. Correct: (0,0) and (1,1)

Solution

Set $y_1 = y_2$: $x^2 = x \Rightarrow x^2 - x = 0 \Rightarrow x(x-1) = 0$

$x = 0 \Rightarrow y = 0$: point $(0,0)$
$x = 1 \Rightarrow y = 1$: point $(1,1)$

∴ (0,0) and (1,1)

8

The area bounded by $x = 4 - y^2$ and the y-axis is found by integrating with respect to:

Correct!

Incorrect. Correct: y

Solution

The curve is $x = 4-y^2$ (expressed as function of y). Boundary is the y-axis ($x=0$).

Use $A = \int_c^d x\,dy$ with limits found from $x = 0 \Rightarrow y^2 = 4 \Rightarrow y = \pm 2$.

∴ y

9

$\int_0^1 e^x\,dx$ equals:

Correct!

Incorrect. Correct: $e-1$

Solution

$$\int_0^1 e^x\,dx = [e^x]_0^1 = e^1 - e^0 = e - 1$$

∴ $e-1$

10

The area between $y = 2$ and $y = 5$ from $x = 0$ to $x = 1$ is:

Correct!

Incorrect. Correct: 3

Solution

Area between two horizontal lines $y=2$ (lower) and $y=5$ (upper):

$$A = \int_0^1(5-2)\,dx = \int_0^1 3\,dx = 3(1-0) = 3 \text{ sq. units}$$

∴ 3

Short Questions (2 Marks Each)

1

State the theorem for the area bounded by a curve and the x-axis.

[2 marks]

Solution

The area bounded by the curve $y = f(x)$, the x-axis, and the ordinates $x = a$ and $x = b$ is:

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

provided $f(x) \ge 0$ throughout $[a, b]$.

∴ $A = \int_a^b y\,dx$

2

Find the area bounded by $y = 4x^3$, x-axis, $x = 1$ to $x = 2$.

[2 marks]

Solution

$$A = \int_1^2 4x^3\,dx = [x^4]_1^2 = 2^4 - 1^4 = 16 - 1 = 15 \text{ sq. units}$$

∴ 15 sq. units

3

Calculate the area under $y = \sin x$ from $x = 0$ to $x = \pi$.

[2 marks]

Solution

On $[0, \pi]$, $\sin x \ge 0$ (above x-axis):

$$A = \int_0^\pi \sin x\,dx = [-\cos x]_0^\pi = -\cos\pi + \cos 0 = -(-1) + 1 = 2 \text{ sq. units}$$

∴ 2 sq. units

4

Find the area bounded by $y^2 = 4x$ and $x = 1$ in the first quadrant.

[2 marks]

Solution

In the first quadrant, $y = 2\sqrt{x}$:

$$A = \int_0^1 2x^{1/2}\,dx = 2\left[\frac{x^{3/2}}{3/2}\right]_0^1 = \frac{4}{3}[x^{3/2}]_0^1 = \frac{4}{3} \text{ sq. units}$$

∴ 4/3 sq. units

5

Find the area bounded by $x = y^2$, $y = 0$, $y = 3$ and the y-axis.

[2 marks]

Solution

$$A = \int_0^3 x\,dy = \int_0^3 y^2\,dy = \left[\frac{y^3}{3}\right]_0^3 = \frac{27}{3} - 0 = 9 \text{ sq. units}$$

∴ 9 sq. units

6

Find the area bounded by $y = \frac{1}{x}$, x-axis, $x = 1$ to $x = e$.

[2 marks]

Solution

$$A = \int_1^e \frac{1}{x}\,dx = [\ln x]_1^e = \ln e - \ln 1 = 1 - 0 = 1 \text{ sq. unit}$$

∴ 1 sq. unit

7

Find the area between $y = x^2$ and $y = -x^2$ from $x = 0$ to $x = 1$.

[2 marks]

Solution

$y_1 = x^2$ is upper, $y_2 = -x^2$ is lower:

$$A = \int_0^1(x^2-(-x^2))\,dx = \int_0^1 2x^2\,dx = \left[\frac{2x^3}{3}\right]_0^1 = \frac{2}{3} \text{ sq. units}$$

∴ 2/3 sq. units

8

Find the area bounded by $y = e^{2x}$, x-axis, $x = 0$ to $x = 1$.

[2 marks]

Solution

$$A = \int_0^1 e^{2x}\,dx = \left[\frac{e^{2x}}{2}\right]_0^1 = \frac{e^2}{2} - \frac{1}{2} = \frac{e^2-1}{2} \text{ sq. units}$$

∴ $(e^2-1)/2$ sq. units

9

Find the area bounded by $y = 2x + 3$, x-axis, $x = 0$, $x = 2$.

[2 marks]

Solution

$$A = \int_0^2(2x+3)\,dx = [x^2+3x]_0^2 = (4+6) - 0 = 10 \text{ sq. units}$$

∴ 10 sq. units

10

Explain why $\int_a^b y\,dx$ can be zero for a non-zero geometric area.

[2 marks]

Solution

If the curve has equal parts above and below the x-axis in $[a, b]$, the positive and negative areas cancel out in the integral.

Example: $\int_0^{2\pi}\sin x\,dx = 0$, but the geometric area is 4 (two humps of area 2 each — one positive, one negative).

Fix: Split at the crossing point and integrate each part separately, negating negative portions.

∴ Positive and negative areas cancel — must integrate each part separately

Long Questions (5 Marks Each)

1

Prove the formula for the area of a circle $x^2 + y^2 = a^2$ using integration.

[5 marks]

Solution

From $x^2+y^2=a^2$, in the first quadrant: $y = \sqrt{a^2-x^2}$.

Quarter Rule: $A = 4\int_0^a \sqrt{a^2-x^2}\,dx$

Substitute $x = a\sin\theta$, $dx = a\cos\theta\,d\theta$. Limits: $x=0 \Rightarrow \theta=0$; $x=a \Rightarrow \theta=\pi/2$.

$\sqrt{a^2-x^2} = \sqrt{a^2-a^2\sin^2\theta} = a\cos\theta$

$$A = 4\int_0^{\pi/2}a\cos\theta \cdot a\cos\theta\,d\theta = 4a^2\int_0^{\pi/2}\cos^2\theta\,d\theta$$

$$= 4a^2\int_0^{\pi/2}\frac{1+\cos 2\theta}{2}\,d\theta = 2a^2\left[\theta+\frac{\sin 2\theta}{2}\right]_0^{\pi/2}$$

$$= 2a^2\left(\frac{\pi}{2}+0\right) = \pi a^2 \quad \blacksquare$$

∴ Area of circle = πa² — proved by integration with trig substitution

2

Find the area bounded by the parabola $y^2 = 16x$ and its latus rectum ($x = 4$).

[5 marks]

Solution

The parabola $y^2 = 16x$ has $4a = 16 \Rightarrow a = 4$. Latus rectum is $x = 4$.

The curve is symmetric about the x-axis. In the first quadrant (upper half): $y = 4\sqrt{x}$.

By symmetry:
$$A = 2\int_0^4 4\sqrt{x}\,dx = 8\int_0^4 x^{1/2}\,dx = 8\left[\frac{x^{3/2}}{3/2}\right]_0^4$$

$$= \frac{16}{3}[x^{3/2}]_0^4 = \frac{16}{3}(4^{3/2}) = \frac{16}{3}(8) = \frac{128}{3} \text{ sq. units}$$

∴ 128/3 sq. units

3

Find the area of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$.

[5 marks]

Solution

Here $a = 5$, $b = 4$. From the ellipse: $y = \frac{4}{5}\sqrt{25-x^2}$.

Quarter Rule:
$$A = 4\int_0^5 \frac{4}{5}\sqrt{25-x^2}\,dx = \frac{16}{5}\int_0^5\sqrt{25-x^2}\,dx$$

Substitute $x = 5\sin\theta$, $dx = 5\cos\theta\,d\theta$:
$$= \frac{16}{5}\int_0^{\pi/2}5\cos\theta \cdot 5\cos\theta\,d\theta = \frac{16}{5}\cdot 25\int_0^{\pi/2}\cos^2\theta\,d\theta$$

$$= 80 \cdot \frac{\pi}{4} = 20\pi \text{ sq. units}$$

Check: $\pi ab = \pi(5)(4) = 20\pi$ ✓

∴ 20π sq. units

4

Find the area between the curves $y^2 = 4ax$ and $x^2 = 4ay$.

[5 marks]

Solution

Find intersections: From $x^2 = 4ay \Rightarrow y = x^2/4a$. Substitute into $y^2 = 4ax$:
$(x^2/4a)^2 = 4ax \Rightarrow x^4/16a^2 = 4ax \Rightarrow x^4 = 64a^3x \Rightarrow x(x^3-64a^3)=0$.
$x = 0$ and $x = 4a$.

Identify curves: Upper: $y_1 = \sqrt{4ax} = 2\sqrt{ax}$ (parabola). Lower: $y_2 = x^2/4a$ (other parabola).
Check at $x=a$: $y_1 = 2a$, $y_2 = a/4$. ✓

$$A = \int_0^{4a}\left(2\sqrt{ax} - \frac{x^2}{4a}\right)dx = \left[\frac{2\sqrt{a} \cdot 2}{3}x^{3/2} - \frac{x^3}{12a}\right]_0^{4a}$$

$$= \left[\frac{4\sqrt{a}}{3}(4a)^{3/2} - \frac{(4a)^3}{12a}\right] = \frac{4\sqrt{a}}{3}(8a\sqrt{a}) - \frac{64a^3}{12a}$$

$$= \frac{32a^2}{3} - \frac{16a^2}{3} = \frac{16a^2}{3} \text{ sq. units}$$

∴ 16a²/3 sq. units

5

Find the area bounded by $y = x^2 - 5x + 6$ and the x-axis.

[5 marks]

Solution

Find x-intercepts: $x^2-5x+6 = 0 \Rightarrow (x-2)(x-3) = 0 \Rightarrow x = 2, 3$.

Check sign: At $x = 2.5$: $y = 6.25-12.5+6 = -0.25 < 0$. Curve is below x-axis in $[2, 3]$.

$$A = -\int_2^3(x^2-5x+6)\,dx = -\left[\frac{x^3}{3}-\frac{5x^2}{2}+6x\right]_2^3$$

At $x=3$: $9 - 22.5 + 18 = 4.5$
At $x=2$: $8/3 - 10 + 12 = 8/3 + 2 = 14/3$

$$A = -\left(\frac{9}{2} - \frac{14}{3}\right) = -\left(\frac{27-28}{6}\right) = -\left(-\frac{1}{6}\right) = \frac{1}{6} \text{ sq. units}$$

∴ 1/6 sq. units

6

Find the area bounded by $y = \ln x$, the x-axis, and the ordinate $x = e$.

[5 marks]

Solution

$y = \ln x$ is zero at $x = 1$ and positive for $x > 1$. So integrate from $x=1$ to $x=e$.

$$A = \int_1^e \ln x\,dx$$

Use integration by parts: $u = \ln x$, $dv = dx$, so $du = 1/x\,dx$, $v = x$:

$$= [x\ln x]_1^e - \int_1^e x\cdot\frac{1}{x}\,dx = [x\ln x]_1^e - \int_1^e 1\,dx$$

$$= [x\ln x - x]_1^e = (e\ln e - e) - (1\cdot\ln 1 - 1) = (e - e) - (0 - 1) = 0 + 1 = 1 \text{ sq. unit}$$

∴ 1 sq. unit

7

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

[5 marks]

Solution

Find intersections: $x^2 = 4 \Rightarrow x = \pm 2$. Region between $x=-2$ and $x=2$.

Identify curves: $y_1 = 4$ (line, upper), $y_2 = x^2$ (parabola, lower). Check at $x=0$: $y_1=4 > y_2=0$ ✓.

By symmetry:
$$A = 2\int_0^2(4-x^2)\,dx = 2\left[4x-\frac{x^3}{3}\right]_0^2$$

$$= 2\left(8-\frac{8}{3}\right) = 2\cdot\frac{16}{3} = \frac{32}{3} \text{ sq. units}$$

∴ 32/3 sq. units

8

Find the area bounded by $y = x^3$, x-axis, $x = -1$ and $x = 2$.

[5 marks]

Solution

$y = x^3$ is negative for $x \in [-1, 0]$ and positive for $x \in [0, 2]$. Must split at $x = 0$.

Part 1 (below x-axis, $x \in [-1, 0]$):
$$A_1 = -\int_{-1}^0 x^3\,dx = -\left[\frac{x^4}{4}\right]_{-1}^0 = -(0 - \frac{1}{4}) = \frac{1}{4}$$

Part 2 (above x-axis, $x \in [0, 2]$):
$$A_2 = \int_0^2 x^3\,dx = \left[\frac{x^4}{4}\right]_0^2 = \frac{16}{4} = 4$$

Total: $A = A_1 + A_2 = \frac{1}{4} + 4 = \frac{17}{4}$ sq. units

∴ 17/4 sq. units

9

Find the area bounded by the y-axis and the curve $x = 4y - y^2$.

[5 marks]

Solution

Find where curve meets y-axis ($x = 0$):
$4y - y^2 = 0 \Rightarrow y(4-y) = 0 \Rightarrow y = 0$ and $y = 4$.

Check: At $y = 2$: $x = 8-4 = 4 > 0$ ✓ (curve is to the right of y-axis).

Integrate with respect to y:
$$A = \int_0^4 x\,dy = \int_0^4(4y-y^2)\,dy = \left[2y^2 - \frac{y^3}{3}\right]_0^4$$

$$= 2(16) - \frac{64}{3} = 32 - \frac{64}{3} = \frac{96-64}{3} = \frac{32}{3} \text{ sq. units}$$

∴ 32/3 sq. units

10

Find the area between the parabola $y = x^2$ and the line $y = 2x + 3$.

[5 marks]

Solution

Find intersections: $x^2 = 2x+3 \Rightarrow x^2-2x-3=0 \Rightarrow (x-3)(x+1)=0$.
$x = -1$ and $x = 3$.

Identify upper/lower: At $x = 0$: line $y = 3$, parabola $y = 0$. Line is upper. ✓

$$A = \int_{-1}^3[(2x+3)-x^2]\,dx = \left[x^2+3x-\frac{x^3}{3}\right]_{-1}^3$$

At $x=3$: $9+9-9 = 9$.
At $x=-1$: $1-3+\frac{1}{3} = -2+\frac{1}{3} = -\frac{5}{3}$.

$$A = 9-\left(-\frac{5}{3}\right) = 9+\frac{5}{3} = \frac{32}{3} \text{ sq. units}$$

∴ 32/3 sq. units

Past NEB Board Questions

1

Find the area between the parabola $y = x^2$ and the line $y = 2x + 3$.

NEB 2079 [5 marks]

Solution

Intersections: $x^2 = 2x+3 \Rightarrow x=-1, 3$.

Line is upper, parabola is lower in $[-1, 3]$.

$$A = \int_{-1}^3(2x+3-x^2)\,dx = \left[x^2+3x-\frac{x^3}{3}\right]_{-1}^3 = 9-\left(-\frac{5}{3}\right) = \frac{32}{3} \text{ sq. units}$$

∴ 32/3 sq. units

2

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

NEB 2078 [5 marks]

Solution

Intersections at $(0,0)$ and $(4a, 4a)$. Upper: $y = 2\sqrt{ax}$. Lower: $y = x^2/4a$.

$$A = \int_0^{4a}\left(2\sqrt{ax}-\frac{x^2}{4a}\right)dx = \frac{16a^2}{3} \text{ sq. units}$$

∴ 16a²/3 sq. units

3

Prove that the area under $y = f(x)$ between ordinates $x = a$ and $x = b$ is $\int_a^b y\,dx$.

NEB 2077 [4 marks]

Solution

Let $A(x)$ = area from starting ordinate to $x$. For small $\Delta x$: $y\Delta x < \Delta A < (y+\Delta y)\Delta x$.

Divide by $\Delta x$: $y < \Delta A/\Delta x < y+\Delta y$. As $\Delta x\to 0$: $dA/dx = y$.

Integrate: $\int_a^b \frac{dA}{dx}\,dx = \int_a^b y\,dx \Rightarrow A(b)-A(a) = \int_a^b y\,dx$.

Since $A(a)=0$: $A = \int_a^b y\,dx$. $\square$

∴ Proved using squeeze on ΔA — see Long Answer Q1 above for full proof

4

Find the area bounded by $y = x^3$, the x-axis, and ordinates $x = -1$ and $x = 2$.

NEB 2076 [5 marks]

Solution

Split at $x=0$ (where $y=x^3$ crosses x-axis).

$A_1 = -\int_{-1}^0 x^3\,dx = \frac{1}{4}$ (below x-axis, negate).

$A_2 = \int_0^2 x^3\,dx = 4$ (above x-axis).

$A = \frac{1}{4} + 4 = \frac{17}{4}$ sq. units.

∴ 17/4 sq. units

Chapter Test — Full Mixed Paper

Attempt in 50 minutes without notes. Remember to write 'sq. units' after every area answer.

1 Section A — MCQs (1 Mark Each)

Mixed Level — MCQ

1

Area of $y^2 = x$ from $x = 0$ to $x = 1$ (first quadrant) is:

Correct!

Incorrect. Correct: 2/3

Solution

First quadrant of $y^2=x$ means $y=\sqrt{x}$:
$A = \int_0^1 \sqrt{x}\,dx = \left[\frac{2}{3}x^{3/2}\right]_0^1 = \frac{2}{3}$ sq. units

∴ 2/3

2

The formula for area bounded by the y-axis is $\int$ _____ $dy$:

Correct!

Incorrect. Correct: x

Solution

Area bounded by the y-axis: $A = \int_c^d x\,dy$ where $x = f(y)$.

∴ x

3

Area of ellipse with $a = 2$, $b = 3$ is:

Correct!

Incorrect. Correct: $6\pi$

Solution

Area of ellipse = $\pi ab = \pi(2)(3) = 6\pi$ sq. units

∴ $6\pi$

2 Section B — Short Questions (2 Marks Each)

1

Find the area under $y = 3x^2 + 2$ from $x = 0$ to $x = 2$.

[2 marks]

Solution

$$A = \int_0^2(3x^2+2)\,dx = [x^3+2x]_0^2 = (8+4)-0 = 12 \text{ sq. units}$$

∴ 12 sq. units

2

Find the area under $y = \sin x$ for $x \in [0, \pi/2]$.

[2 marks]

Solution

$$A = \int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = -\cos(\pi/2)+\cos 0 = 0+1 = 1 \text{ sq. unit}$$

∴ 1 sq. unit

3 Section C — Long Questions (5 Marks Each)

1

Find the area between the parabola $y = x^2$ and the line $y = 2x + 3$.

[5 marks]

Solution

Intersections: $x^2=2x+3 \Rightarrow x=-1, 3$. Upper: $y=2x+3$.

$A=\int_{-1}^3(2x+3-x^2)\,dx=[x^2+3x-x^3/3]_{-1}^3=9+5/3=32/3$ sq. units.

∴ 32/3 sq. units

One-Page Revision Cheat Sheet

1

Core idea: Integration = summing infinite thin rectangles. ∫y dx for vertical strips, ∫x dy for horizontal strips.

2

x-axis boundary (above): $A = \int_a^b y\,dx$ (positive when y ≥ 0).

3

x-axis boundary (below): $A = -\int_a^b y\,dx$ (negate when y < 0). Always positive!

4

Curve crosses x-axis: Split at crossing point, compute each part separately, add magnitudes.

5

y-axis boundary: $A = \int_c^d x\,dy$ — express curve as x = f(y).

6

Between two curves: $A = \int_a^b(y_{\text{upper}}-y_{\text{lower}})\,dx$ — Top Minus Bottom.

7

No limits given? Solve $y_1 = y_2$ to find intersection points, use those as limits.

8

Circle $x^2+y^2=a^2$: Area = $\pi a^2$. Prove: $4\int_0^a \sqrt{a^2-x^2}\,dx$ with $x=a\sin\theta$.

9

Ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$: Area = $\pi ab$. Use Quarter Rule × 4.

10

Parabola $y^2=4ax$ and $x^2=4ay$: Area = $\frac{16a^2}{3}$ — very frequently tested.

11

Proof of area theorem: $y\,\Delta x < \Delta A < (y+\Delta y)\,\Delta x$ → squeeze → $dA/dx = y$ → integrate.

12

Always check: Is the curve above or below the x-axis? Is the stated 'upper' curve truly on top?

13

Always write: Answer in sq. units — mandatory for full marks in NEB.