Solved Examples

How to Use This Section

Twenty fully worked area problems in four batches: straight setups, circle-and-ellipse regions, sign-splitting problems, and mixed harder pieces (horizontal strips, log and exponential boundaries, trig powers).

The five-step ritual for every area problem:

1. Sketch the curve(s) and shade the region.
2. Intercepts and symmetry: mark axis crossings inside the interval; exploit even/odd or both-axes symmetry.
3. Choose the strip: vertical ($y\,dx$) or horizontal ($x\,dy$) — whichever the boundary solves for cleanly.
4. Set up and evaluate the integral(s), splitting at sign changes.
5. Sanity-check against geometry: triangles, rectangles, circle fractions.

Key Point: In board marking schemes, the sketch and correct setup carry marks of their own. Write the strip, the limits and the split before integrating — the calculus itself is Chapter 7 routine.

Batch 1 — Straight Setups (Easy)

Example 1: A trapezium by integration

Find the area bounded by $y = x$, the $x$-axis and the ordinates $x = 2$, $x = 5$.

Solution:

1. Set up: $A = \int_2^5 x\,dx = \frac{x^2}{2}\Big|_2^5 = \frac{25 - 4}{2}$.
2. Check (trapezium): parallel sides $2$ and $5$, width $3$: $\frac{2+5}{2}\cdot3 = \frac{21}{2}$. ✓

Final Answer: $\dfrac{21}{2}$ square units.

Example 2: A shifted parabola

Find the area under $y = x^2 + 1$ from $x = 0$ to $x = 2$.

Solution:

1. Set up: the curve is above the axis: $A = \int_0^2\left(x^2 + 1\right)dx$.
2. Evaluate: $\left[\frac{x^3}{3} + x\right]_0^2 = \frac83 + 2$.

Final Answer: $\dfrac{14}{3}$ square units.

Example 3: A semicircle read at sight

Find the area bounded by $y = \sqrt{25 - x^2}$ and the $x$-axis.

Solution:

1. Recognise: the upper half of the circle $x^2 + y^2 = 25$, radius $5$, spanning $[-5, 5]$.
2. Semicircle area: $\frac12\pi\cdot25$.

Final Answer: $\dfrac{25\pi}{2}$ square units.

Example 4: A sideways parabola, upper branch

Find the area bounded by the curve $y^2 = 9x$ (taking $y \geq 0$), the $x$-axis and the ordinates $x = 1$, $x = 4$.

Solution:

1. Strip height: $y = 3\sqrt x$ (positive branch).
2. Set up: $A = \int_1^4 3\sqrt x\,dx = 3\cdot\frac23 x^{3/2}\Big|_1^4 = 2(8 - 1)$.

Final Answer: $14$ square units.

Example 5: A line against the $y$-axis

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

Solution:

1. Horizontal strips: length $x = 2y + 3$, width $dy$: $A = \int_1^3(2y + 3)dy$.
2. Evaluate: $\left[y^2 + 3y\right]_1^3 = 18 - 4 = 14$.
3. Check (trapezium): parallel sides $x(1) = 5$ and $x(3) = 9$, width $2$: $\frac{5+9}{2}\cdot2 = 14$. ✓

Final Answer: $14$ square units.

Batch 2 — Circles and Ellipses (Easy-Medium)

Example 6: A numerical circle

Using integration, find the area enclosed by the circle $x^2 + y^2 = 16$.

Solution:

1. Symmetry: $A = 4\int_0^4\sqrt{16 - x^2}\,dx$.
2. Quarter-circle value: $\int_0^4\sqrt{16 - x^2}\,dx = \frac{\pi\cdot16}{4} = 4\pi$.
3. Multiply: $A = 16\pi$.

Final Answer: $16\pi$ square units ($= \pi r^2$ with $r = 4$ ✓).

Example 7: A numerical ellipse

Find the area of the region bounded by the ellipse $\dfrac{x^2}{25} + \dfrac{y^2}{16} = 1$.

Solution:

1. Semi-axes: $a = 5$, $b = 4$.
2. Apply $\pi ab$: $A = \pi\cdot5\cdot4$.

Final Answer: $20\pi$ square units.

Example 8: A quadrant of an ellipse

Find the area of the region in the first quadrant bounded by the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ and the coordinate axes.

Solution:

1. Whole ellipse: $\pi ab = \pi\cdot3\cdot2 = 6\pi$.
2. Quarter by symmetry: $\frac{6\pi}{4}$.

Final Answer: $\dfrac{3\pi}{2}$ square units.

Example 9: A partial circular strip

Evaluate the area under $y = \sqrt{4 - x^2}$ from $x = 0$ to $x = 1$.

Solution:

1. Not a neat quarter — the limits stop at $1$, not at the radius $2$: use the root formula.
2. Evaluate: $\left[\frac x2\sqrt{4 - x^2} + 2\sin^{-1}\frac x2\right]_0^1 = \frac{\sqrt3}{2} + 2\cdot\frac{\pi}{6}$.

Final Answer: $\dfrac{\sqrt 3}{2} + \dfrac{\pi}{3}$ square units.

Takeaway: Circle fractions with "nice" limits ($0$ to $a$) are pure geometry; anything else calls the root formula back on stage — keep both tools ready.

Example 10: A circular segment

Find the area of the region bounded by the circle $x^2 + y^2 = 4$ and the line $x = 1$, lying to the right of the line.

Solution:

1. Strips: for $1 \leq x \leq 2$, the strip runs from $y = -\sqrt{4 - x^2}$ to $y = +\sqrt{4 - x^2}$: height $2\sqrt{4 - x^2}$.
2. Set up: $A = 2\int_1^2\sqrt{4 - x^2}\,dx$.
3. Evaluate: $2\left[\frac x2\sqrt{4-x^2} + 2\sin^{-1}\frac x2\right]_1^2 = 2\left[(0 + \pi) - \left(\frac{\sqrt3}{2} + \frac{\pi}{3}\right)\right]$.

Final Answer: $\dfrac{4\pi}{3} - \sqrt 3$ square units.

Takeaway: A chord cuts a circle into two segments; integrating the strip between the two half-circle branches handles either one. This shape is a board favourite.

Batch 3 — Sign Splits (Medium)

Example 11: A downward parabola's bowl

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

Solution:

1. Roots: $x = \pm2$; between them the parabola dips below the axis.
2. Integrate (even function): $\int_{-2}^{2}(x^2 - 4)dx = 2\int_0^2(x^2-4)dx = 2\left[\frac83 - 8\right] = -\frac{32}{3}$.
3. Absolute value: $A = \frac{32}{3}$.

Final Answer: $\dfrac{32}{3}$ square units.

Example 12: One-and-a-half periods of sine

Find the area bounded by $y = \sin x$ and the $x$-axis from $x = 0$ to $x = \frac{3\pi}{2}$.

Solution:

1. Sign chart: positive on $[0, \pi]$, negative on $\left[\pi, \frac{3\pi}{2}\right]$.
2. Pieces: $\int_0^{\pi}\sin x\,dx = 2$; $\left|\int_{\pi}^{3\pi/2}\sin x\,dx\right| = |-1| = 1$.
3. Add: $2 + 1 = 3$.

Final Answer: $3$ square units.

Example 13: A cubic with three roots

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

Solution:

1. Factor: $y = x(x-1)(x-2)$ — roots at $0, 1, 2$; positive on $(0,1)$, negative on $(1,2)$.
2. Pieces: $\int_0^1\left(x^3 - 3x^2 + 2x\right)dx = \frac14$; $\int_1^2 = -\frac14$, absolute value $\frac14$.
3. Add: $\frac14 + \frac14 = \frac12$.

Final Answer: $\dfrac{1}{2}$ square units.

Takeaway: The two lobes of this cubic are congruent (the curve has point symmetry about $(1, 0)$) — the signed total is $0$ but the area is $\frac12$.

Example 14: A parabola crossing inside the interval

Find the area bounded by $y = 1 - x^2$, the $x$-axis and the ordinates $x = 0$, $x = 2$.

Solution:

1. Intercept inside: $1 - x^2 = 0$ at $x = 1$ — split.
2. Above on $[0,1]$: $\int_0^1(1 - x^2)dx = \frac23$.
3. Below on $[1,2]$: $\int_1^2(1 - x^2)dx = -\frac43$, absolute value $\frac43$.
4. Add: $\frac23 + \frac43 = 2$.

Final Answer: $2$ square units.

Example 15: A shifted modulus

Find the area bounded by $y = |x - 1|$, the $x$-axis and the ordinates $x = 0$, $x = 3$.

Solution:

1. Corner: at $x = 1$; branches $1 - x$ (left) and $x - 1$ (right).
2. Pieces: $\int_0^1(1 - x)dx = \frac12$; $\int_1^3(x - 1)dx = 2$.
3. Add: $\frac12 + 2 = \frac52$. (Check: triangles $\frac12\cdot1\cdot1$ and $\frac12\cdot2\cdot2$. ✓)

Final Answer: $\dfrac{5}{2}$ square units.

Batch 4 — Mixed and Harder (Medium-Hard)

Example 16: A sideways parabola against the $y$-axis

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

Solution:

1. Meets the $y$-axis where $x = 0$: $y(y - 4) = 0$, so $y = 0$ and $y = 4$; between them $x = y^2 - 4y < 0$ (the curve bulges left).
2. Horizontal strips: $A = \left|\int_0^4\left(y^2 - 4y\right)dy\right| = \left|\frac{64}{3} - 32\right|$.
3. Evaluate: $\left|-\frac{32}{3}\right| = \frac{32}{3}$.

Final Answer: $\dfrac{32}{3}$ square units.

Takeaway: "Negative strip length" means the curve lies on the negative-$x$ side — same absolute-value treatment as curves below the $x$-axis, rotated $90^\circ$.

Example 17: A logarithmic boundary

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

Solution:

1. Sketch: $\log x \geq 0$ for $x \geq 1$; the region runs from $x = 1$ (where $y = 0$) to $x = e$.
2. Set up: $A = \int_1^e\log x\,dx$ — by parts: $\left[x\log x - x\right]_1^e$.
3. Evaluate: $(e - e) - (0 - 1) = 1$.

Final Answer: $1$ square unit.

Example 18: Parabola and its latus rectum

Find the area of the region bounded by the parabola $y^2 = 4x$ and its latus rectum $x = 1$.

Solution:

1. Symmetry: the parabola is symmetric about the $x$-axis: $A = 2\times$ (area above the axis).
2. Upper branch: $y = 2\sqrt x$: $A = 2\int_0^1 2\sqrt x\,dx = 4\cdot\frac23 x^{3/2}\Big|_0^1$.
3. Evaluate: $\frac83$.

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

Takeaway: For $y^2 = 4ax$, the region cut by the latus rectum $x = a$ has area $\frac{8a^2}{3}$ — a formula-worthy special case ($a = 1$ here).

Example 19: An exponential region

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

Solution:

1. Set up: $e^{2x} > 0$: $A = \int_0^1 e^{2x}dx$.
2. Evaluate: $\frac{e^{2x}}{2}\Big|_0^1 = \frac{e^2 - 1}{2}$.

Final Answer: $\dfrac{e^2 - 1}{2}$ square units.

Example 20: A squared trig boundary

Find the area bounded by $y = \cos^2 x$, the $x$-axis and the ordinates $x = 0$, $x = \pi$.

Solution:

1. Sign: $\cos^2 x \geq 0$ always — no splitting despite $\cos x$ changing sign inside.
2. Power-reduce: $A = \int_0^{\pi}\frac{1 + \cos 2x}{2}dx = \left[\frac x2 + \frac{\sin 2x}{4}\right]_0^{\pi}$.
3. Evaluate: $\frac{\pi}{2} + 0$.

Final Answer: $\dfrac{\pi}{2}$ square units.

Takeaway: Squares and moduli manufacture non-negative curves — the area equals the plain integral again, and the average-value reading ($\cos^2$ averages $\frac12$ over $[0, \pi]$) checks it instantly.