Worked examples — Area between curves — horizontal and vertical slices

Two symbols we lean on the whole page, restated in plain words:

• ${\int }_a^b(\dots )dx$ means "add up infinitely many thin standing-up rectangles from $x=a$ to $x=b$"; the $dx$ is the tiny width of each. (Built from scratch in Definite Integral as Riemann Sum.)
• ${\int }_c^d(\dots )dy$ means the same but with lying-down rectangles from $y=c$ to $y=d$; $dy$ is the tiny height.

The scenario matrix

Every area-between-curves problem falls into one of these cells. Below the table, one example clears each cell.

Ex 1 — Cell A: clean vertical slice

Forecast: guess before computing — is the answer bigger or smaller than $5$? (The region looks like a fat lens.)

1. Find where they meet. Set $4-x^2=x+2$. $4-x^2=x+2\Rightarrow x^2+x-2=0\Rightarrow (x+2)(x-1)=0\Rightarrow x=-2,1.$ Why this step? The limits $a,b$ are exactly where the region pinches shut — the two curves touch there (see Solving Quadratic & Polynomial Equations).
2. Decide top vs bottom. Test $x=0$: parabola $=4$, line $=2$. So the parabola is on top across $[-2,1]$. Why this step? One interior test point fixes the order everywhere between the crossings — they only swap at the endpoints.
3. Integrate top minus bottom. $A={\int }_{-2}^1[(4-x^2)-(x+2)]dx={\int }_{-2}^1(2-x-x^2)dx.$ Why subtract? Each thin standing rectangle has height (top $y$) − (bottom $y$); that vertical gap is the parabola minus the line.
4. Antiderivative (via Fundamental Theorem of Calculus): $[2x-\frac{x^2}{2}-\frac{x^3}{3}{]}_{-2}^1=(2-\frac{1}{2}-\frac{1}{3})-(-4-2+\frac{8}{3})=\frac{7}{6}-(-\frac{10}{3})=\frac{9}{2}.$

Verify: the integrand $2-x-x^2$ is positive at $x=0$ (value $2$) and zero at both ends — no sign flip, so no cancellation worry. Answer $\frac{9}{2}=4.5$, close to our guess of "around 5." ✔

Ex 2 — Cell B: the curves cross inside the interval

Forecast: if you naively do ${\int }_0^{\pi /2}(\cos x-\sin x)dx$ you'll get... something. Guess whether it's too small.

1. Find the crossing inside the interval. $\sin x=\cos x\Rightarrow x=\frac{\pi }{4}$. Why this step? If the curves swap top/bottom inside $[a,b]$, one formula can't cover the whole thing — this is exactly Cell B.
2. See who's on top in each piece. On $[0,\frac{\pi }{4})$: at $x=0$, $\cos 0=1>\sin 0=0$, so cosine is top. On $(\frac{\pi }{4},\frac{\pi }{2}]$: at $x=\frac{\pi }{2}$, $\sin =1>\cos =0$, so sine is top. Why this step? "Top" literally flips at the crossing; each region needs its own (top − bottom).
3. Split and add, always top minus bottom. $A={\int }_0^{\pi /4}(\cos x-\sin x)dx+{\int }_{\pi /4}^{\pi /2}(\sin x-\cos x)dx.$ First piece: $[\sin x+\cos x{]}_0^{\pi /4}=(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2})-(0+1)=\sqrt{2}-1$. Second piece: $[-\cos x-\sin x{]}_{\pi /4}^{\pi /2}=(0-1)-(-\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2})=\sqrt{2}-1$. $A=(\sqrt{2}-1)+(\sqrt{2}-1)=2\sqrt{2}-2\approx \mathrm{0.828.}$

Verify: both pieces came out equal ($\sqrt{2}-1$), which matches the mirror symmetry of $\sin$ and $\cos$ about $x=\frac{\pi }{4}$. Positive answer, as area must be. ✔

Ex 3 — Cell C: forced horizontal slicing

Forecast: would you rather solve for $y$ (getting $\pm \sqrt{\cdot }$) or keep things as $x=f(y)$?

1. Intersect in terms of $y$. $y^2-3=-y^2+5\Rightarrow 2y^2=8\Rightarrow y=\pm 2$. Why this step? Both curves are already written as $x=(\text{function of }y)$, so the natural variable to sweep is $y$ — the crossings live at $y=-2,2$.
2. Which curve is on the right? At $y=0$: right-facing candidate $x=-y^2+5=5$, other is $x=y^2-3=-3$. So $x=-y^2+5$ is the right curve, $x=y^2-3$ the left. Why this step? Horizontal rectangles have width (right $x$) − (left $x$); we need to know which is which.
3. Integrate right minus left in $y$. $A={\int }_{-2}^2[(-y^2+5)-(y^2-3)]dy={\int }_{-2}^2(8-2y^2)dy.$
4. Evaluate (integrand is even, so double the $0\to 2$ part): $A=2{\int }_0^2(8-2y^2)dy=2[8y-\frac{2y^3}{3}{]}_0^2=2(16-\frac{16}{3})=2\cdot \frac{32}{3}=\frac{64}{3}.$

Why horizontal and not vertical? As a function of $x$ each parabola gives two $y$-values (top and bottom halves). Vertical slicing would demand splitting and inverting; horizontal slicing dodges it entirely (this is where Inverse Functions would otherwise bite).

Verify: $\frac{64}{3}\approx 21.3$; the region is roughly a $2$-tall by up-to-$8$-wide lens, plausibly around $20$. Symmetry (even integrand) confirmed the doubling. ✔

Ex 4 — Cell D: the top curve changes mid-sweep

Forecast: how many separate $dx$ integrals do you think this needs?

1. Find all three pairwise crossings.
   • $y=x$ meets $y=3-\frac{x}{2}$: $x=3-\frac{x}{2}\Rightarrow \frac{3x}{2}=3\Rightarrow x=2$ (point $(2,2)$).
   • $y=\frac{x}{4}$ meets $y=3-\frac{x}{2}$: $\frac{x}{4}=3-\frac{x}{2}\Rightarrow \frac{3x}{4}=3\Rightarrow x=4$ (point $(4,1)$).
   • $y=x$ meets $y=\frac{x}{4}$ at the origin $(0,0)$. Why this step? Corners of the region are where boundaries meet; the top boundary can only switch at such a corner.
2. Spot the switch. The bottom is always $y=\frac{x}{4}$. The top is $y=x$ from $x=0$ to $x=2$, then it becomes $y=3-\frac{x}{2}$ from $x=2$ to $x=4$. Why this step? Beyond $x=2$ the line $y=x$ has left the region — the ceiling handed off to the other line. That forces a split (Cell D).
3. Two integrals, each top − bottom. $A={\int }_0^2(x-\frac{x}{4})dx+{\int }_2^4(3-\frac{x}{2}-\frac{x}{4})dx.$ First: ${\int }_0^2\frac{3x}{4}dx=\frac{3}{4}\cdot \frac{x^2}{2}{|}_0^2=\frac{3}{4}\cdot 2=\frac{3}{2}$. Second: ${\int }_2^4(3-\frac{3x}{4})dx=[3x-\frac{3x^2}{8}{]}_2^4=(12-6)-(6-\frac{3}{2})=6-\frac{9}{2}=\frac{3}{2}$. $A=\frac{3}{2}+\frac{3}{2}=3.$

Verify: the region's three corners are $(0,0),(2,2),(4,1)$. Shoelace area of that triangle: $\frac{1}{2}|0(2-1)+2(1-0)+4(0-2)|=\frac{1}{2}|2-8|=3$. Matches. ✔

Ex 5 — Cell E: degenerate / zero area

Forecast: these look like a parabola and a line — surely they trap a normal region?

1. Intersect. $x^2=2x-1\Rightarrow x^2-2x+1=0\Rightarrow (x-1{)}^2=0\Rightarrow x=1$ (double root). Why this step? A repeated root means the curves touch without crossing — the line is tangent to the parabola.
2. Interpret. There is only one intersection point, so $a=b=1$. The limits pinch to the same value. Why this step? A region needs two distinct crossings to have width; here there's none.
3. Compute. $A={\int }_1^1(x^2-(2x-1))dx=0.$

Verify: $(x-1{)}^2\ge 0$ always, so $x^2-(2x-1)\ge 0$ with equality only at $x=1$; the "region" is a single point, area $0$. ✔

Ex 6 — Cell F: limiting behaviour

Forecast: as $k$ shrinks to $0$, the parabola flattens onto the axis — will the area vanish smoothly, or jump?

1. Where does the parabola hit the axis? $k-x^2=0\Rightarrow x=\pm \sqrt{k}$. Why this step? The region lives between these two roots; the limits depend on $k$.
2. Area (even integrand, double the half). $A(k)={\int }_{-\sqrt{k}}^{\sqrt{k}}(k-x^2)dx=2{\int }_0^{\sqrt{k}}(k-x^2)dx=2[kx-\frac{x^3}{3}{]}_0^{\sqrt{k}}.$ $=2(k\sqrt{k}-\frac{k\sqrt{k}}{3})=2\cdot \frac{2}{3}{k}^{3/2}=\frac{4}{3}{k}^{3/2}.$ Why this step? Both the height ($\sim k$) and width ($\sim \sqrt{k}$) shrink, so the area scales like $k^{3/2}$.
3. Take the limit. ${\lim }_{k\to 0^{+}}\frac{4}{3}{k}^{3/2}=0.$

Verify: at $k=1$, $A(1)=\frac{4}{3}$ — matches the classic $y=1-x^2$ arch area. As $k\to 0^{+}$ the arch's area vanishes continuously (no jump), confirming the region collapses smoothly. ✔

Ex 7 — Cell G: real-world word problem (with units)

Forecast: the bed is a downward dip $2$ m deep and $4$ m wide — guess the area before integrating.

1. Identify top and bottom. Top curve: surface $y=0$. Bottom curve: bed $y=-(4-x^2)=x^2-4$. Why this step? Water fills between the flat surface on top and the curved bed below — a clean vertical-slice setup.
2. Limits are the banks. Bed meets surface where $x^2-4=0\Rightarrow x=\pm 2$ (the water's edges). Why this step? Beyond $x=\pm 2$ the bed rises above $y=0$ — no water there.
3. Integrate top − bottom. $A={\int }_{-2}^2[0-(x^2-4)]dx={\int }_{-2}^2(4-x^2)dx=2[4x-\frac{x^3}{3}{]}_0^2=2(8-\frac{8}{3})=\frac{32}{3}.$

Verify: $\frac{32}{3}\approx 10.67{\text{m}}^2$. Units check: (metres) × (metres) integrated gives ${\text{m}}^2$, correct for a cross-sectional area. A $4\text{ m}\times 2\text{ m}$ bounding box has area $8{\text{ m}}^2$; the parabolic dip fills $\frac{2}{3}$ of the box $=\frac{16}{3}\times 2=\frac{32}{3}$. ✔

Ex 8 — Cell H: exam twist (region above and below an axis)

Forecast: this region has a part below the $x$-axis and a part above. Will a single integral get it right?

1. Find all crossings in $[-1,1]$. $x^3=x\Rightarrow x^3-x=0\Rightarrow x(x-1)(x+1)=0\Rightarrow x=-1,0,1$. Why this step? Three crossings mean the top/bottom order can flip — this is really Cell B plus the exam trap of crossing the axis.
2. Determine top on each piece.
   • On $(-1,0)$: at $x=-\frac{1}{2}$, $x^3=-\frac{1}{8}$, $x=-\frac{1}{2}$. Since $-\frac{1}{8}>-\frac{1}{2}$, $x^3$ is on top.
   • On $(0,1)$: at $x=\frac{1}{2}$, $x^3=\frac{1}{8}$, $x=\frac{1}{2}$. Since $\frac{1}{2}>\frac{1}{8}$, $x$ is on top. Why this step? The curves swap at $x=0$; each piece needs its own (top − bottom).
3. Two integrals, always top − bottom. $A={\int }_{-1}^0(x^3-x)dx+{\int }_0^1(x-x^3)dx.$ First: $[\frac{x^4}{4}-\frac{x^2}{2}{]}_{-1}^0=0-(\frac{1}{4}-\frac{1}{2})=\frac{1}{4}$. Second: $[\frac{x^2}{2}-\frac{x^4}{4}{]}_0^1=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$. $A=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$

Verify: by the odd symmetry of the region about the origin, the two pieces must be equal — both gave $\frac{1}{4}$. Total $\frac{1}{2}>0$. ✔

Connections

• Definite Integral as Riemann Sum — every $\int$ above is a limit of thin-rectangle sums.
• Fundamental Theorem of Calculus — evaluates each integral via antiderivatives.
• Inverse Functions — the tool avoided in Ex 3 by slicing horizontally.
• Solving Quadratic & Polynomial Equations — finds the intersection limits in Ex 1, 4, 5, 8.
• Volumes by Slicing & Disks — the same slice-measure-integrate pattern, one dimension up.
• Hinglish version.