Area between curves — horizontal and vertical slices

WHY do we even need "between curves"?

Vertical slices (width $dx$ )

Formula Deriving it from scratch (Riemann sum)

WHAT we do: chop $[a,b]$ into $n$ strips of width $\Delta x = \frac{b-a}{n}$ . HOW: pick a sample point $x_i^*$ in strip $i$ . A thin vertical rectangle there has

height $=$ top minus bottom $= f(x_i^*)-g(x_i^*)$ (Why? top $y$ minus bottom $y$ = vertical gap)
width $=\Delta x$

so its area is $\big(f(x_i^*)-g(x_i^*)\big)\Delta x$ . Total area $\approx$ $\sum_{i=1}^{n}\big(f(x_i^*)-g(x_i^*)\big)\Delta x .$ Let $n\to\infty$ ( $\Delta x\to 0$ ). By definition of the Riemann integral this limit is $A=\int_a^b \big(f(x)-g(x)\big)\,dx.$ Why subtract and not use absolute value inside the integral? Because we chose $f\ge g$ on the whole interval, so $f-g\ge 0$ already. If they cross, split the interval (see mistakes).

Horizontal slices (height $dy$ )

Figure — Area between curves — horizontal and vertical slices

Worked Example 1 — vertical slices

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

Step — find intersections. Set $x^2=2x \Rightarrow x^2-2x=0 \Rightarrow x(x-2)=0$ , so $x=0,2$ . Why? The limits $a,b$ are exactly where the curves meet (the region is pinched shut there).
Step — which is on top? Test $x=1$ : line $2x=2$ , parabola $x^2=1$ . So $2x>x^2$ ; the line is top. Why test a point? Cheaper than reasoning; one interior point fixes the order on the whole interval (they only cross at the endpoints).
Step — integrate top minus bottom. $A=\int_0^2\big(2x-x^2\big)\,dx=\Big[x^2-\tfrac{x^3}{3}\Big]_0^2=\Big(4-\tfrac83\Big)-0=\frac{4}{3}.$ Why this antiderivative? $\frac{d}{dx}(x^2)=2x$ , $\frac{d}{dx}(x^3/3)=x^2$ . ✔

Worked Example 2 — same region, both ways

Region bounded by $y=x$ , $y=0$ , and $x=4$ (a triangle... but let's force the curve $y=\sqrt{x}$ idea). Take the region between $y=\sqrt{x}$ and the line $y=x$ for $0\le x\le 1$ .

Intersections: $\sqrt x = x \Rightarrow x = x^2 \Rightarrow x(x-1)=0$ , so $x=0,1$ .
Top at $x=0.25$ : $\sqrt{0.25}=0.5$ vs $x=0.25$ , so $\sqrt x$ is top.
Vertical: $A=\int_0^1(\sqrt x - x)\,dx=\Big[\tfrac{2}{3}x^{3/2}-\tfrac{x^2}{2}\Big]_0^1=\tfrac23-\tfrac12=\frac{1}{6}.$
Horizontal (switch variables): invert curves. $y=\sqrt x \Rightarrow x=y^2$ (this is the right curve since for fixed $y$ , $y^2 \le y$ ... check). $y=x \Rightarrow x=y$ . For $0\le y\le 1$ , which is right? At $y=0.5$ : $x=y=0.5$ vs $x=y^2=0.25$ , so $x=y$ is on the right, $x=y^2$ on the left. $A=\int_0^1 (y - y^2)\,dy=\Big[\tfrac{y^2}{2}-\tfrac{y^3}{3}\Big]_0^1=\tfrac12-\tfrac13=\frac16.\;\checkmark$ Why both give $1/6$ ? It's the same region; the slicing direction can never change the area.

Worked Example 3 — horizontal slices are much easier

Region bounded by $x=y^2$ and $x=y+2$ .

Intersections in $y$ : $y^2=y+2\Rightarrow y^2-y-2=0\Rightarrow(y-2)(y+1)=0$ , so $y=-1,2$ .
Which curve is on the right? At $y=0$ : $x=y+2=2$ vs $x=y^2=0$ . So $x=y+2$ is right, $x=y^2$ is left. $A=\int_{-1}^{2}\big((y+2)-y^2\big)\,dy=\Big[\tfrac{y^2}{2}+2y-\tfrac{y^3}{3}\Big]_{-1}^{2}.$ At $y=2$ : $2+4-\tfrac83=\tfrac{10}{3}$ . At $y=-1$ : $\tfrac12-2+\tfrac13=-\tfrac76$ . $A=\tfrac{10}{3}-\big(-\tfrac76\big)=\frac{20}{6}+\frac{7}{6}=\frac{27}{6}=\frac{9}{2}.$ Why horizontal here? As a function of $x$ the parabola $x=y^2$ gives two $y$ -values; vertical slicing would need splitting at $x=1$ into two pieces. Horizontal slicing dodges that entirely.

Common mistakes (Steel-man them)

Recall Feynman: explain it to a 12-year-old

Imagine the space between two squiggly lines is a garden, and you want its area. Cut the garden into super-thin sticks. Each standing-up stick is a tiny rectangle: its height is how far the top line is above the bottom line right there, and its width is tiny. Add up all the sticks' areas — that adding-up is the integral. If the garden is taller-than-wide in a tricky way, lay the sticks down sideways instead (height $dy$ ). Same garden, same area — you just changed which way you laid the sticks.

Flashcards

Vertical-slice area formula

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

with

f

the top curve,

g

the bottom.

Horizontal-slice area formula

A=\int_c^d (R(y)-L(y))\,dy

with

R

the right curve,

L

the left.

Where do the integration limits come from?

The intersection points of the two curves (in

x

for vertical, in

y

for horizontal).

When should you prefer horizontal (

dy

) slices?

When right/left curves are constant but top/bottom changes — e.g. when a curve is naturally

x=f(y)

(like

x=y^2

), so vertical slicing would need splitting.

Why subtract instead of taking absolute value at the end?

If curves cross, signed areas cancel; you must split at crossings and use top−bottom on each piece.

Derive the integrand from a Riemann sum

Thin rectangle = (top−bottom)·Δx; sum and let Δx→0 gives

\int(f-g)\,dx

Area between

y=2x

and

y=x^2

\int_0^2(2x-x^2)dx = 4/3

Area between

x=y^2

and

x=y+2

\int_{-1}^2((y+2)-y^2)dy = 9/2

Connections

Definite Integral as Riemann Sum — the engine behind every area formula here.
Fundamental Theorem of Calculus — turns the limit-of-sums into antiderivative evaluation.
Volumes by Slicing & Disks — same "slice, measure, integrate" pattern, one dimension up.
Inverse Functions — needed to rewrite $y=f(x)$ as $x=f^{-1}(y)$ for horizontal slices.
Solving Quadratic & Polynomial Equations — used to find intersection points.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, area between curves ka funda bahut simple hai: tum region ko patli-patli strips mein kaat do aur unke area add kar do — yahi integral kar raha hota hai. Agar strips khade (vertical) kaate, to har strip ki height = top curve minus bottom curve aur width = dx, isliye $A=\int_a^b (f(x)-g(x))\,dx$ . Limits $a,b$ wahan se aate hain jahan dono curves milte hain (intersection points), kyunki wahin region band ho jata hai.

Kabhi-kabhi vertical kaatne mein dikkat aati hai — jaise parabola $x=y^2$ ho, to ek hi $x$ ke do $y$ values aa jate hain, ya top curve beech mein badal jaata hai. Aise case mein apna sir 90 degree ghuma lo: strips ko leta hua (horizontal) kaato, height $dy$ , aur length = right curve minus left curve, yaani $A=\int_c^d (R(y)-L(y))\,dy$ . Yaad rakho: vertical = "top minus bottom, dx"; horizontal = "right minus left, dy".

Sabse common galti: agar curves beech mein cross kar jaayein, to bas $\int(f-g)$ mat karo, kyunki positive aur negative parts cancel ho jaate hain aur answer kam aa jaata hai. Crossing point pe interval ko todo aur har piece pe (upar wala minus neeche wala) lo. Dusri galti: $dy$ integral mein $x$ -limits mat daalo — thickness $dy$ hai to limits bhi $y$ ki range honi chahiye. Bas yeh dhyaan rakhoge to area between curves har baar nikal jaayega.

Area between curves — horizontal and vertical slices

WHY do we even need "between curves"?

Vertical slices (width $dx$ )

Horizontal slices (height $dy$ )

Worked Example 1 — vertical slices

Worked Example 2 — same region, both ways

Worked Example 3 — horizontal slices are much easier

Common mistakes (Steel-man them)

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Calculus II — Integration

Connections