Single-Variable-Calculus · Unit 17 · Video 3 · Interactive Practice

Setting Up the Area Between Two Curves

IKey Formulas

Formula	Meaning	Holds when
$dA \approx \big(f(x) - g(x)\big)\,dx$	One vertical strip: height top minus bottom, base $dx$	$f(x) \ge g(x)$ at that $x$
$A = \displaystyle\int_a^b \big(f(x) - g(x)\big)\,dx$	Area between two curves	$f \ge g$ on $[a, b]$ , with $a$ and $b$ the left and right ends
$A = \displaystyle\int_a^b f(x)\,dx$	Area under one curve: the case $g = 0$	$f \ge 0$ on $[a, b]$
$A = \displaystyle\int_a^c (f_1 - g_1)\,dx + \int_c^b (f_2 - g_2)\,dx$	The top or bottom changes formula at $x = c$	$f_1 \ge g_1$ on $[a, c]$ and $f_2 \ge g_2$ on $[c, b]$

Key Insight: A region has four sides. Its left and right ends become the limits $a$ and $b$ ; its top and bottom become $f(x)$ and $g(x)$ in the integrand. Top minus bottom is the height of a strip whatever the signs, so a bottom curve below the $x$ -axis needs no special treatment.

IIVisualization 1 — Top Minus Bottom, Whatever the Signs

The height of a strip is $f(x) - g(x)$ , even where the bottom curve dips below the $x$ -axis.

Strip at x 2

IIIVisualization 2 — Where the Bottom Switches

Under $y = 3 - \tfrac{x^2}{2}$ the bottom is $y = -x - 1$ , then $y = x - 1$ : two formulas, so two integrals.

Strip at x −2

IVQuiz Questions

Problem 1 · Two Curves Through the Origin

Given: the region enclosed by $y = \sqrt{x}$ and $y = x^2$ — find its area.

✅ Correct! The curves meet at

x = 0

and

x = 1

\sqrt{x}

is on top between them, and

\int_0^1 (\sqrt{x} - x^2)\,dx = \tfrac{2}{3} - \tfrac{1}{3} = \tfrac{1}{3}

❌ That is the area under $y = \sqrt{x}$ alone. The region stops at the bottom curve

y = x^2

, so subtract it: the integrand is top minus bottom,

\sqrt{x} - x^2

❌ The order is reversed.

\int_0^1 (x^2 - \sqrt{x})\,dx

is bottom minus top. On

(0, 1)

\sqrt{x} > x^2

(at

x = \tfrac14

\tfrac12 > \tfrac{1}{16}

), so

\sqrt{x}

is the top, and an area is never negative.

❌ That adds the curves.

\int_0^1 (\sqrt{x} + x^2)\,dx = 1

stacks the two areas under the curves; the strip's height is the gap between them,

\sqrt{x} - x^2

❌ Not quite. Find where

\sqrt{x} = x^2

, decide which curve is on top, then integrate top minus bottom.

Show solution

Ends: $\sqrt{x} = x^2 \Rightarrow x = x^4 \Rightarrow x(x^3 - 1) = 0$ , so $x = 0$ and $x = 1$ .

Top: at $x = \tfrac14$ , $\sqrt{x} = \tfrac12$ and $x^2 = \tfrac{1}{16}$ , so $\sqrt{x}$ is on top on $[0, 1]$ .

A = \int_0^1 \big(\sqrt{x} - x^2\big)\,dx = \Big[\tfrac{2}{3}x^{3/2} - \tfrac{1}{3}x^3\Big]_0^1 = \tfrac{2}{3} - \tfrac{1}{3} = \tfrac{1}{3}

Problem 2 · The Top Changes Sides

Given: the region enclosed by $y = x^3$ and $y = x$ — find its total area.

✅ Correct! The top changes at

x = 0

x^3

is on top on

[-1, 0]

and

x

[0, 1]

, so the area is

\tfrac14 + \tfrac14 = \tfrac12

❌ The two pieces cancelled.

\int_{-1}^{1} (x - x^3)\,dx = 0

because on

[-1, 0]

the integrand

x - x^3

is negative: there

x^3

is the top. Split at

x = 0

and use top minus bottom on each piece.

❌ That is only the right piece.

\int_0^1 (x - x^3)\,dx = \tfrac14

, but the curves also enclose a piece on

[-1, 0]

, where

x^3

is on top.

❌ The cubic was dropped.

2\int_0^1 x\,dx = 1

is the area between the line

y = x

and the

x

-axis on

[-1, 1]

. The region's other side is the curve

y = x^3

, not the axis, so each strip's height is the gap between

x

and

x^3

❌ Not quite. Solve

x^3 = x

and check which curve is on top on each interval between the crossings.

Show solution

Crossings: $x^3 = x \Rightarrow x(x - 1)(x + 1) = 0$ , so $x = -1, 0, 1$ : three points, two pieces.

Top on each piece: at $x = -\tfrac12$ , $x^3 = -\tfrac18 > -\tfrac12 = x$ , so $x^3$ is on top on $[-1, 0]$ ; at $x = \tfrac12$ , $x = \tfrac12 > \tfrac18 = x^3$ , so $x$ is on top on $[0, 1]$ .

A = \int_{-1}^{0} (x^3 - x)\,dx + \int_{0}^{1} (x - x^3)\,dx = \tfrac14 + \tfrac14 = \tfrac12

A single integral $\int_{-1}^{1} (x - x^3)\,dx$ gives $0$ : on the left piece it subtracts the wrong way round.

Problem 3 · Parabola and Line

Given: the region enclosed by $y = 4 - x^2$ and $y = x + 2$ — find where the curves meet, then the area.

Where do the curves meet?

What is the area?

✅ Correct!

\int_{-2}^{1} \big((4 - x^2) - (x + 2)\big)\,dx = \int_{-2}^{1} (2 - x - x^2)\,dx = \tfrac{9}{2}

❌ A sign slipped.

4 - x^2 = x + 2

gives

x^2 + x - 2 = 0

, which factors as

(x + 2)(x - 1)

. The roots

-1

and

2

belong to

x^2 - x - 2

❌ Those are where the parabola meets the $x$ -axis. The region's ends are where the parabola meets the line: set

4 - x^2 = x + 2

❌ That is the area under the parabola alone.

\int_{-2}^{1} (4 - x^2)\,dx = 9

runs down to the

x

-axis; the region's bottom is the line, so subtract

x + 2

❌ The order is reversed. At

x = 0

the parabola is at

4

and the line at

2

, so the parabola is the top: integrate

(4 - x^2) - (x + 2)

❌ Right integrand, wrong limits.

\int_{-1}^{2} (2 - x - x^2)\,dx = \tfrac32

uses the roots of

x^2 - x - 2

; the curves meet at

x = -2

and

x = 1

❌ Not quite. Set the curves equal, then integrate top minus bottom between the crossings.

Show solution

Ends: $4 - x^2 = x + 2 \Rightarrow x^2 + x - 2 = 0 \Rightarrow (x + 2)(x - 1) = 0$ , so $x = -2$ and $x = 1$ .

Top: at $x = 0$ , $4 - x^2 = 4$ and $x + 2 = 2$ , so the parabola is on top on $[-2, 1]$ .

A = \int_{-2}^{1} \big((4 - x^2) - (x + 2)\big)\,dx = \Big[2x - \tfrac{x^2}{2} - \tfrac{x^3}{3}\Big]_{-2}^{1}

= \Big(2 - \tfrac12 - \tfrac13\Big) - \Big(-4 - 2 + \tfrac83\Big) = \tfrac76 + \tfrac{10}{3} = \tfrac{9}{2}

Problem 4 · Find the Height of the Line

Given: the region enclosed by $y = x^2$ and the horizontal line $y = c$ , with $c > 0$ , has area $36$ — find $c$ .

✅ Correct! The area is

\tfrac43 c^{3/2}

, and

\tfrac43 \cdot 9^{3/2} = \tfrac43 \cdot 27 = 36

❌ Wrong root. From

c^{3/2} = 27

, raise both sides to the power

\tfrac23

c = 27^{2/3} = 9

. The cube root alone gives

27^{1/3} = 3 = \sqrt{c}

, not

c

❌ One step short.

27

c^{3/2}

, not

c

: solve

c^{3/2} = 27

for

c

❌ Only half the region. Integrating from

0

\sqrt{c}

gives

\tfrac23 c^{3/2}

; the region runs from

x = -\sqrt{c}

x = \sqrt{c}

❌ Not quite. The curves meet at

x = \pm\sqrt{c}

, and the line is on top.

Show solution

Ends: $x^2 = c \Rightarrow x = -\sqrt{c}$ and $x = \sqrt{c}$ . Top: the line $y = c$ , since $x^2 < c$ between them.

A = \int_{-\sqrt{c}}^{\sqrt{c}} \big(c - x^2\big)\,dx = \Big[cx - \tfrac{x^3}{3}\Big]_{-\sqrt{c}}^{\sqrt{c}} = 2\Big(c^{3/2} - \tfrac13 c^{3/2}\Big) = \tfrac43 c^{3/2}

Set $\tfrac43 c^{3/2} = 36$ : $c^{3/2} = 27$ , so $c = 27^{2/3} = 9$ .

Check: $\int_{-3}^{3} (9 - x^2)\,dx = 54 - 18 = 36$ .

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