Area Between Curves (Functions of x)

Part 1 · Hook

Why this matters

A landscaper wants to know how much sod to buy for a curved lawn that sits between a winding path and a straight fence. Neither edge is the x-axis, so the area under one curve is not the answer. Slice the lawn into thin vertical strips, measure each from the lower edge to the upper edge, and add them up: that is one integral.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\int_0^2 3x^2\,dx$ ?

$8$
$12$
$6$
$24$

Show the answer

An antiderivative is $x^3$ , so the integral is $2^3 - 0^3 = 8$ .

Correct: $8$ :
$12$ :
$6$ :
$24$ :

2. On $[0, 3]$ , $f(x) < 0$ . What sign does $\int_0^3 f(x)\,dx$ have?

Negative
Positive
Zero
It depends on the length of the interval

Show the answer

Every rectangle in a Riemann sum has a negative height, so the definite integral is negative: it counts area below the axis as negative.

Correct: Negative:
Positive:
Zero:
It depends on the length of the interval:

Part 4 · See it

See it first

The region between the parabola y = 6 − x² (on top) and the line y = x (on the bottom), shaded from x = −3 to x = 2, where they meet at (−3, −3) and (2, 2). One thin vertical rectangle of width Δx has height (6 − x²) − x. A side panel shows the limits from 6 − x² = x and the area, the integral from −3 to 2 of [(6 − x²) − x] dx = 125/6. — One thin rectangle between $y = 6 - x^2$ and $y = x$ : its height is top minus bottom, and the region runs between the crossings at $x = -3$ and $x = 2$ . LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Slice the region into thin vertical rectangles of width $\Delta x$ each one reaches from the bottom curve $g$ up to the top curve $f$ , so its height is $f(x) - g(x)$
Each rectangle has area $\left(f(x) - g(x)\right)\Delta x$ adding them is a Riemann sum for $f - g$
Let the rectangles get thinner and thinnerthe sum becomes $\int_a^b \left(f(x) - g(x)\right)\,dx$
The region starts and ends where the curves meetthe limits $a$ and $b$ are the solutions of $f(x) = g(x)$
Top minus bottom is never negative on $[a, b]$ the integral is positive, so a negative answer means the subtraction is backward

Part 6 · Key ideas

Key ideas

Area between curves: if $f(x) \ge g(x)$ on $[a, b]$ , the area is $\int_a^b \left(f(x) - g(x)\right)\,dx$ , top minus bottom.
Find the limits by solving $f(x) = g(x)$ . They are $x$ -values, because the integral is in $dx$ .
Put the bottom function in parentheses before you subtract: $(x + 1) - (x^2 - 1)$ $= x + 2 - x^2$ .
To tell which curve is on top, test one $x$ between the crossings in both functions.
The formula works wherever the curves are, above or below the $x$ -axis: only the difference matters.

Part 7 · Misconception

A common mistake

The wrong idea: The area between two curves is the area under the top curve, or the integral in whatever order you subtract.

What actually happens: Each strip runs from the bottom curve to the top curve, so the integrand is top minus bottom. The area under the top curve alone counts space below the region. Subtracting bottom minus top gives a negative number, and an area is never negative.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The region $R$ between $f(x) = 3 + 2x - x^2$ and $g(x) = x^2 - 2x + 3$

1. Which integral gives the area of $R$ ?

$\int_0^2 \left[(3 + 2x - x^2) - (x^2 - 2x + 3)\right]\,dx$
$\int_0^2 \left[(x^2 - 2x + 3) - (3 + 2x - x^2)\right]\,dx$
$\int_0^2 (3 + 2x - x^2)\,dx$
$\int_2^4 \left[(3 + 2x - x^2) - (x^2 - 2x + 3)\right]\,dx$

Show the answer

On $[0, 2]$ the graph of $f$ is on top, so each vertical rectangle has height $f(x) - g(x)$ , and the curves meet at $x = 0$ and $x = 2$ .

Correct: $\int_0^2 \left[(3 + 2x - x^2) - (x^2 - 2x + 3)\right]\,dx$ : Top minus bottom, from the left crossing to the right one.
$\int_0^2 \left[(x^2 - 2x + 3) - (3 + 2x - x^2)\right]\,dx$ : Bottom minus top: every rectangle gets a negative height, and the integral comes out as $-\frac{8}{3}$ .
$\int_0^2 (3 + 2x - x^2)\,dx$ : This is the area under $f$ all the way down to the $x$ -axis, which includes the space under $g$ .
$\int_2^4 \left[(3 + 2x - x^2) - (x^2 - 2x + 3)\right]\,dx$ : The limits are the $y$ -values 2 and 4. With $dx$ the limits are $x$ -values: the crossings at $x = 0$ and $x = 2$ .

2. Find the area of $R$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f(x) - g(x) = 4x - 2x^2$ , so the area is $\int_0^2 (4x - 2x^2)\,dx$ $= \left[2x^2 - \frac{2x^3}{3}\right]_0^2$ $= 8 - \frac{16}{3}$ $= \frac{8}{3}$ .

Answer: $\frac{8}{3}$ ≈ 2.667

3. A thin vertical rectangle in $R$ stands at $x$ , for $0 \le x \le 2$ . Write its height $h(x)$ and simplify.

Type an expression for $h(x) =$ .

Show the answer

$h(x)$ $= (3 + 2x - x^2) - (x^2 - 2x + 3)$ $= 4x - 2x^2$ . Subtract every term of $g$ : put it in parentheses first.

Answer: $h(x) =$ $4x - 2x^{2}$

4. The line $y = x + 1$ and the parabola $y = x^2 - 1$ bound a region from $x = -1$ to $x = 2$ . Write the integrand (top minus bottom) for its area, simplified.

Type an expression for $\text{integrand} =$ .

Show the answer

$(x + 1) - (x^2 - 1)$ $= x + 2 - x^2$ . The area is $\int_{-1}^{2} (x + 2 - x^2)\,dx$ $= \frac{9}{2}$ .

Answer: $\text{integrand} =$ $x + 2 - x^{2}$

5. Find the area of the region bounded by $y = e^x$ , $y = 1$ and $x = \ln 3$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$e^x = 1$ at $x = 0$ , so the region runs from $x = 0$ to $x = \ln 3$ with $e^x$ on top. Area $= \int_0^{\ln 3} (e^x - 1)\,dx$ $= \left[e^x - x\right]_0^{\ln 3}$ $= (3 - \ln 3) - 1$ $= 2 - \ln 3$ .

Answer: $2 - \ln(3)$ ≈ 0.901

6. A student says the area of the region between $y = x^2 - 4$ and the $x$ -axis is $\int_{-2}^{2} (x^2 - 4)\,dx$ $= -\frac{32}{3}$ . What is the area?

$\frac{32}{3}$ , from $\int_{-2}^{2} \left(0 - (x^2 - 4)\right)\,dx$
$-\frac{32}{3}$ , since the region lies below the axis
$0$ , since the region is symmetric about the $y$ -axis
$\frac{64}{3}$ , the sum of its left and right halves

Show the answer

Here the $x$ -axis ( $y = 0$ ) is on top and the parabola is on the bottom, so the height is $0 - (x^2 - 4) = 4 - x^2$ . $\int_{-2}^{2} (4 - x^2)\,dx$ $= \frac{32}{3}$ .

Correct: $\frac{32}{3}$ , from $\int_{-2}^{2} \left(0 - (x^2 - 4)\right)\,dx$ : Top minus bottom with $y = 0$ on top gives $4 - x^2 \ge 0$ .
$-\frac{32}{3}$ , since the region lies below the axis: A definite integral can be negative, but an area cannot. The student subtracted in the wrong order.
$0$ , since the region is symmetric about the $y$ -axis: Symmetry about the $y$ -axis does not cancel anything here: both halves are below the axis and both add area.
$\frac{64}{3}$ , the sum of its left and right halves: Each half is $\frac{16}{3}$ , so the two halves add to $\frac{32}{3}$ .

Part 9 · Summary

Summary

The area between two curves is a sum of thin vertical rectangles whose heights are top minus bottom. Find where the curves meet for the limits, check which function is on top, put the bottom function in parentheses, and integrate:

A

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

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The region RR between f(x)=3+2x−x2f(x) = 3 + 2x - x^2 and g(x)=x2−2x+3g(x) = x^2 - 2x + 3

Summary

What comes next

Connections

The region $R$ between $f(x) = 3 + 2x - x^2$ and $g(x) = x^2 - 2x + 3$