8.4 - Finding the Area Between Curves Expressed as Functions of x

The Fundamental Concept

To find the area between two curves, subtract the lower function from the upper one and integrate:

\text{Area} = \int_{a}^{b} [f(x) - g(x)]\,dx

where

f(x)

is the upper function,

g(x)

is the lower one, and

[a, b]

is the interval.

Critical requirement: this only works where

f(x) \geq g(x)

throughout. If the curves cross inside your interval, you need separate integrals.

8.4 - Finding the Area Between Curves Expressed as Functions of x

Conceptual understanding: if you found the area under

f(x)

and subtracted the area under

g(x)

, you’d be left with exactly the area between them.

Example 1: Given Boundaries

Problem: Find the area bounded by

y = x^2 + 1

y = 2x

x = 0

, and

x = 2

Step 1: Which function is on top? Test

x = 0.5

Parabola: $0.25 + 1 = 1.25$

Line: $2(0.5) = 1$

x^2 + 1

is on top. (They actually touch at

x = 1

, but the parabola never dips below.)

Step 2: Set up.

\text{Area} = \int_{0}^{2} [(x^2 + 1) - 2x]\,dx = \int_{0}^{2} (x^2 - 2x + 1)\,dx

Step 3: Integrate and evaluate.

= \left[\frac{x^3}{3} - x^2 + x\right]_{0}^{2} = \left(\frac{8}{3} - 4 + 2\right) - 0 = \frac{8}{3} - 2 = \frac{2}{3}

Area:

\frac{2}{3}

square units

Example 2: Finding Intersection Points

Problem: Find the area bounded by

y = 4 - x^2

and

y = x + 2

Step 1: Find where they intersect.

4 - x^2 = x + 2 \quad \Rightarrow \quad x^2 + x - 2 = 0 \quad \Rightarrow \quad (x + 2)(x - 1) = 0

x = -2 \text{ or } x = 1

Step 2: Which is on top? Test

x = 0

Parabola: $4$

Line: $2$

The parabola is on top.

Step 3: Set up and evaluate.

\text{Area} = \int_{-2}^{1} [(4 - x^2) - (x + 2)]\,dx = \int_{-2}^{1} (2 - x^2 - x)\,dx

= \left[2x - \frac{x^3}{3} - \frac{x^2}{2}\right]_{-2}^{1}

x = 1

2 - \frac{1}{3} - \frac{1}{2} = \frac{7}{6}

x = -2

-4 + \frac{8}{3} - 2 = -\frac{10}{3}

= \frac{7}{6} + \frac{10}{3} = \frac{7}{6} + \frac{20}{6} = \frac{27}{6} = \frac{9}{2}

Area:

\frac{9}{2}

square units

When Boundaries Change: Multiple Integrals

Sometimes the upper or lower boundary switches partway through the region. When it does, you have to split the integral at the switch point.

Example 3: Changing Lower Boundary

Problem: Set up (but don’t evaluate) an integral for the area bounded by

y = \sqrt{x}

above, with the lower boundary being

y = 0

from

x = 0

x = 4

and

y = x - 4

from

x = 4

onward.

Upper boundary:

y = \sqrt{x}

the whole way, never changes.

Lower boundary:

From $x = 0$ to $x = 4$ : $y = 0$

From $x = 4$ on: $y = x - 4$

Because the lower boundary changes at

x = 4

, you need two integrals:

\text{Area} = \int_{0}^{4} \sqrt{x}\,dx + \int_{4}^{b} [\sqrt{x} - (x - 4)]\,dx

where

b

is wherever the region ends on the right.

Strategy: track both the top and the bottom boundary. Anywhere either one changes, start a new integral.

Using a Calculator to Find Intersections

When the intersection points are ugly to find by hand, use the graphing calculator.

Example 4: Calculator-Based Problem

Problem: Find the area between

y = \ln(x)

and

y = \frac{1}{3}x - 1

Step 1: Graph both.

Press Y=

Enter Y1 = ln(X)

Enter Y2 = (1/3)X - 1

Press ZOOM → 6:ZStandard

Step 2: Find the intersections.

Press 2nd → TRACE (CALC)

Select 5:intersect

For “First curve?” press ENTER

For “Second curve?” press ENTER

For “Guess?” move the cursor near the left intersection and press ENTER

Result:

x \approx 0.4237

Step 3: Store it.

Press 2nd → MODE (QUIT)

Press STO→

Press ALPHA → A

Press ENTER

This stores the full precision value, not just what’s displayed.

Step 4: Repeat for the second intersection. Result:

x \approx 9.8678

. Store it as B.

Step 5: Compute the area with the stored values.

Press MATH → 9:fnInt(

Enter fnInt(Y1 - Y2, X, A, B)

Get Y1 and Y2 from VARS → Y-VARS → 1:Function

Get A and B with ALPHA then the letter

Press ENTER

Result: area

\approx 6.755

square units

Why store the values? If you round

A

to 0.42 and

B

to 9.87, you introduce rounding error. Storing keeps full precision.

Finding a Line That Divides a Region

A classic AP problem: find the vertical line

x = k

that splits a region into two equal areas.

Example 5: Equal Division

Problem: The region bounded by

y = 2\sqrt{x}

and

y = \frac{x^2}{4}

(which meet at

(0,0)

and

(4,4)

) is divided into two equal areas by the line

x = k

. Find

k

Strategy: set the area from 0 to

k

equal to the area from

k

to 4.

Step 1: Set up.

\int_{0}^{k} \left[2\sqrt{x} - \frac{x^2}{4}\right]dx = \int_{k}^{4} \left[2\sqrt{x} - \frac{x^2}{4}\right]dx

Step 2: Antiderivative.

F(x) = \frac{4x^{3/2}}{3} - \frac{x^3}{12}

Step 3: Evaluate both sides.

Left side:

F(k) - F(0) = \frac{4k^{3/2}}{3} - \frac{k^3}{12}

Right side:

F(4) - F(k)

, where

F(4) = \frac{32}{3} - \frac{16}{3} = \frac{16}{3}

Step 4: Set equal and simplify.

\frac{4k^{3/2}}{3} - \frac{k^3}{12} = \frac{16}{3} - \left(\frac{4k^{3/2}}{3} - \frac{k^3}{12}\right)

Move everything with

k

to the left:

\frac{8k^{3/2}}{3} - \frac{k^3}{6} = \frac{16}{3}

Step 5: Solve with a calculator.

Graph both sides:

Y1 = (8X^(3/2))/3 - X^3/6

Y2 = 16/3

Find the intersection with 2nd → CALC → 5:intersect.

Result:

k \approx 1.764

Answer: the line

x \approx 1.76

divides the region into two equal areas.

Problem-Solving Strategy

Step 1: Identify the functions and boundaries. What are the curves? What are the x-limits?

Step 2: Find intersection points if you need them. Set the functions equal, or use 2nd → CALC → intersect.

Step 3: Figure out which function is on top. Plug a test value from the interval into both. The bigger output wins.

Step 4: Check whether the boundaries change. If the top curve changes, split there. If the bottom curve changes, split there.

Step 5: Set up the integral(s).

\int_{a}^{b} [\text{top} - \text{bottom}]\,dx

Step 6: Evaluate. By hand, or with MATH → 9:fnInt(.

Calculator Workflow for Complex Problems

Store intersection points:

Find the intersection: 2nd CALC → 5

Right after, press 2nd MODE (QUIT)

Store: STO→ → ALPHA → letter → ENTER

Use the stored values in the integral:

MATH → 9

Enter fnInt(Y1 - Y2, X, A, B)

Get Y-variables from VARS → Y-VARS → Function

Get stored letters with ALPHA

Why it matters: full precision (no rounding error), speed (no retyping), and accuracy, especially when the boundaries are irrational.

Key Takeaways

Area between curves: $\int_{a}^{b} [\text{upper} - \text{lower}]\,dx$

Figure out which is on top by plugging in a test value

Find intersections by setting the functions equal, or use the calculator

Split into multiple integrals whenever the top or bottom function changes

Store calculator values with STO→ to keep precision

Multiple integrals just get added together

For equal-division problems, set two integrals equal and solve for $k$

Calculator integration: MATH → 9 → fnInt(function, X, lower, upper)

Store functions in Y= and reference them as Y1, Y2, etc.

Always sanity-check your answer against the graph

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