Area Between Curves

Part 1 of 7 — Area Between Two Curves

The Formula

For two functions where $f(x) \geq g(x)$ on $[a, b]$:

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

Top function minus bottom function.

Worked Example

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

Step 1: Find intersection points. $x^2 = x + 2$ → $x^2 - x - 2 = 0$ → $(x-2)(x+1) = 0$ → $x = -1, 2$

Step 2: Determine which is on top. At $x = 0$: $y = 0$ vs $y = 2$. So $x + 2$ is on top.

Step 3: Integrate. $A = \int_{-1}^{2} [(x+2) - x^2]\,dx = \left[\frac{x^2}{2} + 2x - \frac{x^3}{3}\right]_{-1}^{2}$

$= \left(2 + 4 - \frac{8}{3}\right) - \left(\frac{1}{2} - 2 + \frac{1}{3}\right) = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{27}{6} = \frac{9}{2}$

Area Between Curves 🎯

Key Takeaways — Part 1

1. Area = $\int$ (top curve $-$ bottom curve)
2. Always find intersection points first
3. Check which function is on top by testing a point in the interval

Area Between Curves

Part 2 of 7 — When Curves Switch Position

Splitting the Integral

When the "top" and "bottom" curves switch, split into separate integrals:

$A = \int_a^c [f(x) - g(x)]\,dx + \int_c^b [g(x) - f(x)]\,dx$

Or equivalently: $A = \int_a^b |f(x) - g(x)|\,dx$

Worked Example

Find the area between $y = x^3$ and $y = x$ on $[-1, 1]$.

Intersections: $x^3 = x$ → $x(x^2-1) = 0$ → $x = -1, 0, 1$

• On $[-1, 0]$: $x^3 \geq x$ (test $x = -0.5$: $-0.125 > -0.5$)
• On $[0, 1]$: $x \geq x^3$ (test $x = 0.5$: $0.5 > 0.125$)

$A = \int_{-1}^0 (x^3 - x)\,dx + \int_0^1 (x - x^3)\,dx = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Curves that Cross 🎯

Key Takeaways — Part 2

1. When curves cross, split the integral at each intersection
2. Always use $|\text{top} - \text{bottom}|$ to get positive area
3. Test a point in each subinterval to determine which is on top

Area Between Curves

Part 3 of 7 — Integrating with Respect to $y$

When to Use $dy$

Sometimes it's easier to integrate horizontally (with respect to $y$):

$A = \int_c^d [\text{right}(y) - \text{left}(y)]\,dy$

Worked Example

Find the area between $x = y^2$ and $x = 4$.

Intersect: $y^2 = 4$ → $y = \pm 2$

Right curve: $x = 4$. Left curve: $x = y^2$.

$A = \int_{-2}^{2} (4 - y^2)\,dy = 2\int_0^2 (4-y^2)\,dy = 2\left[4y - \frac{y^3}{3}\right]_0^2 = 2\left(8 - \frac{8}{3}\right) = \frac{32}{3}$

Integrating in $y$ 🎯

Key Takeaways — Part 3

1. Use $\int dy$ when curves are functions of $y$ (like $x = y^2$)
2. Right minus left (instead of top minus bottom)
3. Limits are $y$-values when integrating in $y$

Area Between Curves

Part 4 of 7 — Multiple Regions

When three or more curves define a region, identify which curves bound each piece of the region separately.

Multi-Region Area 🎯

Key Takeaways — Part 4

1. Sketch the region! Identify all intersection points.
2. Break complex regions into simpler sub-regions if needed.

Area Between Curves

Part 5 of 7 — Area with Absolute Value

Using Absolute Value

$\int_a^b |f(x)|\,dx$ gives the total area between $f(x)$ and the $x$-axis (always positive).

Example

$\int_0^{2\pi} |\sin x|\,dx = \int_0^{\pi} \sin x\,dx + \int_{\pi}^{2\pi} (-\sin x)\,dx = 2 + 2 = 4$

Total vs Signed Area 🎯

Key Takeaways — Part 5

• Total area = $\int |f(x)|\,dx$ (split where $f$ changes sign)
• Signed area = $\int f(x)\,dx$ (can be negative)

Area Between Curves

Part 6 of 7 — AP-Style Workshop

AP-Style Area Problems 🎯

Workshop Complete!

Area Between Curves — Review

Part 7 of 7 — Comprehensive Assessment

Final Assessment 🎯

Area Between Curves — Complete! ✅

You have mastered:

• ✅ Area between two curves ($dx$ and $dy$)
• ✅ When curves cross (splitting integrals)
• ✅ Multiple regions and absolute value
• ✅ Choosing $dx$ vs $dy$ integration