💡 Key Idea: Area between curves = (Upper curve) - (Lower curve), integrated over the interval!

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

where $f(x) \geq g(x)$ on $[a, b]$

The Formula (Vertical Slices)

When integrating with respect to x

$A = \int_a^b [\text{top} - \text{bottom}]\,dx$

Steps:

1. Sketch the curves
2. Identify which is on top
3. Find intersection points (if needed for limits)
4. Integrate (top - bottom)

Example 1: Basic Area Between Curves

Find the area between $y = x^2$ and $y = x$ from $x = 0$ to $x = 1$.

Step 1: Sketch the curves

$y = x$ is a line through the origin $y = x^2$ is a parabola

On $[0, 1]$, the line is above the parabola.

Step 2: Set up integral

Top curve: $y = x$ Bottom curve: $y = x^2$

$A = \int_0^1 (x - x^2)\,dx$

Step 3: Integrate

$= \int_0^1 (x - x^2)\,dx$

$= \left[\frac{x^2}{2} - \frac{x^3}{3}\right]_0^1$

$= \left(\frac{1}{2} - \frac{1}{3}\right) - (0 - 0)$

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

Answer: $\frac{1}{6}$ square units

Finding Intersection Points

Often, the limits of integration are intersection points of the curves.

To find intersections: Set the functions equal and solve!

$f(x) = g(x)$

Example 2: Finding Intersections First

Find the area between $y = x^2$ and $y = 2x + 3$.

Step 1: Find intersection points

Set $x^2 = 2x + 3$:

$x^2 = 2x + 3$ $x^2 - 2x - 3 = 0$ $(x - 3)(x + 1) = 0$

So $x = -1$ or $x = 3$

Step 2: Determine which is on top

Test a point between $x = -1$ and $x = 3$, say $x = 0$:

At $x = 0$:

• Line: $y = 2(0) + 3 = 3$
• Parabola: $y = 0^2 = 0$

The line is above the parabola!

Step 3: Set up and evaluate integral

$A = \int_{-1}^3 [(2x + 3) - x^2]\,dx$

$= \int_{-1}^3 (2x + 3 - x^2)\,dx$

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

At $x = 3$: $3^2 + 3(3) - \frac{3^3}{3} = 9 + 9 - 9 = 9$

At $x = -1$: $(-1)^2 + 3(-1) - \frac{(-1)^3}{3} = 1 - 3 + \frac{1}{3} = -\frac{5}{3}$

Step 4: Subtract

$A = 9 - \left(-\frac{5}{3}\right) = 9 + \frac{5}{3} = \frac{27}{3} + \frac{5}{3} = \frac{32}{3}$

Answer: $\frac{32}{3}$ square units

When Curves Switch Position

Sometimes the "top" curve changes!

Strategy: Split the integral at the switching point.

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

where $c$ is where the curves cross

Example 3: Curves That Cross

Find the area between $y = \sin x$ and $y = \cos x$ from $x = 0$ to $x = \frac{\pi}{2}$.

Step 1: Find where they intersect

Set $\sin x = \cos x$:

$\tan x = 1$ $x = \frac{\pi}{4}$ (in the given interval)

Step 2: Determine top/bottom on each piece

On $[0, \frac{\pi}{4}]$: $\cos x > \sin x$ (cosine on top)

On $[\frac{\pi}{4}, \frac{\pi}{2}]$: $\sin x > \cos x$ (sine on top)

Step 3: Set up two integrals

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

Step 4: Integrate each piece

First integral: $\int_0^{\pi/4} (\cos x - \sin x)\,dx = [\sin x + \cos x]_0^{\pi/4}$

$= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1)$

$= \sqrt{2} - 1$

Second integral: $\int_{\pi/4}^{\pi/2} (\sin x - \cos x)\,dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}$

$= (0 - 1) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right)$

$= -1 - (-\sqrt{2}) = \sqrt{2} - 1$

Step 5: Add the areas

$A = (\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$

Answer: $2(\sqrt{2} - 1)$ square units

Horizontal Slices (Integrating with respect to y)

Sometimes it's easier to integrate with respect to $y$!

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

When to use:

• Functions given as $x = f(y)$
• Simpler to express in terms of $y$
• Horizontal rectangles make more sense

Example 4: Horizontal Integration

Find the area between $x = y^2$ and $x = y + 2$ from $y = -1$ to $y = 2$.

Step 1: Identify right and left

Test a point, say $y = 0$:

• Parabola: $x = 0^2 = 0$
• Line: $x = 0 + 2 = 2$

Line is to the right!

Step 2: Set up integral

$A = \int_{-1}^2 [(y + 2) - y^2]\,dy$

Step 3: Integrate

$= \int_{-1}^2 (y + 2 - y^2)\,dy$

$= \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^2$

At $y = 2$: $\frac{4}{2} + 4 - \frac{8}{3} = 2 + 4 - \frac{8}{3} = 6 - \frac{8}{3} = \frac{10}{3}$

At $y = -1$: $\frac{1}{2} - 2 + \frac{1}{3} = \frac{3}{6} - \frac{12}{6} + \frac{2}{6} = -\frac{7}{6}$

Step 4: Subtract

$A = \frac{10}{3} - \left(-\frac{7}{6}\right) = \frac{20}{6} + \frac{7}{6} = \frac{27}{6} = \frac{9}{2}$

Answer: $\frac{9}{2}$ square units

Choosing Between dx and dy

Use $dx$ (vertical slices) when:

• Functions are $y = f(x)$
• Vertical lines cross each curve once
• Natural to think "top minus bottom"

Use $dy$ (horizontal slices) when:

• Functions are $x = f(y)$
• Horizontal lines cross each curve once
• Would need multiple pieces with $dx$

⚠️ Common Mistakes

Mistake 1: Wrong Order

WRONG: Bottom - Top

RIGHT: Top - Bottom (or Right - Left for dy)

Always subtract the lower/left function!

Mistake 2: Forgetting Absolute Value

If you get a negative area, you either:

• Subtracted in wrong order, OR
• Need to split where curves cross

Area should be positive!

Mistake 3: Wrong Limits

Check: Are your limits the actual intersection points?

Set $f(x) = g(x)$ and solve to find where curves meet!

Mistake 4: Not Splitting When Needed

If curves cross in the middle of the interval, you must split the integral at that point!

Area Between Multiple Curves

For three curves where $f(x) \geq g(x) \geq h(x)$:

Area between f and h: $\int_a^b [f(x) - h(x)]\,dx$

Area between f and g only: $\int_a^b [f(x) - g(x)]\,dx$

Area between g and h only: $\int_a^b [g(x) - h(x)]\,dx$

Summary of Steps

1. Sketch the curves
2. Find intersection points (set functions equal)
3. Determine which is on top (or right)
4. Check if curves switch position
5. Set up integral: $\int_a^b [\text{top} - \text{bottom}]\,dx$
6. Evaluate using FTC
7. Check: Is the answer positive?

📝 Practice Strategy

1. Always sketch - visual helps identify top/bottom
2. Find intersections by setting functions equal
3. Test a point to determine which is on top
4. Watch for crossings - split the integral if needed
5. Choose dx or dy based on which is simpler
6. Top minus bottom (or right minus left)
7. Check your answer - area should be positive!

Intersection points: $x = 0$ and $x = 2$

Step 2: Determine which function is on top.

At $x = 1$: $y = 1^2 = 1$ (parabola) and $y = 2(1) = 2$ (line)

The line is above the parabola on $[0, 2]$.

Step 3: Set up and evaluate integral.

Area $= \int_0^2 [\text{top} - \text{bottom}] \, dx$

$= \int_0^2 (2x - x^2) \, dx$

$= \left[x^2 - \frac{x^3}{3}\right]_0^2$

$= \left(4 - \frac{8}{3}\right) - 0$

$= \frac{12 - 8}{3} = \frac{4}{3}$ square units

So $e^x$ is on top.

Step 2: Set up integral

$A = \int_0^1 (e^x - e^{-x})\,dx$

Step 3: Integrate

$= [e^x - (-e^{-x})]_0^1$

$= [e^x + e^{-x}]_0^1$

Step 4: Evaluate

At $x = 1$: $e^1 + e^{-1} = e + \frac{1}{e}$

At $x = 0$: $e^0 + e^0 = 1 + 1 = 2$

Step 5: Subtract

$A = \left(e + \frac{1}{e}\right) - 2 = e + \frac{1}{e} - 2$

Answer: $e + \frac{1}{e} - 2$ square units (or $e + e^{-1} - 2$)

Step 2: Determine which is on top.

On $[0, \pi/4]$: $\cos x > \sin x$ (test $x = 0$: $\cos 0 = 1 > 0 = \sin 0$) On $[\pi/4, \pi/2]$: $\sin x > \cos x$ (test $x = \pi/2$: $\sin(\pi/2) = 1 > 0 = \cos(\pi/2)$)

Step 3: Split into two integrals.

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

First integral:

$\int_0^{\pi/4} (\cos x - \sin x) \, dx = [\sin x + \cos x]_0^{\pi/4}$

$= \left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}\right) - (0 + 1) = \sqrt{2} - 1$

Second integral:

$\int_{\pi/4}^{\pi/2} (\sin x - \cos x) \, dx = [-\cos x - \sin x]_{\pi/4}^{\pi/2}$

$= (0 - 1) - \left(-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}\right) = -1 + \sqrt{2}$

Total area: $(\sqrt{2} - 1) + (\sqrt{2} - 1) = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$ square units

So $x = -1, 0, 1$

Step 2: Determine top/bottom on each interval

On $[-1, 0]$: Test $x = -0.5$:

• Line: $y = -0.5$
• Cubic: $y = (-0.5)^3 = -0.125$

Cubic is above line (less negative).

On $[0, 1]$: Test $x = 0.5$:

• Line: $y = 0.5$
• Cubic: $y = 0.125$

Line is above cubic.

Step 3: Set up two integrals

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

Step 4: Integrate first piece

$\int_{-1}^0 (x^3 - x)\,dx = \left[\frac{x^4}{4} - \frac{x^2}{2}\right]_{-1}^0$

$= (0 - 0) - \left(\frac{1}{4} - \frac{1}{2}\right)$

$= 0 - \left(-\frac{1}{4}\right) = \frac{1}{4}$

Step 5: Integrate second piece

$\int_0^1 (x - x^3)\,dx = \left[\frac{x^2}{2} - \frac{x^4}{4}\right]_0^1$

$= \left(\frac{1}{2} - \frac{1}{4}\right) - (0 - 0)$

$= \frac{1}{4}$

Step 6: Add the areas

$A = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

Answer: $\frac{1}{2}$ square unit

Note: By symmetry, both pieces have the same area!