Applied Integration

🚀 Applied Integration: Work & Average Value

The Big Picture: Integration in the Real World

Imagine you have a magical measuring tape. But instead of just measuring distance, it can measure how much effort goes into pushing something, or what the typical temperature was during a whole day. That’s what applied integration does—it takes tiny pieces of information and adds them up to tell a complete story.

Today, we’re exploring two super-useful applications:

1. Work Applications – How much energy does it take to do something?
2. Average Value of a Function – What’s the “typical” value over a period?

🔧 Part 1: Work Applications

What is “Work” in Physics?

Think of work like effort points in a video game. When you push a heavy box across the floor, you spend effort. In physics:

Work = Force × Distance

But here’s the twist: sometimes the force changes as you move!

The Simple Case: Constant Force

If you push a box with the same strength the whole way:

Work = Force × Distance
W = F × d

Example: Push a cart with 10 Newtons of force for 5 meters.

• Work = 10 × 5 = 50 Joules ✅

The Interesting Case: Variable Force

What if the force changes? Like stretching a rubber band—the more you stretch, the harder it pulls back!

When force varies, we use integration:

$W = \int_a^b F(x) , dx$

This adds up all the tiny bits of work done over every tiny distance.

🌟 Classic Work Problems

1. Spring Work (Hooke’s Law)

Springs follow a simple rule: the more you stretch them, the harder they pull back.

Hooke’s Law: $F(x) = kx$

• $k$ = spring constant (stiffness)
• $x$ = distance stretched

Finding Work to Stretch a Spring:

$W = \int_0^x kx , dx = \frac{1}{2}kx^2$

Example: A spring has $k = 20$ N/m. How much work to stretch it 0.3 m?

$W = \frac{1}{2}(20)(0.3)^2 = \frac{1}{2}(20)(0.09) = 0.9 \text{ Joules}$

graph TD
    A["Spring at Rest"] --> B["Apply Force"]
    B --> C["Spring Stretches"]
    C --> D["Force Increases!"]
    D --> E["More Work Needed"]
    E --> F["Integration Adds It All Up"]

2. Pumping Water Out of a Tank

Imagine you have a swimming pool and need to pump all the water out. Water at the bottom needs to travel farther than water at the top!

The Setup:

• Slice the water into thin horizontal layers
• Each layer has weight and must travel a certain height
• Add up all the work for each layer

Formula Pattern:

$W = \int_a^b \rho g A(y) \cdot d(y) , dy$

Where:

• $\rho$ = water density (1000 kg/m³)
• $g$ = gravity (9.8 m/s²)
• $A(y)$ = cross-section area at height $y$
• $d(y)$ = distance that slice must travel

Example: Pump water from a cylindrical tank (radius 2m, height 4m) over the top.

For a slice at height $y$:

• Area = $\pi(2)^2 = 4\pi$ m²
• Distance to top = $(4 - y)$ m
• Weight = $\rho g \cdot 4\pi , dy$

$W = \int_0^4 (1000)(9.8)(4\pi)(4-y) , dy$

$W = 39200\pi \int_0^4 (4-y) , dy$

$W = 39200\pi \left[4y - \frac{y^2}{2}\right]_0^4$

$W = 39200\pi (16 - 8) = 39200\pi \times 8$

$W \approx 985,203 \text{ Joules}$

That’s a LOT of work! 💪

3. Lifting a Chain or Rope

When you lift a heavy chain, you’re not just lifting—the part still hanging gets lighter as you go!

Example: A 10-meter chain weighs 3 kg/m. Find work to lift it completely.

At position $y$ (meters lifted):

• Length remaining = $(10 - y)$
• Weight of remaining chain = $3(10 - y) \times 9.8$

$W = \int_0^{10} 3 \cdot 9.8 (10-y) , dy$

$W = 29.4 \int_0^{10} (10-y) , dy$

$W = 29.4 \left[10y - \frac{y^2}{2}\right]_0^{10}$

$W = 29.4 (100 - 50) = 29.4 \times 50 = 1470 \text{ Joules}$

📊 Part 2: Average Value of a Function

The Everyday Idea

What if someone asks: “What was the average temperature today?”

You could:

1. Take the temperature every minute (1440 readings!)
2. Add them all up
3. Divide by 1440

But what if temperature changes continuously? That’s where integration shines!

The Magic Formula

The average value of a function $f(x)$ on interval $[a, b]$:

$f_{avg} = \frac{1}{b-a} \int_a^b f(x) , dx$

Think of it like this:

• The integral gives you the total accumulation
• Dividing by $(b-a)$ spreads it evenly across the interval

🧠 Why This Formula Works

Imagine a rectangle with:

• Width = $(b - a)$
• Height = $f_{avg}$

The area of this rectangle equals the area under the curve!

$\text{Rectangle Area} = f_{avg} \times (b-a) = \int_a^b f(x) , dx$

It’s like finding what height a flat line would need to have the same “total” as the curvy function.

graph TD
    A["Curvy Function"] --> B["Find Area Under Curve"]
    B --> C["Divide by Width"]
    C --> D["Get Average Height"]
    D --> E["Same Total Area!"]

✨ Examples

Example 1: Average of a Simple Function

Find the average value of $f(x) = x^2$ on $[0, 3]$.

$f_{avg} = \frac{1}{3-0} \int_0^3 x^2 , dx$

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

$= \frac{1}{3} \times \frac{27}{3} = \frac{1}{3} \times 9 = 3$

Answer: The average value is 3.

Notice: $f(0) = 0$ and $f(3) = 9$. The average isn’t simply $(0+9)/2 = 4.5$ because the function spends more time near lower values!

Example 2: Temperature Over Time

Temperature during a 12-hour period follows $T(t) = 60 + 10\sin(\pi t/12)$ °F.

Find the average temperature from $t = 0$ to $t = 12$ hours.

$T_{avg} = \frac{1}{12} \int_0^{12} \left(60 + 10\sin\frac{\pi t}{12}\right) dt$

$= \frac{1}{12} \left[60t - 10 \cdot \frac{12}{\pi}\cos\frac{\pi t}{12}\right]_0^{12}$

$= \frac{1}{12} \left[(720 + \frac{120}{\pi}) - (0 - \frac{120}{\pi})\right]$

$= \frac{1}{12} \left[720 + \frac{240}{\pi}\right]$

$= 60 + \frac{20}{\pi} \approx 66.4°F$

Example 3: Average Velocity

A car’s velocity is $v(t) = 3t^2 + 2t$ m/s. Find average velocity from $t = 1$ to $t = 4$ seconds.

$v_{avg} = \frac{1}{4-1} \int_1^4 (3t^2 + 2t) , dt$

$= \frac{1}{3} \left[t^3 + t^2\right]_1^4$

$= \frac{1}{3} [(64 + 16) - (1 + 1)]$

$= \frac{1}{3} [80 - 2] = \frac{78}{3} = 26 \text{ m/s}$

🎯 Key Takeaways

Work Applications

Average Value

💡 Pro Tips

1. For work problems: Always ask “Does the force change?” If yes, integrate!

2. For pumping problems: Draw a picture, label the slice, find distance to exit point.

3. For average value: Remember—it’s NOT the same as averaging endpoints!

4. Units matter:

   • Work is in Joules (N·m)
   • Average value has same units as the function

🌈 The Beautiful Connection

Both concepts share a deep truth: Integration lets us find totals when things change continuously.

• Work = total effort when force varies
• Average = total value spread evenly

You now have tools to answer real questions:

• “How much energy to pump this pool?”
• “What was the average pollution level?”
• “How much work to stretch this spring?”

That’s the power of applied integration! 🚀

Remember: Every time you see something changing smoothly, integration might be the answer. It’s not just math—it’s how we measure the continuous, ever-changing world around us.