ApplicationsOfIntegration.Rmd

# Applications of Integration





##  Areas Between Curves 


In single-variable calculus, we first learn to compute the area under a curve using definite integrals. This idea naturally extends to finding the area between two curves. Instead of measuring the distance from a function to the \(x\)-axis, we now measure the vertical distance between two functions across an interval.

Suppose two continuous functions \( f(x) \) and \( g(x) \) are defined on an interval \([a,b]\), and assume  
\[
f(x) \ge g(x) \quad \text{for all } x \in [a,b].
\]
At each point \(x\), the vertical distance between the curves is
\[
f(x) - g(x).
\]
By summing these vertical distances over the interval using rectangles and taking a limit, we define the area between the curves as a definite integral.


<div class="theorem">
Area Formula (Vertical Slices)

If \( f(x) \ge g(x) \) on \([a,b]\), then the area \(A\) between the curves is
\[
A = \int_a^b \big(f(x) - g(x)\big)\,dx.
\]
</div>


**Interpretation:**

- **Top curve:** \( y_T = f(x) \)    
- **Bottom curve:** \( y_B = g(x) \)    
- **Rectangle height:** \( y_T - y_B \)    
- **Rectangle width:** \( dx \)  

So each rectangle has area  
\[
(y_T - y_B)\,dx,
\]
and the integral adds all such rectangles across the interval.



**Geometric Meaning:**

This formula generalizes area under a curve:

- If \( g(x)=0 \), then the formula reduces to
\[
A = \int_a^b f(x)\,dx,
\]
which is exactly the area under \(y=f(x)\).

Thus, **area between curves = area under top curve − area under bottom curve**.


::: {.example}
Simple Polynomial Curves

Find the area between \( y = x^2 \) and \( y = x \) on \([0,1]\).


**Solution:**

**Step 1: Identify top and bottom curves**

On \(0 \le x \le 1\),  
\[
x \ge x^2 \Rightarrow y_T = x,\; y_B = x^2.
\]

**Step 2: Set up the integral**
\[
A = \int_0^1 (x - x^2)\,dx.
\]

**Step 3: Evaluate**
\[
A = \int_0^1 x\,dx - \int_0^1 x^2\,dx
= \left[\frac{x^2}{2}\right]_0^1 - \left[\frac{x^3}{3}\right]_0^1
= \frac{1}{2} - \frac{1}{3}
= \frac{1}{6}.
\]
Therefore,
\[
A = \frac{1}{6}.
\]
:::




::: {.example}
Exponential and Linear Functions

Find the area between \( y = e^x \) and \( y = x \) on \([0,1]\).


**Solution:**

**Top curve:** \( e^x \)  
**Bottom curve:** \( x \)

\[
A = \int_0^1 (e^x - x)\,dx
\]

\[
A = \left[e^x - \frac{x^2}{2}\right]_0^1
= \left(e - \frac{1}{2}\right) - (1 - 0)
= e - \frac{3}{2}.
\]

Therefore,
\[
A = e - \frac{3}{2}.
\]
:::




::: {.example}
Intersection Points First

Find the area enclosed by \( y = x^2 \) and \( y = 2x - x^2 \).


**Solution:**


**Step 1: Find intersections**
\[
x^2 = 2x - x^2 \Rightarrow 2x^2 - 2x = 0 \Rightarrow 2x(x-1)=0
\]
\[
x=0,\; x=1
\]

**Step 2: Identify top and bottom**
\[
y_T = 2x - x^2,\quad y_B = x^2
\]

**Step 3: Set up integral**
\[
A = \int_0^1 \big[(2x - x^2) - x^2\big]\,dx
= \int_0^1 (2x - 2x^2)\,dx
\]

**Step 4: Evaluate**
\[
A = 2\int_0^1 (x - x^2)\,dx
= 2\left(\frac{1}{2} - \frac{1}{3}\right)
= 2\cdot \frac{1}{6}
= \frac{1}{3}.
\]

Therefore,
\[
A = \frac{1}{3}.
\]
:::


---

### When Curves Cross (Absolute Value Form)

If the curves switch which one is on top, the area must be computed using
\[
A = \int_a^b |f(x) - g(x)|\,dx,
\]
which is evaluated by **splitting the interval at intersection points** and computing each region separately.

---

### Horizontal Slices (Integrating with Respect to \(y\))

Sometimes regions are easier to describe using horizontal rectangles.

If the region is bounded by
\[
x = x_R(y), \quad x = x_L(y), \quad y \in [c,d],
\]
then
\[
A = \int_c^d \big(x_R(y) - x_L(y)\big)\,dy.
\]

This is especially useful when:

- the curves are sideways parabolas,  
- the region has complicated vertical boundaries,  
- vertical slicing would require splitting into multiple integrals.  

---

### Putting It All Together



> **Core principle:**  
> Area between curves = accumulation of distance between boundaries.

- Area is computed by **integrating vertical or horizontal distances** between curves.  
- Every area problem follows the same structure:  
  1. Identify the region.   
  2. Determine intersection points.  
  3. Identify top/bottom or left/right curves.  
  4. Subtract functions correctly.  
  5. Integrate over the correct limits.  

**Vertical slicing:**
\[
A = \int_a^b \big(y_T - y_B\big)\,dx
\]

**Horizontal slicing:**
\[
A = \int_c^d \big(x_R - x_L\big)\,dy
\]

---

#### Conceptual Takeaways

- Area is **accumulated distance**, not just height.  
- Integration converts geometry into algebra.  
- Choosing the correct slicing direction (vertical vs horizontal) simplifies computation.  
- Graphing the region is not optional — it is part of the solution method.  
- Absolute value represents **total geometric area**, not signed area.  

---

#### Skills You Should Be Able To Do

After mastering this section, you should be able to:

- Sketch regions between curves    
- Find intersection points algebraically or numerically  
- Identify top/bottom and left/right boundaries  
- Set up correct definite integrals for area  
- Use both \(dx\) and \(dy\) formulations  
- Split integrals when curves cross  
- Interpret area in applied contexts (physics, biology, economics)  
- Translate geometry into integral expressions  


---

####  Problems



























##  Volumes

Just as calculus provides a precise definition of **area** using limits and integrals, it also provides a precise definition of **volume**.  While we have an intuitive sense of what volume means, calculus allows us to compute exact volumes for complex solids by breaking them into thin slices and adding their volumes.  

We begin with simple solids like cylinders, where the volume formula is already known. From there, we generalize the idea by slicing more complicated solids into thin slabs, approximating each slab by a cylinder, and using limits to obtain an exact result.

---

### Volume of a Cylinder

A **right cylinder** is formed by two congruent, parallel bases separated by a fixed height \(h\).  If the base has area \(A\), then the volume is
\[
V = A h.
\]

- Circular cylinder (radius \(r\)):
\[
V = \pi r^2 h
\]  

- Rectangular box (length \(l\), width \(w\), height \(h\)):
\[
V = lwh
\]  

These formulas are special cases of a more general integral definition of volume.

---

### Cross-Sections and Slicing

To compute the volume of a general solid:

1. Slice the solid with planes.  
2. Each slice produces a **cross-section**.  
3. Approximate each slice by a thin cylinder.  
4. Add all slices using integration.  

Let \(A(x)\) be the area of the cross-section at position \(x\).  If the solid lies between \(x=a\) and \(x=b\), then its volume is defined by
\[
V = \int_a^b A(x)\,dx.
\]


**Interpretation**

- Each slice has volume \( A(x)\,dx \)  
- The integral adds all these slices  
- Volume = accumulation of cross-sectional areas  


<div class="theorem">
Fundamental Volume Formula

**Vertical slicing:**
\[
V = \int_a^b A(x)\,dx
\]

**Horizontal slicing:**
\[
V = \int_c^d A(y)\,dy
\]
</div>



::: {.example}
Volume of a Sphere

Find the volume of a sphere of radius \(r\).

**Solution:**

At position \(x\), the cross-section is a circle with radius
\[
y = \sqrt{r^2 - x^2}.
\]

So the area is
\[
A(x) = \pi y^2 = \pi(r^2 - x^2).
\]

**Integral**
\[
V = \int_{-r}^{r} \pi(r^2 - x^2)\,dx
\]

**Evaluation**
\begin{align*}
V &= \pi \int_{-r}^{r} (r^2 - x^2)\,dx\\
&= \pi \left[ r^2 x - \frac{x^3}{3} \right]_{-r}^{r}\\
&= \frac{4}{3}\pi r^3.
\end{align*}

Therefore,
\[
V = \frac{4}{3}\pi r^3.
\]
:::



::: {.example}
Solid of Revolution (Disk Method)

Rotate \(y=\sqrt{x}\) from \(x=0\) to \(x=1\) about the \(x\)-axis.


**Solution:**
The cross-sections are disks with radius \( \sqrt{x} \).

\[
A(x) = \pi(\sqrt{x})^2 = \pi x
\]

\begin{align*}
V &= \int_0^1 \pi x\,dx\\
&= \pi \left[\frac{x^2}{2}\right]_0^1\\
&= \frac{\pi}{2}.
\end{align*}

Therefore,
\[
V = \frac{\pi}{2}.
\]
:::



::: {.example}
Washer Method

Rotate the region between \(y=x\) and \(y=x^2\) on \([0,1]\) about the \(x\)-axis.


**Solution:**

- Outer radius: \(R=x\)  
- Inner radius: \(r=x^2\)

\[
A(x) = \pi(R^2 - r^2) = \pi(x^2 - x^4)
\]

\begin{align*}
V &= \int_0^1 \pi(x^2 - x^4)\,dx\\
&= \pi\left[\frac{x^3}{3} - \frac{x^5}{5}\right]_0^1\\
&= \pi\left(\frac{1}{3}-\frac{1}{5}\right)\\
&= \frac{2\pi}{15}.
\end{align*}

Therefore,
\[
V = \frac{2\pi}{15}.
\]
:::




---

### Solids of Revolution


<div class="definition">
A **solid of revolution** is formed by rotating a region about a line.

**Disk Method**
\[
A = \pi r^2
\]

**Washer Method**
\[
A = \pi(R^2 - r^2)
\]

**Volume Formula**
\[
V = \int A\,dx \quad \text{or} \quad V = \int A\,dy.
\]
</div>


---

### Pulling It All Together

> Volume is accumulated cross-sectional area.

Every volume problem follows the same structure:

1. Identify the slicing direction  
2. Determine the cross-section shape   
3. Express its area as a function  
4. Integrate over the correct limits  

> **Core principle:**  
> Geometry defines the cross-section.  
> Calculus accumulates it.  
> Integration creates volume.

---

#### Conceptual Takeaways

- Volume is defined through **limits and accumulation**
- Slicing turns 3D geometry into 1D integration
- Geometry determines the algebra
- Choosing the right slicing direction simplifies the problem
- All volume formulas come from the same integral principle
- Disks, washers, and general cross-sections are just special cases of \(V=\int A\)

---

#### Skills You Should Be Able To Do

After this section, you should be able to:

- Interpret volume as accumulated area
- Identify appropriate cross-sections
- Choose between vertical and horizontal slicing
- Set up volume integrals correctly
- Use disk and washer methods
- Compute volumes of solids of revolution
- Compute volumes of non-revolution solids
- Translate geometry into area functions
- Sketch solids and cross-sections
- Model real-world volume problems mathematically


---

####  Problems



























##  Volumes by Cylindrical Shells


Some volume problems become algebraically difficult when using the **disk** or **washer** methods, especially when the region must be solved for \( x \) in terms of \( y \) (or vice versa). In many cases, this leads to complicated equations that are hard to invert or integrate.

The **method of cylindrical shells** provides an alternative approach that often avoids these difficulties. Instead of slicing the region perpendicular to the axis of rotation, we slice it parallel to the axis of rotation. When these slices are rotated, they form **thin cylindrical shells**.

This method is especially useful when:

- Solving for one variable is algebraically difficult  
- Washer radii are complicated functions  
- The region is more naturally described using vertical or horizontal strips  



A typical cylindrical shell has:

- **Radius**: distance from the axis of rotation to the slice   
- **Circumference**: \( 2\pi(\text{radius}) \)  
- **Height**: length of the slice (given by the function or difference of functions)  
- **Thickness**: \( dx \) or \( dy \)



The volume of a single shell is given by:
\[
V_{\text{shell}} \;\approx\; (\text{circumference})(\text{height})(\text{thickness})
\]
\[
V_{\text{shell}} \;\approx\; 2\pi r \, h \, \Delta x
\]


<div class="theorem">
Shell Method Formula

**Rotation about the y-axis using vertical slices**

If a region under \( y = f(x) \) from \( x=a \) to \( x=b \) is rotated about the y-axis:
\[
V \;=\; \int_a^b 2\pi x\, f(x)\, dx,
\]
where:

- radius \( = x \)  
- height \( = f(x) \)  
- thickness \( = dx \)  

**Rotation about the x-axis using horizontal slices**

\[
V \;=\; \int_c^d 2\pi y\, (\text{horizontal length})\, dy
\]

where:

- radius \( = y \)  
- height = horizontal length of region  
- thickness \( = dy \)  
</div>




::: {.example}
Rotation about the y-axis

  
Find the volume of the solid obtained by rotating the region bounded by  
\[
y = 2x^2 - x^3 \quad \text{and} \quad y = 0
\]
about the **y-axis**.

**Solution:**

Using cylindrical shells:

- radius: \( x \)  
- height: \( 2x^2 - x^3 \)  
- thickness: \( dx \)  

\begin{align*}
V &= \int_0^2 2\pi x(2x^2 - x^3)\, dx\\
& = 2\pi \int_0^2 (2x^3 - x^4)\, dx\\
& = 2\pi \left[ \frac{1}{2}x^4 - \frac{1}{5}x^5 \right]_0^2\\
& = 2\pi \left( 8 - \frac{32}{5} \right) \\
& = \frac{16\pi}{5}.
\end{align*}
:::


::: {.example}
Region Between Two Curves

 
Rotate the region between \( y = x \) and \( y = x^2 \) about the y-axis.

**Solution:**

- radius: \( x \)  
- height: \( x - x^2 \)

\begin{align*}
V &= \int_0^1 2\pi x(x - x^2)\, dx\\
& = 2\pi \int_0^1 (x^2 - x^3)\, dx\\
& = 2\pi \left[ \frac{x^3}{3} - \frac{x^4}{4} \right]_0^1\\
& = 2\pi\left(\frac{1}{3} - \frac{1}{4}\right)\\
& = \frac{\pi}{6}.
\end{align*}
:::


::: {.example}
Rotation about the x-axis

Rotate the region under \( y=\sqrt{x} \) from \( x=0 \) to \( x=1 \) about the **x-axis**.

**Solution:**

Rewrite as:  
\[
x = y^2
\]

Using shells (horizontal slices):

- radius: \( y \)  
- height: \( 1 - y^2 \)

\begin{align*}
V &= \int_0^1 2\pi y(1 - y^2)\, dy\\
&= 2\pi \int_0^1 (y - y^3)\, dy\\
&= 2\pi \left[ \frac{y^2}{2} - \frac{y^4}{4} \right]_0^1\\
&= 2\pi\left(\frac{1}{2} - \frac{1}{4}\right)\\
&= \frac{\pi}{2}.
\end{align*}
:::


Shells also work when rotating about lines like \( x=c \).

**General Pattern:**

- radius = distance from slice to line of rotation  
- circumference = \( 2\pi(\text{radius}) \)  
- height = top − bottom of region  

---

### Pulling It All Together

**When Shells Are Better Than Washers**

Use shells when:

- Solving for \( x \) in terms of \( y \) is difficult  
- Washer radii require complicated algebra  
- Region is naturally described with vertical or horizontal strips  
- One method gives one integral instead of two  
- One variable leads to simpler bounds  


\[
V = \int (\text{circumference})(\text{height})(\text{thickness})
\]

\[
V = \int 2\pi(\text{radius})(\text{height})\, d(\text{variable})
\]


#### Conceptual Takeaways

The shell method works by:
- slicing **parallel** to the axis of rotation
- rotating each slice into a **cylindrical shell**
- adding up volumes using integration

Instead of finding cross-sectional areas (disks/washers), we use **surface geometry**:
> circumference × height × thickness

This geometric viewpoint makes many problems algebraically simpler and conceptually clearer.


---



#### Skills You Should Be Able To Do

After this section, you should be able to:

- Identify when the shell method is preferable to disks/washers
- Sketch the region and axis of rotation
- Determine:
  - shell radius
  - shell height
  - correct thickness variable
- Set up shell integrals correctly
- Choose correct bounds of integration
- Handle rotations about:
  - x-axis
  - y-axis
  - vertical lines \( x=c \)
  - horizontal lines \( y=c \)
- Compare shell vs washer methods
- Translate geometry into integrals
- Interpret volume using geometric reasoning


---

####  Problems



























##  Work

---

####  Problems



























##  Average Value of a Function


Finding the average of finitely many numbers is straightforward:
\[
y_{\text{avg}} = \frac{y_1 + y_2 + \cdots + y_n}{n}
\]

But many real-world quantities vary continuously:

- temperature over a day  
- velocity during a trip  
- power usage over time  
- density across a region  

In these cases, there are infinitely many values, so we use integration to define an average value for a function over an interval.


<div class="note">
**Deriving the Average Value Formula**

Let \( y = f(x) \) be defined on an interval \( [a,b] \).

1. Divide the interval into \( n \) subintervals of width:
\[
\Delta x = \frac{b-a}{n}
\]

2. Choose a sample point \( x_i^* \) in each subinterval.

3. Average the function values:
\[
\frac{f(x_1^*) + f(x_2^*) + \cdots + f(x_n^*)}{n}
\]

Rewriting using \( \Delta x \) and taking the limit as \( n \to \infty \):

\[
f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx
\]
</div>



**Average Value Formula**

\[
f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx.
\]


**Geometric Interpretation**

For **positive functions**:
\[
\frac{\text{Area under curve}}{\text{Width of interval}} = \text{Average height}
\]
This means the average value is the **height of a rectangle** over \( [a,b] \) that has the **same area** as the region under \( f(x) \).


::: {.example}
Computing an Average Value

 
Find the average value of  
\[
f(x) = 1 + x^2
\]
on the interval \( [-1,2] \).

**Solution:**

\begin{align*}
f_{\text{avg}} &= \frac{1}{2 - (-1)}\int_{-1}^{2} (1 + x^2)\,dx\\
& = \frac{1}{3}\int_{-1}^{2} (1 + x^2)\,dx\\
& = \frac{1}{3}\left[ x + \frac{x^3}{3} \right]_{-1}^{2}\\
& = \frac{1}{3}\Big[(2 + \tfrac{8}{3}) - (-1 - \tfrac{1}{3})\Big]\\
& = \frac{1}{3}(6)\\
& = 2.
\end{align*}
:::


---

### Mean Value Theorem for Integrals


<div class="definition">
**MVT for Integrals**

If \( f \) is **continuous** on \( [a,b] \), then there exists at least one number  
\( c \in [a,b] \) such that:
\[
f(c) = f_{\text{avg}} = \frac{1}{b-a}\int_a^b f(x)\,dx.
\]

Equivalently:
\[
\int_a^b f(x)\,dx = f(c)(b-a).
\]
</div>

Geometrically, there exists a point \( c \) where the function’s height equals the average height over the interval.

> This means: a rectangle of height \( f(c) \) and base \( [a,b] \) has the same area as the area under the curve.


::: {.example}
Finding the Mean Value Point

Using the previous example, we found:
\[
f_{\text{avg}} = 2
\]

Solve:
\[
f(c) = 2
\]

\[
1 + c^2 = 2
\]

\[
c^2 = 1
\]

\[
c = \pm 1
\]

Both \( c=-1 \) and \( c=1 \) lie in the interval \( [-1,2] \), so both satisfy the theorem.
</div>


---

### Application: Average Velocity

Let \( s(t) \) be position and \( v(t) = s'(t) \) be velocity.

**Average velocity (definition):**
\[
v_{\text{avg}} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}
\]

**Average value of velocity function:**
\[
v_{\text{avg}} = \frac{1}{t_2 - t_1}\int_{t_1}^{t_2} v(t)\,dt
\]

Using the Net Change Theorem:
\[
\int_{t_1}^{t_2} v(t)\,dt = s(t_2) - s(t_1)
\]

So:
\[
\text{Average velocity} = \text{Average value of velocity function}
\]

This shows that **physical average velocity** and **integral-based average value** are mathematically identical.


---

### Pulling It All Together

Think:
> **Average value = total accumulation ÷ interval length**

or

> **Area ÷ width = average height**


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

\[
\int_a^b f(x)\,dx = f(c)(b-a)
\]

---

#### Conceptual Takeaways

- Average value of a function is a **continuous analog** of averaging numbers  
- Integration replaces summation  
- Area plays the role of total accumulation  
- Division by interval length produces an average height  
- The Mean Value Theorem for Integrals guarantees that:  
  - the average value is actually **attained** by the function  
- Average value has **geometric meaning** and **physical meaning**  

---

#### Skills You Should Be Able To Do

After this section, you should be able to:

- Explain average value conceptually  
- Derive the average value formula  
- Compute average values using integrals  
- Interpret average value geometrically  
- Apply the Mean Value Theorem for Integrals  
- Find values \( c \) such that \( f(c) = f_{\text{avg}} \)  
- Connect area to physical meaning  
- Apply average value to:  
  - velocity  
  - temperature  
  - accumulation processes  
- Translate between physical averages and integral averages  
- Explain the difference between:  
  - discrete averages  
  - continuous averages  


---

####  Problems