Integrals.Rmd

# Integrals

##  Areas and Distances 

Two fundamental problems motivate the development of integral calculus:

1. **The Area Problem** – finding the area under a curve.  
2. **The Distance Problem** – finding the total distance traveled when velocity varies over time.  

Although these problems appear different, they are solved using the same core idea:  **approximate with rectangles, then take a limit**.  This limiting process leads to the concept of the **definite integral**.

---

### The Area Problem

We want to find the area of a region \( S \) bounded by:

- the graph of a continuous function \( y = f(x) \),  
- the vertical lines \( x = a \) and \( x = b \),  
- and the \(x\)-axis (with \( f(x) \ge 0 \)).  

For simple shapes (rectangles, triangles, polygons), area formulas are exact.  For curved boundaries, we instead:

1. Divide the interval \([a,b]\) into small subintervals.  
2. Approximate the region using rectangles.  
3. Add the rectangle areas.  
4. Take a limit as the number of rectangles increases.  

This process creates increasingly accurate approximations.


<div class="note">
**Riemann Sum Approximation**

Let the interval \([a,b]\) be divided into \(n\) equal subintervals.

- Width of each subinterval:
\[
\Delta x = \frac{b-a}{n}
\]

- Sample points in each subinterval:  
  \( x_1^*, x_2^*, \dots, x_n^* \)

- Rectangle areas:
\[
f(x_i^*)\Delta x
\]

**General Area Approximation**
\[
A \approx \sum_{i=1}^n f(x_i^*)\,\Delta x
\]

This sum is called a **Riemann sum**.
</div>


---

### Types of Riemann Sums

- **Left-endpoint sum**:
\[
L_n = \sum_{i=1}^n f(x_{i-1})\Delta x
\]

- **Right-endpoint sum**:
\[
R_n = \sum_{i=1}^n f(x_i)\Delta x
\]

- **Midpoint sum**:
\[
M_n = \sum_{i=1}^n f(x_i^*)\Delta x
\]
(where \( x_i^* \) is the midpoint of each subinterval)


<div class="definition">
**Definition of Area (Integral Definition)**

The exact area is defined as a limit:
\[
A = \lim_{n \to \infty} \sum_{i=1}^n f(x_i^*) \Delta x.
\]
This limit (when it exists) is the **definite integral**:
\[
A = \int_a^b f(x)\,dx.
\]
</div>


::: {.example}
Find the area Under \( y = x^2 \) on \([0,1]\)

**Solution:**

**Step 1: Define the partition**
\[
\Delta x = \frac{1-0}{n} = \frac{1}{n}
\]

Right endpoints:
\[
x_i = \frac{i}{n}
\]

**Step 2: Build the Riemann sum**
\[
R_n = \sum_{i=1}^n \left(\frac{i}{n}\right)^2 \frac{1}{n}
= \frac{1}{n^3} \sum_{i=1}^n i^2
\]

**Step 3: Use the sum formula**
\[
\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}
\]

So:
\[
R_n = \frac{1}{n^3}\cdot \frac{n(n+1)(2n+1)}{6}
= \frac{(n+1)(2n+1)}{6n^2}
\]

**Step 4: Take the limit**
\[
\lim_{n\to\infty} R_n = \frac{1}{3}
\]

Therefore,
\[
A = \int_0^1 x^2\,dx = \frac{1}{3}.
\]
:::



::: {.example}
Exponential Function

Find the area under \( f(x)=e^{-2x} \) on \([0,2]\) using a limit expression.

**Solution:**
\[
\Delta x = \frac{2}{n}, \quad x_i = \frac{2i}{n}
\]

\[
A = \lim_{n\to\infty} \sum_{i=1}^n e^{-2x_i}\Delta x
= \lim_{n\to\infty} \frac{2}{n}\sum_{i=1}^n e^{-4i/n}
\]

This is a Riemann sum representation of the integral:
\[
\int_0^2 e^{-2x}\,dx
\]
:::



---

### The Distance Problem

If velocity varies with time, distance cannot be computed by  
\[
\text{distance} = \text{velocity} \times \text{time}.
\]

Instead, we approximate:

1. Break time into small intervals \(\Delta t\).  
2. Assume velocity is constant on each interval.  
3. Add distances:  
\[
\text{distance} \approx \sum v(t_i^*)\Delta t
\]

Taking the limit gives:
\[
d = \lim_{n\to\infty} \sum_{i=1}^n v(t_i^*)\Delta t
= \int_a^b v(t)\,dt
\]


Both the area and distance problems lead to the same structure:

\[
\text{Accumulated quantity} = \lim_{n\to\infty} \sum (\text{rate})\times(\text{small interval})
\]

This defines integration as **accumulation of continuous change**.

---

### Putting It All Together


> Integration is the mathematical process of approximating accumulation with rectangles and taking a limit to obtain exact total change.

#### Conceptual Takeaways

- Area and distance are both accumulation problems.  
- Curved regions are handled using rectangle approximations.  
- Increasing the number of rectangles improves accuracy.  
- Exact values come from taking limits.  
- The definite integral is defined as a limit of Riemann sums.  
- Integration measures **total accumulation**, not just geometric area.  

---

#### Skills You Should Be Able to Do

By the end of this section, you should be able to:

- Construct left, right, and midpoint Riemann sums  
- Interpret \( \Delta x \) and \( \Delta t \) correctly  
- Translate word problems into summation form  
- Write definite integrals as limits of sums  
- Recognize integrals as accumulation processes  
- Connect geometry (area) with physical meaning (distance, work, flow, output)  
- Use sigma notation for compact summation expressions  
- Explain integration conceptually, not just computationally  


---

####  Problems





























##  The Definite Integral 

Many problems in calculus lead to limits of sums of the form
\[
\lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\,\Delta x,
\]
where an interval \([a,b]\) is divided into many small subintervals and values of a function are multiplied by small widths and added together.

This structure appears in:

- area under curves,  
- distance traveled,  
- work,  
- volumes,  
- arc length,  
- mass and center of mass,  
- fluid pressure,  
- and many other physical and geometric applications.  

Because this type of limit occurs so frequently, it is given a special name: **the definite integral**.


<div class="definition">

Let \( f \) be defined on \( [a,b] \). Divide the interval into \( n \) subintervals of equal width
\[
\Delta x = \frac{b-a}{n}.
\]
Let \( x_i^* \) be any sample point in the \(i\)-th subinterval.  

The **definite integral** of \( f \) from \( a \) to \( b \) is defined by
\[
\int_a^b f(x)\,dx \;=\; \lim_{n\to\infty} \sum_{i=1}^n f(x_i^*)\,\Delta x,
\]
provided the limit exists and is the same for all choices of sample points.
</div>



If this limit exists, we say that **\( f \) is integrable on \([a,b]\)**.


<div class="note">
Terminology and Notation

\[
\int_a^b f(x)\,dx
\]

- \( \int \) : integral sign (represents a limit of sums)  
- \( f(x) \) : integrand  
- \( a \) : lower limit of integration  
- \( b \) : upper limit of integration  
- \( dx \) : indicates the variable of integration  

The entire expression \( \int_a^b f(x)\,dx \) represents a number, not a function.
</div>

---

### Riemann Sums


<div class="definition">

The sum
\[
\sum_{i=1}^n f(x_i^*)\,\Delta x
\]
is called a **Riemann sum**.
</div>

The definite integral is the **limit of Riemann sums**.

Geometrically,

**Case 1: \( f(x) \ge 0 \) on \([a,b]\)**
\[
\int_a^b f(x)\,dx = \text{area under } y=f(x) \text{ from } a \text{ to } b
\]

**Case 2: \( f(x) \) changes sign**
\[
\int_a^b f(x)\,dx = A_{\text{above}} - A_{\text{below}}
\]

This is called **net area**.


<div class="theorem">
Integrability Theorem

If \( f \) is:

- continuous on \([a,b]\), or   
- has only finitely many jump discontinuities,

then \( f \) is integrable on \([a,b]\), and the definite integral exists.
</div>



If right endpoints are chosen as sample points:
\[
x_i = a + i\Delta x,
\quad \Delta x = \frac{b-a}{n},
\]
then
\[
\int_a^b f(x)\,dx
= \lim_{n\to\infty} \sum_{i=1}^n f(x_i)\,\Delta x.
\]

This form is especially useful for computations.


::: {.example}
Converting a Limit into an Integral

Express the following limit as an integral
\[
\lim_{n\to\infty} \sum_{i=1}^n (x_i^3 + x_i\sin x_i)\,\Delta x
\quad \text{on } [0,\pi]
\]

**Solution:**
Identify the function:
\[
f(x) = x^3 + x\sin x.
\]

Then the limit becomes:
\[
\int_0^{\pi} (x^3 + x\sin x)\,dx
\]
:::



::: {.example}
Integral as Net Area

Sketch the area and determine what the integral represents
\[
\int_0^3 (x^3 - 6x)\,dx
\]

**Solution:**

Since the function takes both positive and negative values, the integral represents:
\[
\text{net area} = A_{\text{above}} - A_{\text{below}}.
\]

This is not purely geometric area — it is **signed accumulation**.
:::



::: {.example}
Integral as a Limit of Sums

Set up
\[
\int_1^3 e^x\,dx
\]
as a limit.

**Solution:**


\[
\Delta x = \frac{3-1}{n} = \frac{2}{n}, 
\quad x_i = 1 + \frac{2i}{n}
\]

\[
\int_1^3 e^x\,dx
=
\lim_{n\to\infty}
\sum_{i=1}^n
e^{\,1+\frac{2i}{n}}\;\frac{2}{n}
\]
:::

### The Midpoint Rule

<div class="theorem">
The Midpoint Rule (Numerical Approximation)

Instead of using endpoints, choose **midpoints**:
\[
x_i^* = \frac{x_{i-1}+x_i}{2}.
\]

Then
\[
\int_a^b f(x)\,dx \;\approx\; 
\sum_{i=1}^n f(x_i^*)\,\Delta x
\]

This is called the **Midpoint Rule** and usually gives better accuracy than left or right sums.
</div>



::: {.example}
Midpoint Rule Approximation

Approximate
\[
\int_1^2 \frac{1}{x}\,dx
\quad \text{with } n=5.
\]

**Solution:**

\[
\Delta x = \frac{2-1}{5} = \frac{1}{5}
\]

Midpoints:
\[
1.1,\; 1.3,\; 1.5,\; 1.7,\; 1.9
\]

\[
\int_1^2 \frac{1}{x}\,dx \approx
\frac{1}{5}\left(
\frac{1}{1.1} + \frac{1}{1.3} + \frac{1}{1.5} + \frac{1}{1.7} + \frac{1}{1.9}
\right)
\approx 0.692
\]
:::

### Properties of the Definite Integral

<div class="theorem">
Properties of the Definite Integral

For continuous functions \( f \) and \( g \):

1. **Constant Rule**
\[
\int_a^b c\,dx = c(b-a)
\]

2. **Sum Rule**
\[
\int_a^b [f(x)+g(x)]dx
=
\int_a^b f(x)dx + \int_a^b g(x)dx
\]

3. **Constant Multiple Rule**
\[
\int_a^b cf(x)\,dx = c\int_a^b f(x)\,dx
\]

4. **Difference Rule**
\[
\int_a^b [f(x)-g(x)]dx
=
\int_a^b f(x)dx - \int_a^b g(x)dx
\]

5. **Additivity Over Intervals**
\[
\int_a^c f(x)dx + \int_c^b f(x)dx
=
\int_a^b f(x)dx
\]
</div>



<div class="theorem">
Comparison Properties

If \( a<b \):

- If \( f(x) > 0 \), then
\[
\int_a^b f(x)\,dx > 0
\]

- If \( f(x) > g(x) \), then
\[
\int_a^b f(x)\,dx > \int_a^b g(x)\,dx
\]

- If \( m \le f(x) \le M \), then
\[
m(b-a) \le \int_a^b f(x)\,dx \le M(b-a)
\]

</div>


::: {.example}
Bounding an Integral

Estimate
\[
\int_0^1 e^{-2x^2}\,dx.
\]

**Solution:**

Since \( e^{-2x^2} \) is decreasing on \([0,1]\):

\[
m = e^{-2}, \quad M = 1
\]

So:
\[
e^{-2} < \int_0^1 e^{-2x^2}\,dx < 1
\]

Numerically:
\[
0.367 < \int_0^1 e^{-2x^2}\,dx < 1
\]
:::

---

### Putting It All Together


> A definite integral is the mathematical limit of accumulation: infinitely many tiny contributions summed to produce total change.


#### Conceptual Takeaways

- A definite integral is a **limit of sums**  
- Integration is **accumulation of continuous change**  
- Definite integrals produce **numbers**, not functions   
- Riemann sums are approximations of integrals  
- Integrals represent:  
  - area,  
  - net area,  
  - distance,  
  - work,  
  - mass,  
  - energy,  
  - total change  
- The sign of a function affects the meaning of the integral  
- Integration is fundamentally about **adding infinitely many infinitesimal contributions**  

---

#### Skills You Should Be Able to Do

By the end of this section, you should be able to:

- Construct Riemann sums from a function  
- Interpret definite integrals geometrically and physically  
- Convert limits of sums into integrals  
- Write integrals as limits of sums  
- Identify when an integral represents area vs net area  
- Use the Midpoint Rule for approximation  
- Apply integral properties to simplify expressions  
- Estimate integrals using bounds  
- Compare integrals using inequalities  
- Interpret integrals as accumulation processes  
- Recognize integrals in applied contexts (distance, work, growth, flow, mass)  



---

####  Problems



























##  The Fundamental Theorem of Calculus 

The **Fundamental Theorem of Calculus** establishes the deep connection between the two main branches of calculus:

- **Differential calculus** (rates of change, slopes, tangents)  
- **Integral calculus** (accumulation, area, total change)  

Although these areas originated from different problems (tangent lines vs. area), the theorem shows that they are inverse processes of each other.  

This insight transforms calculus from a collection of techniques into a coherent mathematical system, making it possible to compute areas, accumulated quantities, and total change efficiently—without relying on complicated limits of sums.

---

### FTC Part 1: Derivative of an Integral (FTC1)


<div class="theorem">
Let \( f \) be continuous on \([a,b]\), and define a function
\[
T(x) = \int_a^x f(t)\,dt.
\]
Then:
\[
T'(x) = f(x)
\]
</div>

This means:

> **The derivative of an accumulation function is the original function.**

**Interpretation:**

- \(T(x)\) represents **accumulated area** (or total change) from \(a\) to \(x\).  
- The derivative \(T'(x)\) represents the **rate at which that accumulation is changing**.  
- That rate is exactly the value of the original function \(f(x)\).  

So:
- accumulation → derivative → original function

Geometrically,

If \( f(x) > 0 \), then:

- \(T(x)\) increases  
- \(T'(x) > 0\)  

If \( f(x) < 0 \), then:

- \(T(x)\) decreases  
- \(T'(x) < 0\)  

The graph of \(T'(x)\) has the same shape as the graph of \(f(x)\).


::: {.example}
Area Function

Let
\[
T(x) = \int_0^x f(t)\,dt.
\]
If \(f\) is positive up to \(x=3\) and negative after, describe the behaviour of $T(x)$.

**Solution:**

By the FTC1,
\[
T'(x) = f(x)
\]


If \(f\) is positive up to \(x=3\) and negative after, then:

- \(T(x)\) increases until \(x=3\)  
- \(T(x)\) has a maximum at \(x=3\)  
- \(T(x)\) decreases after \(x=3\)  
:::


::: {.example}
Direct Application of FTC1

Find $T^{\prime}(x)$ given
\[
T(x) = \int_0^x (1+t^2)\,dt
\]

**Solution:**
\[
T'(x) = 1 + x^2
\]
:::


::: {.example}
Chain Rule + FTC1

Find $T^{\prime}(x)$ given
\[
T(x) = \int_1^{x^4} \sec t\,dt
\]

**Solution:**

Let \(u = x^4\). Then:

\[
\frac{d}{dx}\int_1^{x^4} \sec t\,dt
= \sec(x^4)\cdot \frac{d}{dx}(x^4)
\]

So,
\[
T'(x) = 4x^3\sec(x^4).
\]
:::


---

### FTC Part 2: Evaluating Definite Integrals (FTC2)


<div class="theorem">
If \( f \) is continuous on \([a,b]\) and \(F\) is any antiderivative of \(f\) (so \(F' = f\)), then:
\[
\int_a^b f(x)\,dx = F(b) - F(a).
\]
</div>

This replaces **limits of sums** with **simple subtraction**.

**Interpretation:**

> **To compute total accumulation, evaluate an antiderivative at the endpoints and subtract.**

This is the computational engine of calculus.


::: {.example}
Polynomial Integral

Evaluate:
\[
\int_1^3 e^x\,dx
\]

**Solution:**
An antiderivative is \(F(x)=e^x\).  Therefore,
\[
\int_1^3 e^x\,dx = e^3 - e.
\]
:::



::: {.example}
Area Under a Curve

Find the area under \( y=x^2 \) from 0 to 1:
\[
A = \int_0^1 x^2\,dx
\]

**Solution:**
\[
A = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}
\]
:::



::: {.example}
Logarithmic Integral

Evaluate:
\[
\int_3^6 \frac{1}{x}\,dx
\]

**Solution:**

\[
\int_3^6 \frac{1}{x}\,dx = [\ln x]_3^6 = \ln 6 - \ln 3 = \ln 2
\]
:::


::: {.example}
Trigonometric Area

Evaluate
\[
\int_0^{\pi/2} \cos x\,dx
\]

**Solution:**
\[
\int_0^{\pi/2} \cos x\,dx = [\sin x]_0^{\pi/2} = 1.
\]
:::


**Differentiation and Integration are Inverse Processes**

The FTC shows:

-  Integrate → Differentiate  
\[
\frac{d}{dx}\left(\int_a^x f(t)\,dt\right) = f(x)
\]  

-  Differentiate → Integrate  
\[
\int_a^b F'(x)\,dx = F(b)-F(a)
\]

So:
\[
\text{Differentiation and integration undo each other.}
\]

---

### Putting It All Together



> The Fundamental Theorem of Calculus reveals that **change and accumulation are two sides of the same process**. Differentiation measures change, integration measures accumulation, and each operation undoes the other. This single idea transforms calculus into a unified theory of continuous change.

#### Conceptual Takeaways

- Accumulation creates functions  
- Derivatives measure accumulation rates  
- Integration measures total change  
- Area is a special case of accumulation  
- Definite integrals depend only on endpoint values  
- The entire structure of calculus rests on this theorem  
- Differentiation and integration are inverse operations  
- Motion interpretation:  
  - velocity → integrate → position  
  - position → differentiate → velocity  
- FTC is the bridge between geometry, physics, and analysis  
- Complex accumulation problems reduce to simple subtraction  

---

#### Skills You Should Be Able to Do

After this section, you should be able to:

- Interpret integrals as accumulation functions  
- Differentiate functions defined by integrals  
- Apply FTC1 directly  
- Use the Chain Rule with FTC1  
- Recognize area functions  
- Evaluate definite integrals using antiderivatives  
- Apply FTC2 efficiently  
- Translate between physical and mathematical interpretations  
- Identify when FTC applies (continuity conditions)  
- Compute total change from rate functions  
- Interpret integrals in motion problems  
- Understand accumulation graphs  
- Connect slope behavior to integral behavior  
- Move fluently between graphs, formulas, and meaning  


---

####  Problems



























##  Indefinite Integrals and the Net Change Theorem 



The Fundamental Theorem of Calculus (FTC) creates a powerful link between derivatives, antiderivatives, and definite integrals. This connection allows us to compute areas, accumulated change, and physical quantities using antiderivatives rather than limits of sums.

Two core ideas drive this section:

1. **Indefinite integrals** provide a systematic notation for families of antiderivatives.  
2. **The Net Change Theorem** interprets definite integrals as total accumulated change of a quantity.  

These ideas form the foundation for applications in physics, biology, economics, and engineering.

---

### Indefinite Integrals


<div class="definition">
An **indefinite integral** represents the family of all antiderivatives of a function.
\[
\int f(x)\,dx = F(x) + C
\quad \text{means} \quad
F'(x) = f(x).
\]
</div>

Here:

- \( F(x) \) is any antiderivative of \( f(x) \)  
- \( C \) is the **constant of integration**, representing infinitely many antiderivatives  


An indefinite integral is:

- **Not a number**  
- A **family of functions**  
- A symbolic way to represent all antiderivatives of a function  

In contrast:

- A **definite integral** is a **number**  
- An **indefinite integral** is a **function (family of functions)**  

---

### Fundamental Connection to Definite Integrals

If \( f \) is continuous on \([a,b]\) and \( F' = f \), then:
\[
\int_a^b f(x)\,dx = F(b) - F(a).
\]

This formula allows us to compute definite integrals by:

1. Finding an antiderivative \( F(x) \)  
2. Evaluating it at the endpoints  
3. Subtracting  



<div class="definition">
Rules for Indefinite Integrals

**Linearity**
\[
\int [af(x) + bg(x)]\,dx = a\int f(x)\,dx + b\int g(x)\,dx
\]

**Power Rule**
\[
\int x^n\,dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1)
\]

\[
\int \frac{1}{x}\,dx = \ln|x| + C
\]

**Exponential**
\[
\int e^x\,dx = e^x + C
\qquad
\int b^x\,dx = \frac{b^x}{\ln b} + C
\]

**Trigonometric**
\[
\int \sin x\,dx = -\cos x + C
\qquad
\int \cos x\,dx = \sin x + C
\]

\[
\int \sec^2 x\,dx = \tan x + C
\qquad
\int \csc^2 x\,dx = -\cot x + C
\]
</div>



::: {.example}
Polynomial and Trigonometric Terms

Evaluate
\[
\int \left(10x^4 - 2\sec^2 x\right)\,dx
\]

**Solution**

\begin{align*}
\int \left(10x^4 - 2\sec^2 x\right)\,dx &= 10\int x^4 dx - 2\int \sec^2 x dx\\
&= 10\left(\frac{x^5}{5}\right) - 2(\tan x) + C\\
&= 2x^5 - 2\tan x + C.
\end{align*}
:::


::: {.example}
Trigonometric Simplification

Evaluate
\[
\int \frac{\cos \theta}{\sin^2 \theta}\,d\theta
\]

**Solution**

Rewrite:

\[
\frac{\cos \theta}{\sin^2 \theta} = \csc \theta \cot \theta
\]

So:

\[
\int \csc \theta \cot \theta\, d\theta = -\csc \theta + C
\]
:::



::: {.example}
Definite Integral Using FTC

Evaluate
\[
\int_0^3 (x^3 - 6x)\,dx
\]

**Solution**

Find antiderivative:

\[
F(x) = \frac{x^4}{4} - 3x^2
\]

Apply FTC:

\[
F(3) - F(0) = \left(\frac{81}{4} - 27\right) - 0 = -\frac{27}{4}
\]
:::


---

### The Net Change Theorem


<div class="theorem">  
**Net Change Theorem**


If \( F'(x) = f(x) \), then:
\[
\int_a^b f(x)\,dx = F(b) - F(a)
\]
</div>


> **The integral of a rate of change is the total (net) change.**

This means:

If a function represents a **rate**, then integrating it gives the **accumulated quantity**.


---

### Applications of Net Change

| Rate Function | Integral Represents |
|------|------|
| Velocity \( v(t) \) | Displacement |
| Speed \( |v(t)| \) | Distance traveled |
| Flow rate | Total volume |
| Population growth rate | Population change |
| Marginal cost | Total cost change |
| Reaction rate | Concentration change |
| Power | Energy used |
| Acceleration | Change in velocity |




::: {.example}
Motion Interpretation

Given that
\[
\text{Displacement} = \int_{t_1}^{t_2} v(t)\,dt,
\]
the distance traveled is
\[
\text{Distance} = \int_{t_1}^{t_2} |v(t)|\,dt.
\]
:::

> Displacement measures **net movement**  
> Distance measures **total path length**


---

### Putting It All Together



Indefinite integrals provide a symbolic language for antiderivatives, while the Fundamental Theorem of Calculus allows definite integrals to be evaluated through these antiderivatives. The Net Change Theorem gives integration its physical and applied meaning: **integration accumulates change**. Together, these ideas transform integration from a computational tool into a universal modeling framework for real-world systems.



---

#### Conceptual Takeaways

- Indefinite integrals represent **families of antiderivatives**  
- The constant \( C \) reflects infinitely many vertical shifts  
- Definite integrals measure **accumulated change**  
- FTC connects **rates → accumulation**  
- Integration converts:  
  - velocity → displacement  
  - rate → total quantity  
  - marginal change → net change  

---

#### Skills You Should Be Able to Do

After this section, you should be able to:

- Distinguish between **definite** and **indefinite** integrals  
- Compute basic indefinite integrals  
- Use the constant of integration correctly  
- Apply FTC to evaluate definite integrals  
- Interpret integrals as accumulated change  
- Model physical meaning of integrals (distance, mass, energy, cost, population)  
- Separate displacement and distance using absolute value  
- Translate real-world rates into integrals  
- Use integration to compute net change in applied contexts  

---

####  Problems



























##  The Substitution Rule 


The Fundamental Theorem of Calculus allows us to evaluate integrals using antiderivatives, but many integrals cannot be evaluated directly using basic antidifferentiation formulas. In these cases, we use a strategy called **substitution**, also known as **u-substitution**.

The core idea is simple:

> Replace a complicated expression inside an integral with a new variable to transform the integral into a simpler form.

> This method is the integration counterpart of the Chain Rule from differentiation.


Consider an integral like:
\[
\int 2x\sqrt{1+x^2}\,dx
\]

This cannot be evaluated directly using basic rules.  However, the expression \(1+x^2\) appears inside the square root, and its derivative \(2x\) also appears in the integrand.  This structure suggests a substitution.



<div class="theorem">
The Substitution Rule (Indefinite Integrals)

If \( u = g(x) \) is differentiable and \( f \) is continuous, then:
\[
\int f(g(x))\,g'(x)\,dx = \int f(u)\,du
\]
</div>


Since \( du = g'(x)\,dx \), the integral becomes:
\[
\int f(g(x))g'(x)\,dx = \int f(u)\,du
\]


> We treat \( dx \) and \( du \) like differentials and change variables to simplify the structure of the integral.



<div class="note">
**Strategy for Choosing a Substitution**

1. Look for a **composite function** (an "inside function")  
2. Choose:  
   \[
   u = \text{inner expression}
   \]  
3. Check whether \( du \) (or a constant multiple of it) appears in the integrand  
4. Rewrite the entire integral in terms of \( u \)  
5. Integrate  
6. Substitute back to the original variable  
</div>

> Choosing a good substitution is partly skill and partly pattern recognition.


::: {.example}
Trigonometric Composite Function

Evaluate
\[
\int x^3 \cos(x^4+2)\,dx
\]

**Solution:**

Let:
\[
u = x^4 + 2 \quad \Rightarrow \quad du = 4x^3 dx
\]

So:
\[
x^3 dx = \frac{1}{4}du
\]

Substitute:
\[
\int x^3 \cos(x^4+2)\,dx
= \frac{1}{4}\int \cos u\,du
= \frac{1}{4}\sin u + C
\]

Return to \( x \):
\[
= \frac{1}{4}\sin(x^4+2) + C
\]
:::



::: {.example}
Radical Expression

Evaluate
\[
\int \sqrt{2x+1}\,dx
\]

**Solution**

Let:
\[
u = 2x+1 \Rightarrow du = 2dx \Rightarrow dx = \frac{1}{2}du
\]

\[
= \frac{1}{2}\int u^{1/2}\,du
= \frac{1}{2}\cdot \frac{2}{3}u^{3/2} + C
= \frac{1}{3}u^{3/2} + C
\]

Substitute back:
\[
= \frac{1}{3}(2x+1)^{3/2} + C
\]
:::



::: {.example}
Rational with Radical

Evaluate
\[
\int \frac{x}{\sqrt{1+4x^2}}\,dx
\]

**Solution**

Let:
\[
u = 1+4x^2 \Rightarrow du = 8x dx \Rightarrow xdx = \frac{1}{8}du.
\]

Then,
\begin{align*}
\int \frac{x}{\sqrt{1+4x^2}}\,dx &= \frac{1}{8}\int u^{-1/2}\,du\\
&= \frac{1}{8}\cdot 2u^{1/2} + C \\
&= \frac{1}{4}\sqrt{u} + C \\
&= \frac{1}{4}\sqrt{1+4x^2} + C
\end{align*}
:::



::: {.example}
Exponential Function

Evaluate
\[
\int e^{5x}\,dx
\]

**Solution**

Let:
\[
u = 5x \Rightarrow du = 5dx \Rightarrow dx = \frac{1}{5}du
\]

\begin{align*}
\int e^{5x}\,dx &= \frac{1}{5}\int e^u\,du \\
& = \frac{1}{5}e^u + C \\
& = \frac{1}{5}e^{5x} + C
\end{align*}
:::



::: {.example}
Trigonometric Ratio

Evaluate
\[
\int \tan x\,dx
\]

**Solution**

Rewrite:
\[
\tan x = \frac{\sin x}{\cos x}
\]

Let:
\[
u = \cos x \Rightarrow du = -\sin x\,dx
\]

\begin{align*}
\int \tan x\,dx &= -\int \frac{1}{u}\,du\\
&= -\ln|u| + C \\
&= -\ln|\cos x| + C
\end{align*}

Equivalent form:
\[
= \ln|\sec x| + C
\]
:::


<div class="theorem">
Substitution with Definite Integrals

If \( u = g(x) \), then:
\[
\int_a^b f(g(x))g'(x)\,dx = \int_{u(a)}^{u(b)} f(u)\,du
\]
</div>


**Key Idea**

> When changing variables, change the limits too — and do not switch back to \( x \).


::: {.example}
Definite Integral with Substitution

Evaluate
\[
\int_1^e \frac{\ln x}{x}\,dx
\]

**Solution**

Let:
\[
u = \ln x \Rightarrow du = \frac{1}{x}dx
\]

Change limits:

- \( x=1 \Rightarrow u=0 \)  
- \( x=e \Rightarrow u=1 \)  

\begin{align*}
\int_1^e \frac{\ln x}{x}\,dx &= \int_0^1 u\,du\\
&= \frac{u^2}{2}\Big|_0^1\\
&= \frac{1}{2}
\end{align*}
:::



<div class="note">
**Symmetry in Definite Integrals**


**Even Functions**

If \( f(-x)=f(x) \):
\[
\int_{-a}^{a} f(x)\,dx = 2\int_0^a f(x)\,dx
\]

**Odd Functions**

If \( f(-x)=-f(x) \):
\[
\int_{-a}^{a} f(x)\,dx = 0
\]
</div>


::: {.example}
Symmetry

**Even Function**
\[
\int_{-2}^{2} (x^6+1)\,dx = 2\int_0^2 (x^6+1)\,dx
\]

**Odd Function**
\[
\int_{-1}^{1} \frac{\tan x}{1+x^2+x^4}\,dx = 0
\]
:::

---

###  Putting It All Together

The Substitution Rule transforms difficult integrals into simpler ones by changing variables. It is the inverse operation of the Chain Rule and provides a systematic way to handle composite functions, radicals, exponentials, and trigonometric expressions. When combined with definite integrals and symmetry principles, substitution becomes one of the most powerful tools in single-variable calculus for both computation and modeling.

---

#### Conceptual Takeaways

- Substitution is **reverse Chain Rule**  
- Integration becomes easier by changing variables  
- Good substitutions simplify structure  
- \( du \) must appear (up to a constant factor)  
- Substitution turns complicated integrals into basic ones  
- Definite integrals require **changing limits**  
- Symmetry can eliminate computation entirely  

---

#### Skills You Should Be Able to Do

After this section, you should be able to:

- Recognize composite functions in integrals  
- Choose effective substitutions  
- Compute \( du \) correctly  
- Rewrite integrals fully in terms of \( u \)  
- Evaluate substituted integrals  
- Return to original variables correctly  
- Apply substitution to definite integrals  
- Change limits of integration properly  
- Identify even and odd symmetry  
- Use symmetry to simplify definite integrals  
- Interpret substitution as reverse Chain Rule  


---

####  Problems