# Question 1: Evaluate the following sums

## (a)
$$
\sum_{i=1}^{5} (2i - 1) =
$$

<details> <summary> Example: </summary>

We are tasked with solving the sum:

$$
\sum_{i=1}^{4} (3i - 2)
$$

### Step 1: Create a Table

We will evaluate $3i - 2$ for each value of $i$ from 1 to 4 and then sum the results. Below is the table with the first column as $i$ and the second column as $3i - 2$.

| $i$ | $3i - 2$ |
| --- | -------- |
| 1 | $3(1) - 2 = 1$ |
| 2 | $3(2) - 2 = 4$ |
| 3 | $3(3) - 2 = 7$ |
| 4 | $3(4) - 2 = 10$ |

### Step 2: Sum the Results

Now, we sum the values in the second column:

$$
1 + 4 + 7 + 10 = 22
$$

### Final Answer:

The sum is:

$$
\sum_{i=1}^{4} (3i - 2) = 22
$$

</details>

## (b)
$$
\sum_{j=1}^{6} \frac{1}{j} =
$$

<details> <summary> Example: </summary>

We are tasked with solving the sum:

$$
\sum_{j=1}^{4} \frac{2}{j}
$$

### Step 1: Create a Table

We will evaluate $\frac{2}{j}$ for each value of $j$ from 1 to 4 and then sum the results. Below is the table with the first column as $j$ and the second column as $\frac{2}{j}$ evaluated at each $j$.

| $j$ | $\frac{2}{j}$ |
| --- | ------------- |
| 1 | $\frac{2}{1} = 2$ |
| 2 | $\frac{2}{2} = 1$ |
| 3 | $\frac{2}{3}$ |
| 4 | $\frac{2}{4} = \frac{1}{2}$ |

### Step 2: Sum the Results

Now, we sum the values in the second column:

$$
2 + 1 + \frac{2}{3} + \frac{1}{2}
$$

We will first express everything as a fraction with a common denominator. The least common denominator (LCD) for $1$, $\frac{2}{3}$, and $\frac{1}{2}$ is 6. Rewriting the terms:

$$
2 = \frac{12}{6}, \quad 1 = \frac{6}{6}, \quad \frac{2}{3} = \frac{4}{6}, \quad \frac{1}{2} = \frac{3}{6}
$$

Now sum the fractions:

$$
\frac{12}{6} + \frac{6}{6} + \frac{4}{6} + \frac{3}{6} = \frac{25}{6}
$$

### Final Answer:

Thus, the exact sum is:

$$
\sum_{j=1}^{4} \frac{2}{j} = \frac{25}{6} \approx 4.1667
$$

</details>

## (c)
$$
\sum_{i=1}^{6} \dfrac{(-1)^i}{i} =
$$

<details> <summary> Example: </summary>

We are tasked with solving the sum:

$$
\sum_{i=1}^{5} \dfrac{(-1)^{i+1}}{i}
$$

### Step 1: Create a Table

We will evaluate $\dfrac{(-1)^{i+1}}{i}$ for each value of $i$ from 1 to 5 and then sum the results. Below is the table with the first column as $i$ and the second column as $\dfrac{(-1)^{i+1}}{i}$ evaluated at each $i$.

| $i$ | $\frac{(-1)^{i+1}}{i}$ |
| --- | --------------------- |
| 1 | $\dfrac{(-1)^{1+1}}{1} = \dfrac{1}{1} = 1$ |
| 2 | $\dfrac{(-1)^{2+1}}{2} = \dfrac{-1}{2}$ |
| 3 | $\dfrac{(-1)^{3+1}}{3} = \dfrac{1}{3}$ |
| 4 | $\dfrac{(-1)^{4+1}}{4} = \dfrac{-1}{4}$ |
| 5 | $\dfrac{(-1)^{5+1}}{5} = \dfrac{1}{5}$ |

### Step 2: Sum the Results

Now, we sum the values in the second column:

$$
1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5}
$$

We will find a common denominator for all terms. The least common denominator (LCD) for $1$, $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{4}$, and $\frac{1}{5}$ is 60. Rewriting the terms:

$$
1 = \frac{60}{60}, \quad \frac{1}{2} = \frac{30}{60}, \quad \frac{1}{3} = \frac{20}{60}, \quad \frac{1}{4} = \frac{15}{60}, \quad \frac{1}{5} = \frac{12}{60}
$$

Now sum the fractions:

$$
\frac{60}{60} - \frac{30}{60} + \frac{20}{60} - \frac{15}{60} + \frac{12}{60}
$$

Simplifying:

$$
\frac{60 - 30 + 20 - 15 + 12}{60} = \frac{47}{60}
$$

### Final Answer:

Thus, the exact sum is:

$$
\sum_{i=1}^{5} \frac{(-1)^{i+1}}{i} = \frac{47}{60} \approx 0.7833
$$

</details>

# Question 2: Find the following derivatives

## (a)
For the function:
$$
f(x) = \frac{x - 1}{x + 1}
$$
Find:
$$
f'(x) =
$$

<details> <summary> Example: </summary>

We are tasked with finding the derivative of the function:

$$
f(x) = \frac{x-2}{x+2}
$$

To differentiate this, we will use the **quotient rule** for derivatives. The quotient rule states:

$$
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}
$$

Here, $u(x)$ is the **numerator** and $v(x)$ is the **denominator**. For our function $f(x) = \frac{x-2}{x+2}$, we have:
- $u(x) = x - 2$
- $v(x) = x + 2$

### Step 1: Identify the Inside and Outside Functions

To make the application of the quotient rule easier, we'll organize our work in a table where we identify the "outside" and "inside" functions, their original forms, and their derivatives.

| | Numerator ($u(x)$) | Denominator ($v(x)$) |
| -------- | ------------------------ | ----------------------- |
| Original | $x - 2$ | $x + 2$ |
| Derivative | $1$ | $1$ |

### Step 2: Apply the Quotient Rule

Using the quotient rule formula:

$$
f'(x) = \frac{(x + 2) \cdot (1) - (x - 2) \cdot (1)}{(x + 2)^2}
$$

### Step 3: Simplify the Expression

Now, simplify the numerator:

$$
f'(x) = \frac{(x + 2) - (x - 2)}{(x + 2)^2}
$$

Simplifying further:

$$
f'(x) = \frac{x + 2 - x + 2}{(x + 2)^2} = \frac{4}{(x + 2)^2}
$$

### Final Answer:

Thus, the derivative of the function is:

$$
f'(x) = \frac{4}{(x + 2)^2}
$$

</details>

## (b)
For the function:
$$
g(x) = \frac{2x + 1}{1 - x}
$$
Find:
$$
g'(x) =
$$

<details> <summary> Example: </summary>

We are tasked with finding the derivative of the function:

$$
g(x) = \frac{3x+2}{4-x}
$$

To differentiate this, we will use the **quotient rule** for derivatives. The quotient rule states:

$$
\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2}
$$

Here, $u(x)$ is the **numerator** and $v(x)$ is the **denominator**. For our function $g(x) = \frac{3x+2}{4-x}$, we have:
- $u(x) = 3x + 2$
- $v(x) = 4 - x$

### Step 1: Identify the Inside and Outside Functions

To make the application of the quotient rule easier, we'll organize our work in a table where we identify the "outside" and "inside" functions, their original forms, and their derivatives.

| | Numerator ($u(x)$) | Denominator ($v(x)$) |
| -------- | ------------------------ | ----------------------- |
| Original | $3x + 2$ | $4 - x$ |
| Derivative | $3$ | $-1$ |

### Step 2: Apply the Quotient Rule

Using the quotient rule formula:

$$
g'(x) = \frac{(4 - x) \cdot (3) - (3x + 2) \cdot (-1)}{(4 - x)^2}
$$

### Step 3: Simplify the Expression

First, simplify the numerator:

$$
g'(x) = \frac{3(4 - x) + (3x + 2)}{(4 - x)^2}
$$

Now distribute and simplify:

$$
g'(x) = \frac{12 - 3x + 3x + 2}{(4 - x)^2} = \frac{12 + 2}{(4 - x)^2} = \frac{14}{(4 - x)^2}
$$

### Final Answer:

Thus, the derivative of the function is:

$$
g'(x) = \frac{14}{(4 - x)^2}
$$

</details>

## (c)
For the function:
$$
h(x) = e^x \ln(x)
$$
Find:
$$
h'(x) =
$$

<details> <summary> Example: </summary>

We are tasked with finding the derivative of the function:

$$
f(x) = e^{2x} \ln(x)
$$

To differentiate this, we will use both the **chain rule** and the **product rule**.

### Step 1: Apply the Chain Rule for $e^{2x}$

The first part of the product rule involves the derivative of $e^{2x}$. This requires the **chain rule**, which states:

$$
\frac{d}{dx} \left( f(u) \right) = f'(u) \cdot u'
$$

The Outside function is $e^u$. The inside function is $2x$. Taking derivatives of each and putting them in the table gives:

| | (Outside function) | (Inside function) |
| -------- | --------------------------- | ----------------------- |
| Original | $e^{u}$ | $2x$ |
| Derivative | $e^{u}$ | $2$ |

So, applying the chain rule:

$$
\frac{d}{dx} \left( e^{2x} \right) =e^u \cdot 2 = 2e^{2x}
$$

### Step 2: Apply the Product Rule

Now that we know how to differentiate $e^{2x}$, we can apply the **product rule**. The product rule states:

$$
\frac{d}{dx} \left( u(x) \cdot v(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
$$

For $f(x) = e^{2x} \ln(x)$, we identify:

- $u(x) = e^{2x}$
- $v(x) = \ln(x)$

We already computed that $u'(x) = 2e^{2x}$, and we know the derivative of $\ln(x)$ is $\frac{1}{x}$.

| | $e^{2x}$ (Outside function) | $\ln(x)$ (Inside function) |
| -------- | --------------------------- | ------------------------- |
| Original | $e^{2x}$ | $\ln(x)$ |
| Derivative | $2e^{2x}$ | $\frac{1}{x}$ |

### Step 3: Combine Using the Product Rule

Now, apply the product rule:

$$
f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)
$$

Substitute the values we computed:

$$
f'(x) = 2e^{2x} \cdot \ln(x) + e^{2x} \cdot \frac{1}{x}
$$

### Step 4: Simplify

We can factor out $e^{2x}$ from both terms:

$$
f'(x) = e^{2x} \left( 2 \ln(x) + \frac{1}{x} \right)
$$

### Final Answer:

Thus, the derivative of the function is:

$$
f'(x) = e^{2x} \left( 2 \ln(x) + \frac{1}{x} \right)
$$

</details>

# Question 3



## a.
Divide the interval $[1, 4]$ into three equal subintervals. Draw in left and right rectangles for the function $v(x)$. Compute $L_3$ and $R_3$.

<details> <summary> Left Rectangle Example: </summary>

Let's say we do the same to a similar graph on the interval $[1,3]$ with two left rectangles:


The left rectangles would look like:

Then the area would be the area of the first rectangle, $5 \cdot 1=5$ plus the area of the second rectangle, $4 \cdot 1=4$.

Then $L_2=5+4=9$.

</details>

<details> <summary> Right Rectangle Example: </summary>

Let's say we do the same to a similar graph on the interval $[1,3]$ with two right rectangles:


The right rectangles would look like:

Then the area would be the area of the first rectangle, $4 \cdot 1=4$ plus the area of the second rectangle, $2 \cdot 1=2$.

Then $R_2=4+2=6$.

</details>

---

## b.
Replace the question marks with $L_3$ and $R_3$ as appropriate. Explain your choice.

$$
? \leq \int_1^4 v(x) \, dx \leq ?
$$

<details> <summary> Example: </summary>

If we use the example in the previous part a:

$$
R_2 \leq \int_1^3 v(x) \, dx \leq L_2
$$
</details>

---

# Question 4 In the following problems calculate the definite integral by referring to the figure with the indicated areas.


## a. $\displaystyle \int_a^c f(x)dx$

<details> <summary> Example: </summary>

Say instead that the areas are instead:

| Area | Value |
|--------|---------------------------|
| A | 4.372 |
| B | 2.894 |
| C | 5.146 |
| D | 3.521 |

Then
$$
\int_a^c f(x)dx =A-B+C=4.372-2.894+5.146=6.624
$$

</details>

## b. $\displaystyle \int_b^d f(x)dx$

<details> <summary> Example: </summary>

Say instead that the areas are instead:

| Area | Value |
|--------|---------------------------|
| A | 4.372 |
| B | 2.894 |
| C | 5.146 |
| D | 3.521 |

Then
$$
\int_b^d f(x)dx =-B+C-D=-2.894+5.146-3.521=-1.269
$$

</details>

## c. $\displaystyle \int_a^d f(x)dx$

<details> <summary> Example: </summary>

Say instead that the areas are instead:

| Area | Value |
|--------|---------------------------|
| A | 4.372 |
| B | 2.894 |
| C | 5.146 |
| D | 3.521 |

Then
$$
\int_a^d f(x)dx =A-B+C-D=4.372-2.894+5.146-3.521=3.103
$$

</details>

## d. $\displaystyle \int_0^d f(x)dx$

<details> <summary> Example: </summary>

Say instead that the areas are instead:

| Area | Value |
|--------|---------------------------|
| A | 4.372 |
| B | 2.894 |
| C | 5.146 |
| D | 3.521 |

Then
$$
\int_0^d f(x)dx =C-D=5.146-3.521=1.625
$$

</details>

# Question 5:
Calculate the indicated Riemann sum $S_n$ for the function $f(x) = 25 - 3x^2$. Partition $[-2, 8]$ into five subintervals of equal length, and for each subinterval $[x_{k-1}, x_k]$, let $c_k = \dfrac{x_{k-1} + x_k}{2}$, $k = 1, 2, 3, 4, 5$.

<details> <summary> Example: </summary>

We are tasked with calculating the Riemann sum $S_n$ for the function:

$$
f(x) = 16 - 2x^2
$$

over the interval $[-4, 6]$, partitioned into five subintervals of equal length. For each subinterval $[x_{k-1}, x_k]$, the midpoint $c_k$ is given by:

$$
c_k = \frac{x_{k-1} + x_k}{2}, \quad k = 1, 2, 3, 4, 5
$$

### Step 1: Determine the Length of Each Subinterval

The interval $[-4, 6]$ has a total length:

$$
6 - (-4) = 10
$$

Since we are partitioning the interval into 5 subintervals, the length of each subinterval is:

$$
\Delta x = \frac{10}{5} = 2
$$

### Step 2: Identify the Partition Points

The partition points are:

| $k$ | $x_{k-1}$ | $x_k$ |
|------|-----------|-------|
| 1 | $-4$ | $-2$ |
| 2 | $-2$ | $0$ |
| 3 | $0$ | $2$ |
| 4 | $2$ | $4$ |
| 5 | $4$ | $6$ |

### Step 3: Compute the Midpoints

The midpoints $c_k$ for each subinterval are calculated as:

$$
c_k = \frac{x_{k-1} + x_k}{2}
$$

| $k$ | $x_{k-1}$ | $x_k$ | Midpoint $c_k$ |
|------|-----------|-------|----------------|
| 1 | $-4$ | $-2$ | $-3$ |
| 2 | $-2$ | $0$ | $-1$ |
| 3 | $0$ | $2$ | $1$ |
| 4 | $2$ | $4$ | $3$ |
| 5 | $4$ | $6$ | $5$ |

### Step 4: Evaluate the Function at Each Midpoint

We now evaluate $f(x) = 16 - 2x^2$ at each midpoint $c_k$:

$$
f(x) = 16 - 2x^2
$$

| $c_k$ | $f(c_k) = 16 - 2c_k^2$ |
|-------|------------------------|
| $-3$ | $16 - 2(-3)^2 = -2$ |
| $-1$ | $16 - 2(-1)^2 = 14$ |
| $1$ | $16 - 2(1)^2 = 14$ |
| $3$ | $16 - 2(3)^2 = -2$ |
| $5$ | $16 - 2(5)^2 = -34$ |

### Step 5: Compute the Riemann Sum

The Riemann sum $S_n$ is given by:

$$
S_n = \sum_{k=1}^{5} f(c_k) \cdot \Delta x
$$

Using $\Delta x = 2$ and the values of $f(c_k)$ computed above:

$$
S_n = (-2) \cdot 2 + 14 \cdot 2 + 14 \cdot 2 + (-2) \cdot 2 + (-34) \cdot 2
$$

Simplifying:

$$
S_n = -4 + 28 + 28 - 4 - 68 = -20
$$

### Final Answer:

The Riemann sum is:

$$
S_n = -20
$$

</details>