# Question 1: Evaluate the expressions

## (a) $$\left( x^2 - 2x \right) \Bigg|_0^5 =$$

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

$$\left( x^3 - 3x \right) \Bigg|_0^2 =$$

We are tasked with evaluating the expression $\left( x^3 - 3x \right) \Bigg|_0^2$. This means we will compute the difference $f(2) - f(0)$, where $f(x) = x^3 - 3x$.

### Step 1: Evaluate at the upper limit $x = 2$
Substitute $x = 2$ into the expression $x^3 - 3x$:
$$
f(2) = 2^3 - 3(2) = 8 - 6 = 2
$$

### Step 2: Evaluate at the lower limit $x = 0$
Substitute $x = 0$ into the expression $x^3 - 3x$:
$$
f(0) = 0^3 - 3(0) = 0
$$

### Step 3: Subtract the lower limit from the upper limit
Now subtract the values:
$$
f(2) - f(0) = 2 - 0 = 2
$$

Thus, the evaluation of the expression is:
$$
\left( x^3 - 3x \right) \Bigg|_0^2 = 2
$$

</details>

## (b) $$\left( (x - 1)e^x \right) \Bigg|_0^3 =$$

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

$$\left( (x - 2)e^x \right) \Bigg|_0^4 =$$

We are tasked with evaluating the expression $\left( (x - 2)e^x \right) \Bigg|_0^4$. This means we will compute the difference $f(4) - f(0)$, where $f(x) = (x - 2)e^x$.

### Step 1: Evaluate at the upper limit $x = 4$
Substitute $x = 4$ into the expression $(x - 2)e^x$:
$$
f(4) = (4 - 2)e^4 = 2e^4
$$

### Step 2: Evaluate at the lower limit $x = 0$
Substitute $x = 0$ into the expression $(x - 2)e^x$:
$$
f(0) = (0 - 2)e^0 = (-2)(1) = -2
$$

### Step 3: Subtract the lower limit from the upper limit
Now subtract the values:
$$
f(4) - f(0) = 2e^4 - (-2) = 2e^4 + 2
$$

Thus, the evaluation of the expression is:
$$
\left( (x - 2)e^x \right) \Bigg|_0^4 = 2e^4 + 2
$$

</details>

## (c) $$\ln(9x^2 + 5) \Bigg|_0^3 =$$

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

$$\ln(5x^2 + 4) \Bigg|_0^3 =$$

We are tasked with evaluating the expression $\ln(5x^2 + 4) \Bigg|_0^3$. This means we will compute the difference $f(3) - f(0)$, where $f(x) = \ln(5x^2 + 4)$.

### Step 1: Evaluate at the upper limit $x = 3$
Substitute $x = 3$ into the expression $\ln(5x^2 + 4)$:
$$
f(3) = \ln(5(3)^2 + 4) = \ln(5(9) + 4) = \ln(45 + 4) = \ln(49)
$$

### Step 2: Evaluate at the lower limit $x = 0$
Substitute $x = 0$ into the expression $\ln(5x^2 + 4)$:
$$
f(0) = \ln(5(0)^2 + 4) = \ln(4)
$$

### Step 3: Subtract the lower limit from the upper limit
Now subtract the values:
$$
f(3) - f(0) = \ln(49) - \ln(4)
$$

Using the logarithmic property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$:
$$
\ln(49) - \ln(4) = \ln\left(\frac{49}{4}\right)
$$

Thus, the evaluation of the expression is:
$$
\ln(5x^2 + 4) \Bigg|_0^3 = \ln\left(\frac{49}{4}\right)
$$

</details>

## (d) $$\left( -\frac{1}{x^2} \right) \Bigg|_1^5$$

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

$$\left( -\frac{1}{x^3} \right) \Bigg|_1^4$$

We are tasked with evaluating the expression $\left( -\frac{1}{x^3} \right) \Bigg|_1^4$. This means we will compute the difference $f(4) - f(1)$, where $f(x) = -\frac{1}{x^3}$.

### Step 1: Evaluate at the upper limit $x = 4$
Substitute $x = 4$ into the expression $-\frac{1}{x^3}$:
$$
f(4) = -\frac{1}{4^3} = -\frac{1}{64}
$$

### Step 2: Evaluate at the lower limit $x = 1$
Substitute $x = 1$ into the expression $-\frac{1}{x^3}$:
$$
f(1) = -\frac{1}{1^3} = -1
$$

### Step 3: Subtract the lower limit from the upper limit
Now subtract the values:
$$
f(4) - f(1) = -\frac{1}{64} - (-1) = -\frac{1}{64} + 1 = 1 - \frac{1}{64} = \frac{64}{64} - \frac{1}{64} = \frac{63}{64}
$$

Thus, the evaluation of the expression is:
$$
\left( -\frac{1}{x^3} \right) \Bigg|_1^4 = \frac{63}{64}
$$

</details>

# Question 2:
Find the following derivatives

## (a)
For the function:
$$
f(x) = \frac{x + 2}{x^2 + 1}
$$
Find:
$$
f'(x) =
$$

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

For the function:
$$
f(x) = \frac{x + 3}{x^3 + 2}
$$
Find:
$$
f'(x) =
$$

### Derivative Table

| | Numerator (First Function) | Denominator (Second Function) |
| --- | --- | --- |
| **Original** | $x + 3$ | $x^3 + 2$ |
| **Derivative** | $1$ | $3x^2$ |

### Applying the quotient rule:
$$
f'(x) = \frac{(1)(x^3 + 2) - (x + 3)(3x^2)}{(x^3 + 2)^2}
$$

### Simplifying:
$$
f'(x) = \frac{x^3 + 2 - (3x^3 + 9x^2)}{(x^3 + 2)^2}
$$
Simplify further:
$$
f'(x) = \frac{-2x^3 - 9x^2 + 2}{(x^3 + 2)^2}
$$

Thus, the derivative of the function is:
$$
f'(x) = \frac{-2x^3 - 9x^2 + 2}{(x^3 + 2)^2}
$$

</details>

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

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

For the function:
$$
g(x) = \frac{3x - 2}{2 + x}
$$
Find:
$$
g'(x) =
$$

We are tasked with finding the derivative of the function:
$$
g(x) = \frac{3x - 2}{2 + x}
$$
To do this, we will use the quotient rule.

### Quotient Rule
The quotient rule states that for two functions $u(x)$ and $v(x)$, the derivative of their quotient is:
$$
g'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}
$$
Let:
- $u(x) = 3x - 2$
- $v(x) = 2 + x$

### Step 1: Differentiate $u(x)$ and $v(x)$
- $u'(x) = 3$
- $v'(x) = 1$

### Derivative Table

| | Numerator (First Function) | Denominator (Second Function) |
| --- | --- | --- |
| **Original** | $3x - 2$ | $2 + x$ |
| **Derivative** | $3$ | $1$ |

### Step 2: Apply the quotient rule
Substitute $u(x)$, $u'(x)$, $v(x)$, and $v'(x)$ into the quotient rule:
$$
g'(x) = \frac{(3)(2 + x) - (3x - 2)(1)}{(2 + x)^2}
$$

### Step 3: Simplify
Expand the numerator:
$$
g'(x) = \frac{3(2 + x) - (3x - 2)}{(2 + x)^2}
$$
$$
g'(x) = \frac{6 + 3x - 3x + 2}{(2 + x)^2} = \frac{8}{(2 + x)^2}
$$

Thus, the derivative of the function is:
$$
g'(x) = \frac{8}{(2 + x)^2}
$$

</details>

## (c)
For the function:
$$
h(x) = \ln(5x + x^2)
$$
Find:
$$
h'(x) =
$$

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

For the function:
$$
h(x) = \ln(4x + x^3)
$$
Find:
$$
h'(x) =
$$

We are tasked with finding the derivative of the function:
$$
h(x) = \ln(4x + x^3)
$$
To do this, we will use the chain rule and the derivative of the natural logarithm.

### Chain Rule and Logarithmic Differentiation
The derivative of $\ln(u(x))$ is:
$$
h'(x) = \frac{1}{u(x)} \cdot u'(x)
$$
Where $u(x) = 4x + x^3$.

### Step 1: Differentiate $u(x)$
First, we find the derivative of $u(x)$:
$$
u(x) = 4x + x^3
$$
$$
u'(x) = 4 + 3x^2
$$

### Derivative Table

| | Argument of $\ln$ | Derivative of Argument |
| --- | --- | --- |
| **Original** | $4x + x^3$ | $4 + 3x^2$ |

### Step 2: Apply the chain rule
Substitute $u(x)$ and $u'(x)$ into the chain rule:
$$
h'(x) = \frac{1}{4x + x^3} \cdot (4 + 3x^2)
$$

Thus, the derivative of the function is:
$$
h'(x) = \frac{4 + 3x^2}{4x + x^3}
$$

</details>

# Question 3:
## Evaluate the integrals

### (a) $$\int_1^4 (5x + 3) \, dx = $$

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

$$\int_1^3 (4x + 2) \, dx = $$

We are tasked with evaluating the definite integral:
$$
\int_1^3 (4x + 2) \, dx
$$
We will break it down into 4 steps.

### Step 1: Find the antiderivative
The antiderivative of $4x + 2$ is:
$$
\int (4x + 2) \, dx = 2x^2 + 2x + C
$$

### Step 2: Evaluate the antiderivative at the upper bound $x = 3$
Substitute $x = 3$ into the antiderivative:
$$
F(3) = 2(3)^2 + 2(3) = 2(9) + 6 = 18 + 6 = 24
$$

### Step 3: Evaluate the antiderivative at the lower bound $x = 1$
Substitute $x = 1$ into the antiderivative:
$$
F(1) = 2(1)^2 + 2(1) = 2(1) + 2 = 2 + 2 = 4
$$

### Step 4: Subtract the lower bound evaluation from the upper bound evaluation
Now subtract the values:
$$
F(3) - F(1) = 24 - 4 = 20
$$

Thus, the value of the definite integral is:
$$
\int_1^3 (4x + 2) \, dx = 20
$$

</details>

### (b) $$\int_1^2 \frac{1}{x} \, dx = $$

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

$$\int_2^5 \frac{1}{x} \, dx = $$

We are tasked with evaluating the definite integral:
$$
\int_2^5 \frac{1}{x} \, dx
$$
We will break it down into 4 steps.

### Step 1: Find the antiderivative
The antiderivative of $\frac{1}{x}$ is:
$$
\int \frac{1}{x} \, dx = \ln|x| + C
$$

### Step 2: Evaluate the antiderivative at the upper bound $x = 5$
Substitute $x = 5$ into the antiderivative:
$$
F(5) = \ln|5| = \ln(5)
$$

### Step 3: Evaluate the antiderivative at the lower bound $x = 2$
Substitute $x = 2$ into the antiderivative:
$$
F(2) = \ln|2| = \ln(2)
$$

### Step 4: Subtract the lower bound evaluation from the upper bound evaluation
Now subtract the values:
$$
F(5) - F(2) = \ln(5) - \ln(2)
$$
Using the logarithmic property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$:
$$
\ln(5) - \ln(2) = \ln\left(\frac{5}{2}\right)
$$

Thus, the value of the definite integral is:
$$
\int_2^5 \frac{1}{x} \, dx = \ln\left(\frac{5}{2}\right)
$$

</details>

### (c) $$\int_1^4 3\sqrt{x} \, dx = $$

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

$$\int_1^9 2\sqrt{x} \, dx = $$

We are tasked with evaluating the definite integral:
$$
\int_1^9 2\sqrt{x} \, dx
$$
We will break it down into 4 steps.

### Step 1: Find the antiderivative
First, rewrite the integrand as $2x^{1/2}$ and then find the antiderivative:
$$
\int 2\sqrt{x} \, dx = \int 2x^{1/2} \, dx
$$
The antiderivative of $2x^{1/2}$ is:
$$
\frac{2}{3}x^{3/2} + C
$$

### Step 2: Evaluate the antiderivative at the upper bound $x = 9$
Substitute $x = 9$ into the antiderivative:
$$
F(9) = \frac{2}{3}(9)^{3/2} = \frac{2}{3}(27) = 18
$$

### Step 3: Evaluate the antiderivative at the lower bound $x = 1$
Substitute $x = 1$ into the antiderivative:
$$
F(1) = \frac{2}{3}(1)^{3/2} = \frac{2}{3}(1) = \frac{2}{3}
$$

### Step 4: Subtract the lower bound evaluation from the upper bound evaluation
Now subtract the values:
$$
F(9) - F(1) = 18 - \frac{2}{3} = \frac{54}{3} - \frac{2}{3} = \frac{52}{3}
$$

Thus, the value of the definite integral is:
$$
\int_1^9 2\sqrt{x} \, dx = \frac{52}{3}
$$

</details>

# Question 4:
## Cost. A company manufactures mountain bikes. The research department produced the marginal cost function $$C'(x) = 500 - \frac{x}{3}, \, 0 \leq x \leq 900$$ where $C'(x)$ is in dollars and $x$ is the number of bikes produced per month. Compute the increase in cost going from a production level of 300 bikes per month to 900 bikes per month. Set up a definite integral and evaluate it.

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

A company manufactures mountain bikes. The research department produced the marginal cost function $$C'(x) = 400 - \frac{x}{2}, \, 0 \leq x \leq 800$$ where $C'(x)$ is in dollars and $x$ is the number of bikes produced per month. Compute the increase in cost going from a production level of 200 bikes per month to 800 bikes per month. Set up a definite integral and evaluate it.

We are tasked with finding the increase in cost when the production level goes from 200 bikes per month to 800 bikes per month. The marginal cost function is given by:
$$
C'(x) = 400 - \frac{x}{2}, \quad 0 \leq x \leq 800
$$
To compute the increase in cost, we need to evaluate the definite integral of $C'(x)$ from $x = 200$ to $x = 800$.

### Step 1: Set up the definite integral
The increase in cost is represented by the definite integral:
$$
\int_{200}^{800} C'(x) \, dx = \int_{200}^{800} \left(400 - \frac{x}{2}\right) \, dx
$$

### Step 2: Find the antiderivative
The antiderivative of $400 - \frac{x}{2}$ is:
$$
\int \left(400 - \frac{x}{2}\right) \, dx = 400x - \frac{x^2}{4} + C
$$

### Step 3: Evaluate the antiderivative at the upper bound $x = 800$
Substitute $x = 800$ into the antiderivative:
$$
F(800) = 400(800) - \frac{800^2}{4} = 320,000 - \frac{640,000}{4} = 320,000 - 160,000 = 160,000
$$

### Step 4: Evaluate the antiderivative at the lower bound $x = 200$
Substitute $x = 200$ into the antiderivative:
$$
F(200) = 400(200) - \frac{200^2}{4} = 80,000 - \frac{40,000}{4} = 80,000 - 10,000 = 70,000
$$

### Step 5: Subtract the lower bound evaluation from the upper bound evaluation
Now subtract the values:
$$
F(800) - F(200) = 160,000 - 70,000 = 90,000
$$

Thus, the increase in cost going from a production level of 200 bikes per month to 800 bikes per month is:
$$
\int_{200}^{800} \left(400 - \frac{x}{2}\right) \, dx = 90,000 \, \text{dollars}
$$

</details>