---
# System prepended metadata

title: Math 132 Practice Exam 3
tags: [exam 3, practice exam]

---

[TOC]

# Question 1: A company manufactures electrical drills, the price–demand and cost function are given by
$$x = 500 - \dfrac{p}{2}$$
$$C(x) = 10,000 + 200x$$

## a. Find the revenue function R(x) and its domain.
<details> 
<summary> Solution: </summary>

The revenue function is given by the product of price per unit, $p$, and the number of units sold, $x$:
- First, solve for $p$ from the demand equation:

$$x = 500 - \dfrac{p}{2}$$

$$p = 1000 - 2x$$

- The revenue function $R(x)$ is:
$$R(x) = p \cdot x = (1000 - 2x) \cdot x$$
$$R(x) = 1000x - 2x^2$$

### Domain:
Since the demand function is linear, the maximum value of $x$ occurs when $p = 0$. So, set $p = 0$ in the price–demand equation:
$$0 = 1000 - 2x \quad \Rightarrow \quad x = 500$$

Thus, the domain of $R(x)$ is $[0, 500]$.

</details>

## b. Find the profit function P(x) and its domain.
<details> 
<summary> Solution: </summary>

The profit function is given by:
$$P(x) = R(x) - C(x)$$

- We already have $R(x) = 1000x - 2x^2$ and $C(x) = 10,000 + 200x$, so:
$$P(x) = (1000x - 2x^2) - (10,000 + 200x)$$
$$P(x) = 1000x - 2x^2 - 10,000 - 200x$$
$$P(x) = -2x^2 + 800x - 10,000$$

### Domain:
The domain of $P(x)$ is the same as the domain of $R(x)$, which is $[0, 500]$.

</details>

## c. Find the production level that maximizes the profit, the maximum profit, and the price that the company should charge for an electrical drill.
<details> 
<summary> Solution: </summary>

To maximize profit, we need to find the critical points of $P(x)$ by taking its derivative and setting it equal to zero.

- The derivative of $P(x)$ is:
$$P'(x) = -4x + 800$$

- Set $P'(x) = 0$ to find the critical points:
$$-4x + 800 = 0$$
$$x = 200$$

- To confirm that this value of $x$ maximizes profit, take the second derivative:
$$P''(x) = -4$$

Since $P''(x) < 0$, $x = 200$ is indeed a maximum.

### Maximum profit:
Substitute $x = 200$ into the profit function $P(x)$:
$$P(200) = -2(200)^2 + 800(200) - 10,000$$
$$P(200) = -80,000 + 160,000 - 10,000$$
$$P(200) = 70,000$$

### Price per drill:
Substitute $x = 200$ into the price-demand equation:
$$p = 1000 - 2(200)$$
$$p = 1000 - 400$$
$$p = 600$$

Thus, the company should produce 200 units, the maximum profit is $70,000, and the price per drill should be $600.

</details>



# Question 2: A car rental agency rents 400 cars per day at a rate of $80 per day. For each $10 price increase, 30 less cars are rented.
At what daily rate should cars be rented to maximimize revenue. What is the maximum daily revenue?
<details> 
<summary> Solution: </summary>

Let the rate per day be represented by $x$, where the initial rate is $80. Then we can analyze the problem in the following way:

1. **Define the daily rate and cars rented as functions of $x$:**

   - The initial daily rate is $80, so for every increase of $10, the rate becomes $80 + 10x$, where $x$ is the number of $10 increments.
   - Initially, 400 cars are rented. For every $10 increase, 30 fewer cars are rented, so the number of cars rented per day is:
     $$400 - 30x$$

2. **Revenue function:**

   The revenue $R(x)$ can be calculated by multiplying the rate per day by the number of cars rented:
   $$R(x) = (80 + 10x)(400 - 30x)$$

3. **Expand and simplify $R(x)$:**

   Expanding and simplifying the expression, we get:
   $$R(x) = 32000 + 4000x - 2400x - 300x^2$$
   $$R(x) = -300x^2 + 1600x + 32000$$

4. **Take the derivative of $R(x)$ with respect to $x$ and set it equal to zero:**

   To find the maximum revenue, we take the derivative $R'(x)$ and set it equal to zero:
   $$R'(x) = -600x + 1600$$
   Setting $R'(x) = 0$:
   $$-600x + 1600 = 0$$
   $$600x = 1600$$
   $$x = \frac{1600}{600} = \frac{8}{3} \approx 2.67$$

5. **Calculate the rate and maximum revenue:**

   - Substitute $x = 2.67$ into the rate function:
     $$\text{Rate} = 80 + 10 \cdot 2.67 \approx 106.7$$
   - Substitute $x = 2.67$ into the revenue function to get the maximum revenue:
     $$R(2.67) \approx 34133.33$$

Therefore, the optimal daily rate to maximize revenue is approximately $106.67, and the maximum daily revenue is about $34,133.33.


</details>

# Question 3: Calculate the following definite integrals

## a. 
$$ \displaystyle \int_1^3 x \, dx $$

<details> 
<summary> Solution: </summary>


To solve the definite integral:

$$\int_1^3 x \, dx$$

1. Antiderivative:
   $$F(x) = \frac{x^2}{2}$$

2. Evaluate $F(b)$ at $b = 3$:
   $$F(3) = \frac{3^2}{2} = \frac{9}{2}$$

3. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = \frac{1^2}{2} = \frac{1}{2}$$

4. Subtract $F(b) - F(a)$:
   $$F(3) - F(1) = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4$$

Final Answer:
$$\int_1^3 x \, dx = 4$$


</details>

## b. 
$$ \displaystyle \int_{-1}^2 x(1 - x) \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_{-1}^2 x(1 - x) \, dx$$

1. Expand the integrand:
   $$x(1 - x) = x - x^2$$

2. Antiderivative:
   $$F(x) = \frac{x^2}{2} - \frac{x^3}{3}$$

3. Evaluate $F(b)$ at $b = 2$:
   $$F(2) = \frac{2^2}{2} - \frac{2^3}{3} = \frac{4}{2} - \frac{8}{3} = 2 - \frac{8}{3} = \frac{6}{3} - \frac{8}{3} = -\frac{2}{3}$$

4. Evaluate $F(a)$ at $a = -1$:
   $$F(-1) = \frac{(-1)^2}{2} - \frac{(-1)^3}{3} = \frac{1}{2} - \frac{-1}{3} = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}$$

5. Subtract $F(b) - F(a)$:
   $$F(2) - F(-1) = -\frac{2}{3} - \frac{5}{6} = -\frac{4}{6} - \frac{5}{6} = -\frac{9}{6} = -\frac{3}{2}$$

Final Answer:
$$\int_{-1}^2 x(1 - x) \, dx = -\frac{3}{2}$$


</details>

## c. 
$$ \displaystyle \int_{-1}^5 (7x - 3x^3) \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_{-1}^5 (7x - 3x^3) \, dx$$

1. Antiderivative:
   $$F(x) = \frac{7x^2}{2} - \frac{3x^4}{4}$$

2. Evaluate $F(b)$ at $b = 5$:
   $$F(5) = \frac{7(5)^2}{2} - \frac{3(5)^4}{4} = \frac{7 \cdot 25}{2} - \frac{3 \cdot 625}{4} = \frac{175}{2} - \frac{1875}{4}$$
   $$F(5) = \frac{350}{4} - \frac{1875}{4} = \frac{-1525}{4}$$

3. Evaluate $F(a)$ at $a = -1$:
   $$F(-1) = \frac{7(-1)^2}{2} - \frac{3(-1)^4}{4} = \frac{7 \cdot 1}{2} - \frac{3 \cdot 1}{4} = \frac{7}{2} - \frac{3}{4}$$
   $$F(-1) = \frac{14}{4} - \frac{3}{4} = \frac{11}{4}$$

4. Subtract $F(b) - F(a)$:
   $$F(5) - F(-1) = \frac{-1525}{4} - \frac{11}{4} = \frac{-1525 - 11}{4} = \frac{-1536}{4} = -384$$

Final Answer:
$$\int_{-1}^5 (7x - 3x^3) \, dx = -384$$


</details>

## d. 
$$ \displaystyle \int_{-2}^2 (3x + 4x^3) \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_{-2}^2 (3x + 4x^3) \, dx$$

1. Antiderivative:
   $$F(x) = \frac{3x^2}{2} + \frac{4x^4}{4} = \frac{3x^2}{2} + x^4$$

2. Evaluate $F(b)$ at $b = 2$:
   $$F(2) = \frac{3(2)^2}{2} + (2)^4 = \frac{3 \cdot 4}{2} + 16 = 6 + 16 = 22$$

3. Evaluate $F(a)$ at $a = -2$:
   $$F(-2) = \frac{3(-2)^2}{2} + (-2)^4 = \frac{3 \cdot 4}{2} + 16 = 6 + 16 = 22$$

4. Subtract $F(b) - F(a)$:
   $$F(2) - F(-2) = 22 - 22 = 0$$

Final Answer:
$$\int_{-2}^2 (3x + 4x^3) \, dx = 0$$


</details>

## e. 
$$ \displaystyle \int_1^3 \frac{1}{x} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_1^3 \frac{1}{x} \, dx$$

1. Antiderivative:
   $$F(x) = \ln|x|$$

2. Evaluate $F(b)$ at $b = 3$:
   $$F(3) = \ln|3| = \ln(3)$$

3. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = \ln|1| = \ln(1) = 0$$

4. Subtract $F(b) - F(a)$:
   $$F(3) - F(1) = \ln(3) - 0 = \ln(3)$$

Final Answer:
$$\int_1^3 \frac{1}{x} \, dx = \ln(3)$$


</details>

## f. 
$$ \displaystyle \int_1^3 (-2x^2) \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_1^3 (-2x^2) \, dx$$

1. Antiderivative:
   $$F(x) = -\frac{2x^3}{3}$$

2. Evaluate $F(b)$ at $b = 3$:
   $$F(3) = -\frac{2(3)^3}{3} = -\frac{2 \cdot 27}{3} = -18$$

3. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = -\frac{2(1)^3}{3} = -\frac{2 \cdot 1}{3} = -\frac{2}{3}$$

4. Subtract $F(b) - F(a)$:
   $$F(3) - F(1) = -18 - \left(-\frac{2}{3}\right) = -18 + \frac{2}{3} = -\frac{54}{3} + \frac{2}{3} = -\frac{52}{3}$$

Final Answer:
$$\int_1^3 (-2x^2) \, dx = -\frac{52}{3}$$


</details>

## g. 
$$ \displaystyle \int_1^2 \frac{1}{x^3} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_1^2 \frac{1}{x^3} \, dx$$

1. Rewrite the integrand:
   $$\frac{1}{x^3} = x^{-3}$$

2. Antiderivative:
   $$F(x) = \frac{x^{-2}}{-2} = -\frac{1}{2x^2}$$

3. Evaluate $F(b)$ at $b = 2$:
   $$F(2) = -\frac{1}{2(2)^2} = -\frac{1}{2 \cdot 4} = -\frac{1}{8}$$

4. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = -\frac{1}{2(1)^2} = -\frac{1}{2}$$

5. Subtract $F(b) - F(a)$:
   $$F(2) - F(1) = -\frac{1}{8} - \left(-\frac{1}{2}\right) = -\frac{1}{8} + \frac{1}{2} = -\frac{1}{8} + \frac{4}{8} = \frac{3}{8}$$

Final Answer:
$$\int_1^2 \frac{1}{x^3} \, dx = \frac{3}{8}$$


</details>

## h. 
$$ \displaystyle \int_1^9 \sqrt{x} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

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

1. Rewrite the integrand:
   $$\sqrt{x} = x^{1/2}$$

2. Antiderivative:
   $$F(x) = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}$$

3. Evaluate $F(b)$ at $b = 9$:
   $$F(9) = \frac{2}{3}(9)^{3/2} = \frac{2}{3} \cdot (9^{1/2})^3 = \frac{2}{3} \cdot (3)^3 = \frac{2}{3} \cdot 27 = 18$$

4. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = \frac{2}{3}(1)^{3/2} = \frac{2}{3} \cdot 1 = \frac{2}{3}$$

5. Subtract $F(b) - F(a)$:
   $$F(9) - F(1) = 18 - \frac{2}{3} = \frac{54}{3} - \frac{2}{3} = \frac{52}{3}$$

Final Answer:
$$\int_1^9 \sqrt{x} \, dx = \frac{52}{3}$$


</details>

## i. 
$$ \displaystyle \int_1^9 \frac{1}{\sqrt{x}} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_1^9 \frac{1}{\sqrt{x}} \, dx$$

1. Rewrite the integrand:
   $$\frac{1}{\sqrt{x}} = x^{-1/2}$$

2. Antiderivative:
   $$F(x) = \frac{x^{1/2}}{1/2} = 2x^{1/2} = 2\sqrt{x}$$

3. Evaluate $F(b)$ at $b = 9$:
   $$F(9) = 2\sqrt{9} = 2 \cdot 3 = 6$$

4. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = 2\sqrt{1} = 2 \cdot 1 = 2$$

5. Subtract $F(b) - F(a)$:
   $$F(9) - F(1) = 6 - 2 = 4$$

Final Answer:
$$\int_1^9 \frac{1}{\sqrt{x}} \, dx = 4$$


</details>

## j. 
$$ \displaystyle \int_1^{16} \frac{3 - x}{\sqrt{x}} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral:

$$\int_1^{16} \frac{3 - x}{\sqrt{x}} \, dx$$

1. Rewrite the integrand:
   $$\frac{3 - x}{\sqrt{x}} = \frac{3}{\sqrt{x}} - \frac{x}{\sqrt{x}} = 3x^{-1/2} - x^{1/2}$$

2. Antiderivative:
   For $3x^{-1/2}$:
   $$\int 3x^{-1/2} \, dx = 3 \cdot 2x^{1/2} = 6\sqrt{x}$$
   
   For $x^{1/2}$:
   $$\int x^{1/2} \, dx = \frac{2}{3}x^{3/2}$$

   Combine:
   $$F(x) = 6\sqrt{x} - \frac{2}{3}x^{3/2}$$

3. Evaluate $F(b)$ at $b = 16$:
   $$F(16) = 6\sqrt{16} - \frac{2}{3}(16)^{3/2}$$
   $$F(16) = 6 \cdot 4 - \frac{2}{3}(4^3)$$
   $$F(16) = 24 - \frac{2}{3} \cdot 64$$
   $$F(16) = 24 - \frac{128}{3}$$
   $$F(16) = \frac{72}{3} - \frac{128}{3} = \frac{-56}{3}$$

4. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = 6\sqrt{1} - \frac{2}{3}(1)^{3/2}$$
   $$F(1) = 6 \cdot 1 - \frac{2}{3} \cdot 1$$
   $$F(1) = 6 - \frac{2}{3}$$
   $$F(1) = \frac{18}{3} - \frac{2}{3} = \frac{16}{3}$$

5. Subtract $F(b) - F(a)$:
   $$F(16) - F(1) = \frac{-56}{3} - \frac{16}{3}$$
   $$F(16) - F(1) = \frac{-56 - 16}{3} = \frac{-72}{3} = -24$$

Final Answer:
$$\int_1^{16} \frac{3 - x}{\sqrt{x}} \, dx = -24$$



</details>




# Question 4: Use the method of substitution to calculate the following definite integrals.


## a.
$$ \displaystyle \int_0^1 (x + 1)^5 \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral using $u$-substitution:

$$\int_0^1 (x + 1)^5 \, dx$$

1. $u = x + 1$

2. $\frac{du}{dx} = 1$

3. $du = dx$

4. $dx = du$

5. Substitute and simplify:
   $$\int_0^1 (x + 1)^5 \, dx = \int (u)^5 \, du$$

6. Integrate:
   $$\int u^5 \, du = \frac{u^6}{6}$$

7. Plug back in for $u$ and find the antiderivative $F(x)$:
   $$F(x) = \frac{(x + 1)^6}{6}$$

8. Evaluate $F(b)$ at $b = 1$:
   $$F(1) = \frac{(1 + 1)^6}{6} = \frac{2^6}{6} = \frac{64}{6} = \frac{32}{3}$$

9. Evaluate $F(a)$ at $a = 0$:
   $$F(0) = \frac{(0 + 1)^6}{6} = \frac{1^6}{6} = \frac{1}{6}$$

10. Subtract $F(b) - F(a)$:
    $$F(1) - F(0) = \frac{32}{3} - \frac{1}{6} = \frac{64}{6} - \frac{1}{6} = \frac{63}{6} = \frac{21}{2}$$

Final Answer:
$$\int_0^1 (x + 1)^5 \, dx = \frac{21}{2}$$


</details>

## b.
$$ \displaystyle \int_{-1}^2 \frac{1}{x + 2} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral using $u$-substitution:

$$\int_{-1}^2 \frac{1}{x + 2} \, dx$$

1. $u = x + 2$

2. $\frac{du}{dx} = 1$

3. $du = dx$

4. $dx = du$

5. Substitute and simplify:
   $$\int_{-1}^2 \frac{1}{x + 2} \, dx = \int \frac{1}{u} \, du$$

6. Integrate:
   $$\int \frac{1}{u} \, du = \ln|u|$$

7. Plug back in for $u$ and find the antiderivative $F(x)$:
   $$F(x) = \ln|x + 2|$$

8. Evaluate $F(b)$ at $b = 2$:
   $$F(2) = \ln|2 + 2| = \ln(4)$$

9. Evaluate $F(a)$ at $a = -1$:
   $$F(-1) = \ln|-1 + 2| = \ln(1) = 0$$

10. Subtract $F(b) - F(a)$:
    $$F(2) - F(-1) = \ln(4) - 0 = \ln(4)$$

Final Answer:
$$\int_{-1}^2 \frac{1}{x + 2} \, dx = \ln(4)$$


</details>

## c.
$$  \displaystyle \int_0^5 (1 + 2x^2)^2 x \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral using $u$-substitution:

$$\int_0^5 (1 + 2x^2)^2 x \, dx$$

1. $u = 1 + 2x^2$

2. $\frac{du}{dx} = 4x$

3. $du = 4x \, dx$

4. $dx = \frac{du}{4x}$

5. Substitute and simplify:
   $$\int_0^5 (1 + 2x^2)^2 x \, dx = \int u^2 \cdot \frac{du}{4}$$
   $$= \frac{1}{4} \int u^2 \, du$$

6. Integrate:
   $$\int u^2 \, du = \frac{u^3}{3}$$
   $$\frac{1}{4} \int u^2 \, du = \frac{1}{4} \cdot \frac{u^3}{3} = \frac{u^3}{12}$$

7. Plug back in for $u$ and find the antiderivative $F(x)$:
   $$F(x) = \frac{(1 + 2x^2)^3}{12}$$

8. Evaluate $F(b)$ at $b = 5$:
   $$F(5) = \frac{(1 + 2(5)^2)^3}{12} = \frac{(1 + 2 \cdot 25)^3}{12} = \frac{(1 + 50)^3}{12} = \frac{51^3}{12}$$
   $$51^3 = 132651 \quad \text{so} \quad F(5) = \frac{132651}{12} = 11054.25$$

9. Evaluate $F(a)$ at $a = 0$:
   $$F(0) = \frac{(1 + 2(0)^2)^3}{12} = \frac{(1 + 0)^3}{12} = \frac{1}{12}$$

10. Subtract $F(b) - F(a)$:
    $$F(5) - F(0) = 11054.25 - \frac{1}{12}$$
    $$= 11054.25 - 0.0833 = 11054.1667$$

Final Answer:
$$\int_0^5 (1 + 2x^2)^2 x \, dx = 11054.1667$$


</details>

## d.
$$ \displaystyle \int_0^2 e^{x^2}x \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral using $u$-substitution:

$$\int_0^2 e^{x^2}x \, dx$$

1. $u = x^2$

2. $\frac{du}{dx} = 2x$

3. $du = 2x \, dx$

4. $dx = \frac{du}{2x}$

5. Substitute and simplify:
   $$\int_0^2 e^{x^2}x \, dx = \int e^u \cdot \frac{1}{2} \, du$$
   $$= \frac{1}{2} \int e^u \, du$$

6. Integrate:
   $$\int e^u \, du = e^u$$
   $$\frac{1}{2} \int e^u \, du = \frac{1}{2} e^u$$

7. Plug back in for $u$ and find the antiderivative $F(x)$:
   $$F(x) = \frac{1}{2} e^{x^2}$$

8. Evaluate $F(b)$ at $b = 2$:
   $$F(2) = \frac{1}{2} e^{2^2} = \frac{1}{2} e^4$$

9. Evaluate $F(a)$ at $a = 0$:
   $$F(0) = \frac{1}{2} e^{0^2} = \frac{1}{2} e^0 = \frac{1}{2}$$

10. Subtract $F(b) - F(a)$:
    $$F(2) - F(0) = \frac{1}{2} e^4 - \frac{1}{2}$$
    $$= \frac{1}{2} (e^4 - 1)$$

Final Answer:
$$\int_0^2 e^{x^2}x \, dx = \frac{1}{2} (e^4 - 1)$$


</details>

## e.
$$ \displaystyle \int_1^3 x \sqrt{x + 2} \, dx $$

<details> 
<summary> Solution: </summary>

To solve the definite integral using $u$-substitution:

$$\int_1^3 x \sqrt{x + 2} \, dx$$

1. $u = x + 2$

2. $\frac{du}{dx} = 1$

3. $du = dx$

4. Rewrite $x$ in terms of $u$:
   $$x = u - 2$$

5. Substitute and simplify:
   $$\int_1^3 x \sqrt{x + 2} \, dx = \int (u - 2) \sqrt{u} \, du$$
   $$= \int u^{3/2} \, du - \int 2u^{1/2} \, du$$

6. Integrate each term:
   For $\int u^{3/2} \, du$:
   $$\int u^{3/2} \, du = \frac{2}{5} u^{5/2}$$

   For $\int 2u^{1/2} \, du$:
   $$\int 2u^{1/2} \, du = 2 \cdot \frac{2}{3} u^{3/2} = \frac{4}{3} u^{3/2}$$

   Combine:
   $$\int (u - 2) \sqrt{u} \, du = \frac{2}{5} u^{5/2} - \frac{4}{3} u^{3/2}$$

7. Plug back in for $u$:
   $$F(x) = \frac{2}{5} (x + 2)^{5/2} - \frac{4}{3} (x + 2)^{3/2}$$

8. Evaluate $F(b)$ at $b = 3$:
   $$F(3) = \frac{2}{5} (3 + 2)^{5/2} - \frac{4}{3} (3 + 2)^{3/2} =\dfrac{10\sqrt{5}}{3}$$

9. Evaluate $F(a)$ at $a = 1$:
   $$F(1) = \frac{2}{5} (1 + 2)^{5/2} - \frac{4}{3} (1 + 2)^{3/2}=-\dfrac{2\sqrt{3}}{5}$$


10. Subtract $F(b) - F(a)$:
    $$F(3) - F(1) = \dfrac{10\sqrt{5}}{3}- \left(-\dfrac{2\sqrt{3}}{5}\right)=\dfrac{10\sqrt{5}}{3}+\dfrac{2\sqrt{3}}{5}$$

  

Final Answer:
$$\int_1^3 x \sqrt{x + 2} \, dx = \dfrac{10\sqrt{5}}{3}+\dfrac{2\sqrt{3}}{5}$$


</details>


## f.
$$ \displaystyle \int_1^3 \frac{(\ln(x))^3}{x} \, dx $$


<details> 
<summary> Solution: </summary>




1. Let $u = \ln(x)$.

2. $\frac{du}{dx} = \frac{1}{x}$.

3. $du = \frac{1}{x} dx$.

4. $dx = x \, du = du$.

5. Substitute into the integral:
   $$ \int_1^3 \frac{(\ln(x))^3}{x} \, dx = \int (\ln(x))^3 \cdot \frac{1}{x} dx = \int u^3 \, du $$

6. Integrate:
   $$ \int u^3 \, du = \frac{u^4}{4} $$

7. Rewrite the antiderivative in terms of $x$:
   $$ \frac{u^4}{4} = \frac{(\ln(x))^4}{4} $$

8. Plug in the upper bound $x = 3$:
   $$ F(3) = \frac{(\ln(3))^4}{4} $$

9. Plug in the lower bound $x = 1$:
   $$ F(1) = \frac{(\ln(1))^4}{4} = \frac{0^4}{4} = 0 $$

10. Subtract:
    $$ F(3) - F(1) = \frac{(\ln(3))^4}{4} - 0 $$

11. Final Answer:
    $$ \int_1^3 \frac{(\ln(x))^3}{x} \, dx = \frac{(\ln(3))^4}{4} $$


</details>


# Question 5 : The graph of $f(x)$ is given below with the areas.

![image](https://hackmd.io/_uploads/SJxLcZtGyg.png)


## a. Find $\displaystyle \int_{-3}^{-1} f(x) \, dx$



<details> 
<summary> Solution: </summary>

$\displaystyle \int_{-3}^{-1} f(x) \, dx=5.33$

</details>

## b. Find $\displaystyle \int_{-1}^2 f(x) \, dx$



<details> 
<summary> Solution: </summary>

$\displaystyle \int_{-1}^2 f(x) \, dx=-15.75$

</details>

## c. Find $\displaystyle \int_{2}^3 f(x) \, dx$



<details> 
<summary> Solution: </summary>

$\displaystyle \int_{2}^3 f(x) \, dx=10.42$

</details>

## d. Find $\displaystyle \int_{-3}^2 f(x) \, dx$




<details> 
<summary> Solution: </summary>

$\displaystyle \int_{-3}^2 f(x) \, dx=5.33-15.75=-10.42$

</details>

## e. Find $\displaystyle \int_{-3}^{-1} 4 f(x) \, dx$




<details> 
<summary> Solution: </summary>

$\displaystyle \int_{-3}^{-1} 4 f(x) \, dx=$\displaystyle 4\int_{-3}^{-1}  f(x) \, dx=4(5.33)=21.32$$

</details>

## f. Find $\displaystyle \int_{-1}^{-3}  f(x) \, dx$




<details> 
<summary> Solution: </summary>

$\displaystyle \int_{-1}^{-3}  f(x) \, dx=\displaystyle -\int_{-3}^{-1}  f(x) \, dx=-5.33$$

</details>


# Question 6: The marginal cost to produce x tennis shoes is
$$C'(x)=20+\dfrac{1000}{x+1}$$
If the fixed cost are $100,000, find the cost function and the cost to produce 5,000 shoes.

<details> 
<summary> Solution: </summary>


The marginal cost function is:
$$
C'(x) = 20 + \frac{1000}{x+1}
$$

The fixed costs are $100,000.

### Step-by-Step Solution

1. Rewrite the marginal cost function for integration:
   $$ C'(x) = 20 + \frac{1000}{x+1} $$

2. Integrate $C'(x)$ to find the cost function $C(x)$:
   $$ C(x) = \int \left( 20 + \frac{1000}{x+1} \right) dx $$

3. Separate the terms:
   $$ C(x) = \int 20 \, dx + \int \frac{1000}{x+1} \, dx $$

4. Compute each integral:
   - For $\int 20 \, dx$:
     $$ \int 20 \, dx = 20x $$
   - For $\int \frac{1000}{x+1} \, dx$:
     $$ \int \frac{1000}{x+1} \, dx = 1000 \ln|x+1| $$

5. Combine the results:
   $$ C(x) = 20x + 1000 \ln|x+1| + C_0 $$

6. Use the fixed costs to solve for $C_0$:
   At $x = 0$, $C(0) = 100,000$. Substituting:
   $$ 100,000 = 20(0) + 1000 \ln|0+1| + C_0 $$
   $$ 100,000 = 0 + 1000 \ln(1) + C_0 $$
   $$ 100,000 = 0 + C_0 $$
   $$ C_0 = 100,000 $$

7. The cost function is:
   $$ C(x) = 20x + 1000 \ln|x+1| + 100,000 $$

8. Calculate the cost to produce 5,000 shoes ($x = 5000$):
   $$ C(5000) = 20(5000) + 1000 \ln(5000+1) + 100,000 $$
   $$ C(5000) = 100,000 + 1000 \ln(5001) + 100,000 $$
   $$ C(5000) = 200,000 + 1000 \ln(5001) $$

9. Approximate $\ln(5001)$:
   $$ \ln(5001) \approx 8.517 $$
   $$ C(5000) = 200,000 + 1000(8.517) $$
   $$ C(5000) = 200,000 + 8,517 $$
   $$ C(5000) = 208,517 $$

### Final Results:

The cost function is:
$$
C(x) = 20x + 1000 \ln|x+1| + 100,000
$$

The cost to produce 5,000 shoes is:
$$
C(5000) = 208,517 \, \text{dollars.}
$$


</details>

# Question 7: The marginal cost to produce x tennis shoes is
$$C'(x)=20+\dfrac{1000}{(x+1)^2}$$
Find the cost function if it costs $30,000 to produce 999 tennis shoes.



<details> 
<summary> Solution: </summary>

### Question 7: Solve for the Cost Function

The marginal cost function is:
$$
C'(x) = 20 + \frac{1000}{(x+1)^2}
$$

We are given that it costs $30,000 to produce 999 tennis shoes.

### Step-by-Step Solution

1. Rewrite the marginal cost function for integration:
   $$
   C'(x) = 20 + \frac{1000}{(x+1)^2}
   $$

2. Integrate $C'(x)$ to find the cost function $C(x)$:
   $$
   C(x) = \int \left( 20 + \frac{1000}{(x+1)^2} \right) dx
   $$

3. Separate the terms:
   $$
   C(x) = \int 20 \, dx + \int \frac{1000}{(x+1)^2} \, dx
   $$

4. Compute each integral:
   - For $\int 20 \, dx$:
     $$
     \int 20 \, dx = 20x
     $$
   - For $\int \frac{1000}{(x+1)^2} \, dx$:
     Use the rule $\int \frac{1}{(x+1)^2} \, dx = -\frac{1}{x+1}$:
     $$
     \int \frac{1000}{(x+1)^2} \, dx = -\frac{1000}{x+1}
     $$

5. Combine the results:
   $$
   C(x) = 20x - \frac{1000}{x+1} + C_0
   $$

6. Use the given information to solve for $C_0$:
   At $x = 999$, $C(999) = 30,000$. Substituting:
   $$
   30,000 = 20(999) - \frac{1000}{999+1} + C_0
   $$
   $$
   30,000 = 19,980 - \frac{1000}{1000} + C_0
   $$
   $$
   30,000 = 19,980 - 1 + C_0
   $$
   $$
   30,000 = 18,979 + C_0
   $$
   $$
   C_0 = 30,000 - 18,979 = 11021
   $$

7. The cost function is:
   $$
   C(x) = 20x - \frac{1000}{x+1} + 11,021
   $$

### Final Result

The cost function is:
$$
C(x) = 20x - \frac{1000}{x+1} + 11,021
$$


</details>


# Question 8: The marginal cost to produce x tennis shoes is
$$C'(x)=30+\dfrac{1000}{x+1}$$

Find the increase in production cost if the production is increased from 500 to 600 shoes.



<details> 
<summary> Solution: </summary>

### Question: Increase in Production Cost

The marginal cost function is:
$$
C'(x) = 30 + \frac{1000}{x+1}
$$

We need to find the increase in production cost if production increases from 500 to 600 shoes.

### Step-by-Step Solution

1. The increase in production cost is given by the definite integral of the marginal cost function:
   $$
   \Delta C = \int_{500}^{600} C'(x) \, dx
   $$
   Substitute $C'(x) = 30 + \frac{1000}{x+1}$:
   $$
   \Delta C = \int_{500}^{600} \left(30 + \frac{1000}{x+1}\right) \, dx
   $$

2. Separate the terms:
   $$
   \Delta C = \int_{500}^{600} 30 \, dx + \int_{500}^{600} \frac{1000}{x+1} \, dx
   $$

3. Compute each integral:
   - For $\int_{500}^{600} 30 \, dx$:
     $$
     \int_{500}^{600} 30 \, dx = 30 \int_{500}^{600} 1 \, dx = 30[x]_{500}^{600}
     $$
     $$
     = 30(600 - 500) = 30 \cdot 100 = 3000
     $$

   - For $\int_{500}^{600} \frac{1000}{x+1} \, dx$:
     Use the rule $\int \frac{1}{x+1} \, dx = \ln|x+1|$:
     $$
     \int_{500}^{600} \frac{1000}{x+1} \, dx = 1000 \int_{500}^{600} \frac{1}{x+1} \, dx
     $$
     $$
     = 1000 \left[\ln|x+1|\right]_{500}^{600}
     $$
     $$
     = 1000 \left(\ln(600+1) - \ln(500+1)\right)
     $$
     $$
     = 1000 \left(\ln(601) - \ln(501)\right)
     $$
     Simplify using the logarithmic property $\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)$:
     $$
     = 1000 \ln\left(\frac{601}{501}\right)
     $$

4. Combine the results:
   $$
   \Delta C = 3000 + 1000 \ln\left(\frac{601}{501}\right)
   $$

5. Approximate $\ln\left(\frac{601}{501}\right)$:
   $$
   \frac{601}{501} \approx 1.1996 \quad \text{and} \quad \ln(1.1996) \approx 0.1822
   $$
   Substitute this value:
   $$
   \Delta C = 3000 + 1000(0.1822)
   $$
   $$
   \Delta C = 3000 + 182.2 = 3182.2
   $$

### Final Result

The increase in production cost is approximately:
$$
\Delta C = 3182.2 \, \text{dollars.}
$$


</details>

