---
# System prepended metadata

title: In-Class Activity 6.3
tags: ['6.3', in-class-activity]

---

# Question 1:
Write each $H(x)$ as $H(x) = f(x)g'(x)$, then calculate $f'(x)g(x)$, for an antiderivative $g(x)$. Do this in two different ways.

## a. 
$$H(x) = xe^x$$

::: spoiler

<summary> Example: </summary>

$H(x)=x^3 e^x$

We are given the function $H(x) = x^3 e^x=f(x)g'(x)$, and we want to express this in two different ways as $f(x) g'(x)$, then calculate $f'(x) g(x)$, where $g(x)$ is an antiderivative of $g'(x)$.

### Method 1:
1. **Let $f(x) = x^3$ and $g'(x) = e^x$.**
2. The antiderivative of $g'(x) = e^x$ is $g(x) = e^x$.
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = 3x^2$
   - $g(x) = e^x$
   - So, $f'(x) g(x) = 3x^2 e^x$.

### Method 2:
1. **Let $f(x) = e^x$ and $g'(x) = x^3$.**
2. The antiderivative of $g'(x) = x^3$ is $g(x) = \frac{x^4}{4}$.
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = e^x$
   - $g(x) = \frac{x^4}{4}$
   - So, $f'(x) g(x) = \frac{x^4 e^x}{4}$.

### Summary:

Here’s a table summarizing the two methods:

| **Method**                         | **Expression for $f(x)$** | **Expression for $g'(x)$** | **Result of $f'(x)g(x)$**           |
|------------------------------------|----------------------------|-----------------------------|--------------------------------------|
| **Method 1**: $f(x) = x^3$, $g'(x) = e^x$ | $f(x) = x^3$                | $g'(x) = e^x$               | $3x^2 e^x$                          |
| **Method 2**: $f(x) = e^x$, $g'(x) = x^3$ | $f(x) = e^x$                | $g'(x) = x^3$               | $\frac{x^4 e^x}{4}$                 |

This table shows how we express the function in two ways and the results of $f'(x) g(x)$ for each method.



:::

## b.
$$H(x) = x^2 e^x$$

::: spoiler

<summary> Example: </summary>

We are given the function $H(x) = x^4 e^x$, and we want to express this in two different ways as $f(x) g'(x)$, then calculate $f'(x) g(x)$, where $g(x)$ is an antiderivative of $g'(x)$.

### Method 1:
1. **Let $f(x) = x^4$ and $g'(x) = e^x$.**
2. The antiderivative of $g'(x) = e^x$ is $g(x) = e^x$.
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = 4x^3$
   - $g(x) = e^x$
   - So, $f'(x) g(x) = 4x^3 e^x$.

### Method 2:
1. **Let $f(x) = e^x$ and $g'(x) = x^4$.**
2. The antiderivative of $g'(x) = x^4$ is $g(x) = \frac{x^5}{5}$.
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = e^x$
   - $g(x) = \frac{x^5}{5}$
   - So, $f'(x) g(x) = \frac{x^5 e^x}{5}$.

### Summary:

Here’s a table summarizing the two methods:

| **Method**                         | **Expression for $f(x)$** | **Expression for $g'(x)$** | **Result of $f'(x)g(x)$**           |
|------------------------------------|----------------------------|-----------------------------|--------------------------------------|
| **Method 1**: $f(x) = x^4$, $g'(x) = e^x$ | $f(x) = x^4$                | $g'(x) = e^x$               | $4x^3 e^x$                          |
| **Method 2**: $f(x) = e^x$, $g'(x) = x^4$ | $f(x) = e^x$                | $g'(x) = x^4$               | $\frac{x^5 e^x}{5}$                 |

This table shows how we express the function in two ways and the results of $f'(x) g(x)$ for each method.


:::

## c.
$$H(x) = x \ln(x)$$

::: spoiler

<summary> Example: </summary>


We are given the function $H(x) = x^2 \ln(x)$, and we want to express this in two different ways as $f(x) g'(x)$, then calculate $f'(x) g(x)$, where $g(x)$ is an antiderivative of $g'(x)$.

### Method 1:
1. **Let $f(x) = x^2$ and $g'(x) = \ln(x)$.**
2. The antiderivative of $g'(x) = \ln(x)$ is $g(x) = x \ln(x) - x$ (since $\int \ln(x) dx = x \ln(x) - x$).
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = 2x$
   - $g(x) = x \ln(x) - x$
   - So, $f'(x) g(x) = 2x \left(x \ln(x) - x\right) = 2x^2 \ln(x) - 2x^2$.

### Method 2:
1. **Let $f(x) = \ln(x)$ and $g'(x) = x^2$.**
2. The antiderivative of $g'(x) = x^2$ is $g(x) = \frac{x^3}{3}$.
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = \frac{1}{x}$
   - $g(x) = \frac{x^3}{3}$
   - So, $f'(x) g(x) = \frac{1}{x} \cdot \frac{x^3}{3} = \frac{x^2}{3}$.

### Summary:

Here’s a table summarizing the two methods:

| **Method**                         | **Expression for $f(x)$** | **Expression for $g'(x)$** | **Result of $f'(x)g(x)$**               |
|------------------------------------|----------------------------|-----------------------------|-----------------------------------------|
| **Method 1**: $f(x) = x^2$, $g'(x) = \ln(x)$ | $f(x) = x^2$                | $g'(x) = \ln(x)$               | $2x^2 \ln(x) - 2x^2$                   |
| **Method 2**: $f(x) = \ln(x)$, $g'(x) = x^2$ | $f(x) = \ln(x)$                | $g'(x) = x^2$               | $\frac{x^2}{3}$                         |

This table shows how we express the function in two ways and the results of $f'(x) g(x)$ for each method.


:::

## d.
$$H(x) = x^3 \ln(x)$$

::: spoiler

<summary> Example: </summary>

We are given the function $H(x) = x^4 \ln(x)$, and we want to express this in two different ways as $f(x) g'(x)$, then calculate $f'(x) g(x)$, where $g(x)$ is an antiderivative of $g'(x)$.

### Method 1:
1. **Let $f(x) = x^4$ and $g'(x) = \ln(x)$.**
2. The antiderivative of $g'(x) = \ln(x)$ is $g(x) = x \ln(x) - x$ (since $\int \ln(x) dx = x \ln(x) - x$).
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = 4x^3$
   - $g(x) = x \ln(x) - x$
   - So, $f'(x) g(x) = 4x^3 \left(x \ln(x) - x\right) = 4x^4 \ln(x) - 4x^4$.

### Method 2:
1. **Let $f(x) = \ln(x)$ and $g'(x) = x^4$.**
2. The antiderivative of $g'(x) = x^4$ is $g(x) = \frac{x^5}{5}$.
3. Now, calculate $f'(x)$ and multiply by $g(x)$:
   - $f'(x) = \frac{1}{x}$
   - $g(x) = \frac{x^5}{5}$
   - So, $f'(x) g(x) = \frac{1}{x} \cdot \frac{x^5}{5} = \frac{x^4}{5}$.

### Summary:

Here’s a table summarizing the two methods:

| **Method**                         | **Expression for $f(x)$** | **Expression for $g'(x)$** | **Result of $f'(x)g(x)$**               |
|------------------------------------|----------------------------|-----------------------------|-----------------------------------------|
| **Method 1**: $f(x) = x^4$, $g'(x) = \ln(x)$ | $f(x) = x^4$                | $g'(x) = \ln(x)$               | $4x^4 \ln(x) - 4x^4$                   |
| **Method 2**: $f(x) = \ln(x)$, $g'(x) = x^4$ | $f(x) = \ln(x)$                | $g'(x) = x^4$               | $\frac{x^4}{5}$                         |

This table shows how we express the function in two ways and the results of $f'(x) g(x)$ for each method.

:::


# Question 2: Find the following derivatives

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

::: spoiler

<summary> Example: </summary>

Find the derivative of $\dfrac{x}{x^3+2}$




We are given the function $\frac{x}{x^3 + 2}$, and we need to find the derivative using the quotient rule.

The quotient rule for a function $\frac{f(x)}{g(x)}$ is given by:

$$
\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}
$$

In this case:
- $f(x) = x$ (the top function)
- $g(x) = x^3 + 2$ (the bottom function)

### Table:

| **Original / Derivative** | **Top Function**           | **Bottom Function**        |
|----------------------------|----------------------------|----------------------------|
| **Original**               | $f(x) = x$                 | $g(x) = x^3 + 2$           |
| **Derivative**             | $f'(x) = 1$                | $g'(x) = 3x^2$             |

### Step-by-Step Derivative Calculation:

Using the quotient rule:

$$
\frac{d}{dx}\left(\frac{x}{x^3 + 2}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2}
$$

Substitute $f(x) = x$, $f'(x) = 1$, $g(x) = x^3 + 2$, and $g'(x) = 3x^2$:

$$
= \frac{1(x^3 + 2) - x(3x^2)}{(x^3 + 2)^2}
$$

Simplify the numerator:

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

Combine like terms:

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

Finally, factor out the common term in the numerator:

$$
= \frac{2(1 - x^3)}{(x^3 + 2)^2}
$$

### Final Result:

$$
\frac{d}{dx}\left(\frac{x}{x^3 + 2}\right) = \frac{2(1 - x^3)}{(x^3 + 2)^2}
$$


:::

---

## b. 
For the function 
$$g(x) = (x^2+1)^8$$, find:
$$g'(x) = $$

::: spoiler

<summary> Example: </summary>

We are given the function $g(x) = (x^3 + 2)^7$, and we need to find its derivative using the chain rule.

The chain rule states that if $g(x) = f(u)$ and $u = h(x)$, then:

$$
\frac{d}{dx}g(x) = f'(u) \cdot u'(x)
$$

In this case:
- The outer function is $f(u) = u^7$
- The inner function is $u(x) = x^3 + 2$

### Table:

| **Original / Derivative** | **Outer Function $f(u)$**  | **Inner Function $u(x)$**  |
|----------------------------|----------------------------|----------------------------|
| **Original**               | $f(u) = u^7$               | $u(x) = x^3 + 2$           |
| **Derivative**             | $f'(u) = 7u^6$             | $u'(x) = 3x^2$             |

### Step-by-Step Derivative Calculation:

Using the chain rule:

$$
\frac{d}{dx}(x^3 + 2)^7 = f'(u) \cdot u'(x)
$$

Substitute $f'(u) = 7u^6$ and $u'(x) = 3x^2$:

$$
= 7(x^3 + 2)^6 \cdot 3x^2
$$

Simplify:

$$
= 21x^2 (x^3 + 2)^6
$$

### Final Result:

The derivative of $g(x) = (x^3 + 2)^7$ is:

$$
g'(x) = 21x^2 (x^3 + 2)^6
$$

:::

---

## c. 
For the function 
$$h(x) = \sqrt{1-x^2}$$ find
$$h'(x) = $$

::: spoiler

<summary> Example: </summary>

We are given the function $h(x) = \sqrt{2 - x^3}$, and we need to find its derivative using the chain rule.

First, rewrite the function as:

$$
h(x) = (2 - x^3)^{\frac{1}{2}}
$$

The chain rule states that if $h(x) = f(u)$ and $u = g(x)$, then:

$$
\frac{d}{dx}h(x) = f'(u) \cdot u'(x)
$$

In this case:
- The outer function is $f(u) = u^{\frac{1}{2}}$
- The inner function is $u(x) = 2 - x^3$

### Table:

| **Original / Derivative** | **Outer Function $f(u)$**          | **Inner Function $u(x)$**  |
|----------------------------|------------------------------------|----------------------------|
| **Original**               | $f(u) = u^{\frac{1}{2}}$           | $u(x) = 2 - x^3$           |
| **Derivative**             | $f'(u) = \frac{1}{2} u^{-\frac{1}{2}}$  | $u'(x) = -3x^2$            |

### Step-by-Step Derivative Calculation:

Using the chain rule:

$$
\frac{d}{dx}(2 - x^3)^{\frac{1}{2}} = f'(u) \cdot u'(x)
$$

Substitute $f'(u) = \frac{1}{2} u^{-\frac{1}{2}}$ and $u'(x) = -3x^2$:

$$
= \frac{1}{2}(2 - x^3)^{-\frac{1}{2}} \cdot (-3x^2)
$$

Simplify:

$$
= \frac{-3x^2}{2\sqrt{2 - x^3}}
$$

### Final Result:

The derivative of $h(x) = \sqrt{2 - x^3}$ is:

$$
h'(x) = \frac{-3x^2}{2\sqrt{2 - x^3}}
$$

:::

---

# Question 3:
Integrate by parts to find each integral

## a.
$$
\int x^3 \ln(x) \, dx = \quad
$$

::: spoiler

<summary> Example: </summary>

We are tasked with finding $\displaystyle \int x^4 \ln(x) \, dx$ using integration by parts.

The integration by parts formula is:

$$
\int u \, dv = uv - \int v \, du
$$

### Step-by-Step Solution:

We choose:
- $u = \ln(x)$
- $dv = x^4 \, dx$

Now, differentiate $u$ and integrate $dv$:

| **$u$ and $du$**                  | **$dv$ and $v$**             |
|-----------------------------------|------------------------------|
| $u = \ln(x)$                      | $dv = x^4 \, dx$             |
| $du = \frac{1}{x} \, dx$          | $v = \frac{x^5}{5}$          |

Now apply the integration by parts formula:

$$
\int x^4 \ln(x) \, dx = \ln(x) \cdot \frac{x^5}{5} - \int \frac{x^5}{5} \cdot \frac{1}{x} \, dx
$$

Simplify:

$$
= \frac{x^5 \ln(x)}{5} - \frac{1}{5} \int x^4 \, dx
$$

Now, integrate $x^4$:

$$
\int x^4 \, dx = \frac{x^5}{5}
$$

Substitute this back into the equation:

$$
= \frac{x^5 \ln(x)}{5} - \frac{1}{5} \cdot \frac{x^5}{5}
$$

Simplify:

$$
= \frac{x^5 \ln(x)}{5} - \frac{x^5}{25}
$$

Thus, the integral of $x^4 \ln(x)$ is:

$$
\int x^4 \ln(x) \, dx = \frac{x^5 \ln(x)}{5} - \frac{x^5}{25} + C
$$

Where $C$ is the constant of integration.

### Final Answer:

$$
\int x^4 \ln(x) \, dx = \frac{x^5 \ln(x)}{5} - \frac{x^5}{25} + C
$$


:::

---

## b.
$$
\int \sqrt{x} \ln(x) \, dx = 
$$

::: spoiler

<summary> Example: </summary>

We are tasked with finding $\displaystyle \int \sqrt[3]{x} \ln(x) \,  dx$ using integration by parts.

First, rewrite the cube root as a power:

$$
\int \sqrt[3]{x} \ln(x) \, dx = \int x^{\frac{1}{3}} \ln(x) \, dx
$$

The integration by parts formula is:

$$
\int u \, dv = uv - \int v \, du
$$

### Step-by-Step Solution:

We choose:
- $u = \ln(x)$
- $dv = x^{\frac{1}{3}} \, dx$

Now, differentiate $u$ and integrate $dv$:

| **$u$ and $du$**                  | **$dv$ and $v$**             |
|-----------------------------------|------------------------------|
| $u = \ln(x)$                      | $dv = x^{\frac{1}{3}} \, dx$ |
| $du = \frac{1}{x} \, dx$          | $v = \frac{3}{4} x^{\frac{4}{3}}$ |

Now apply the integration by parts formula:

$$
\int x^{\frac{1}{3}} \ln(x) \, dx = \ln(x) \cdot \frac{3}{4} x^{\frac{4}{3}} - \int \frac{3}{4} x^{\frac{4}{3}} \cdot \frac{1}{x} \, dx
$$

Simplify:

$$
= \frac{3}{4} x^{\frac{4}{3}} \ln(x) - \frac{3}{4} \int x^{\frac{1}{3}} \, dx
$$

Now, integrate $x^{\frac{1}{3}}$:

$$
\int x^{\frac{1}{3}} \, dx = \frac{3}{4} x^{\frac{4}{3}}
$$

Substitute this back into the equation:

$$
= \frac{3}{4} x^{\frac{4}{3}} \ln(x) - \frac{9}{16} x^{\frac{4}{3}}
$$

Thus, the integral of $\sqrt[3]{x} \ln(x)$ is:

$$
\int x^{\frac{1}{3}} \ln(x) \, dx = \frac{3}{4} x^{\frac{4}{3}} \ln(x) - \frac{9}{16} x^{\frac{4}{3}} + C
$$

Where $C$ is the constant of integration.

### Final Answer:

$$
\int x^{\frac{1}{3}} \ln(x) \, dx = \frac{3}{4} x^{\frac{4}{3}} \ln(x) - \frac{9}{16} x^{\frac{4}{3}} + C
$$


:::

---

# Question 4:
Integrate as indicated. Use integration by parts.

## a.
$$
\int_0^1 (x+1) e^x \, dx = \quad
$$

::: spoiler

<summary> Example: </summary>

We are tasked with evaluating the definite integral:

$$
\int_0^2 (x+2)e^x \, dx
$$

### Step-by-Step Solution:

We'll use integration by parts. The integration by parts formula is:

$$
\int u \, dv = uv - \int v \, du
$$

We choose:
- $u = (x+2)$
- $dv = e^x \, dx$

Now, differentiate $u$ and integrate $dv$:

| **$u$ and $du$**                  | **$dv$ and $v$**             |
|-----------------------------------|------------------------------|
| $u = x+2$                         | $dv = e^x \, dx$             |
| $du = dx$                         | $v = e^x$                    |

Now apply the integration by parts formula:

$$
\int (x+2) e^x \, dx = (x+2) e^x - \int e^x \, dx
$$

The second integral is straightforward:

$$
\int e^x \, dx = e^x
$$

Thus, the integral becomes:

$$
(x+2) e^x - e^x
$$

Now, evaluate this from 0 to 2:

$$
\left[ (x+2) e^x - e^x \right]_0^2
$$

At the upper limit $x = 2$:

$$
\left( (2+2) e^2 - e^2 \right) = (4e^2 - e^2) = 3e^2
$$

At the lower limit $x = 0$:

$$
\left( (0+2) e^0 - e^0 \right) = (2 \cdot 1 - 1) = 1
$$

Thus, the final result is:

$$
3e^2 - 1
$$

### Final Answer:

$$
\int_0^2 (x+2)e^x \, dx = 3e^2 - 1
$$


:::

---

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

::: spoiler

<summary> Example: </summary>

We are tasked with evaluating the definite integral:

$$
\int_2^e \frac{\ln x}{x^3} \, dx
$$

### Step-by-Step Solution:

We'll first find the **antiderivative** of the function \( \frac{\ln x}{x^3} \) using **integration by parts**.

The integration by parts formula is:

$$
\int u \, dv = uv - \int v \, du
$$

### Step 1: Choosing $u$ and $dv$

Let:
- $u = \ln x$, so $du = \frac{1}{x} \, dx$
- $dv = \frac{1}{x^3} \, dx$, so $v = -\frac{1}{2x^2}$ (since $\int \frac{1}{x^3} \, dx = -\frac{1}{2x^2}$)

### Step 2: Applying the integration by parts formula

Now, apply the integration by parts formula:

$$
\int \frac{\ln x}{x^3} \, dx = \ln x \cdot \left( -\frac{1}{2x^2} \right) - \int \left( -\frac{1}{2x^2} \right) \cdot \frac{1}{x} \, dx
$$

This simplifies to:

$$
\int \frac{\ln x}{x^3} \, dx = -\frac{\ln x}{2x^2} + \frac{1}{2} \int \frac{1}{x^3} \, dx
$$

### Step 3: Solving the remaining integral

The remaining integral is:

$$
\int \frac{1}{x^3} \, dx = -\frac{1}{2x^2}
$$

So, the full antiderivative is:

$$
-\frac{\ln x}{2x^2} + \frac{1}{2} \left( -\frac{1}{2x^2} \right) = -\frac{\ln x}{2x^2} - \frac{1}{4x^2}
$$

### Step 4: Evaluate the antiderivative at the bounds

Now we evaluate the antiderivative from $x = 2$ to $x = e$.

At $x = e$:

$$
-\frac{\ln e}{2e^2} - \frac{1}{4e^2} = -\frac{1}{2e^2} - \frac{1}{4e^2} = -\frac{3}{4e^2}
$$

At $x = 2$:

$$
-\frac{\ln 2}{2 \cdot 2^2} - \frac{1}{4 \cdot 2^2} = -\frac{\ln 2}{8} - \frac{1}{16}
$$

### Step 5: Subtract to find the definite integral

Now subtract the values at $x = e$ and $x = 2$:

$$\left( -\frac{3}{4e^2} \right) - \left( -\frac{\ln 2}{8} - \frac{1}{16} \right)$$

Simplifying:

$$= -\frac{3}{4e^2} + \frac{\ln 2}{8} + \frac{1}{16}$$

### Final Answer:

$$\int_2^e \frac{\ln x}{x^3} \, dx = - \frac{3}{4e^2} + \frac{\ln 2}{8} + \frac{1}{16}$$



:::
