Question 17

Given the function $f(x) = (x+3)(x^2+6x+8)^4$ and $F(x) = \int f(x) \, dx$ . If $F(-3) = 2$ , then the function $F(x)$ is ....

$\frac{1}{10}(x^2+6x+8)^5 + \frac{1}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{11}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{21}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{31}{10}$

$\frac{1}{10}(x^2+6x+8)^5 + \frac{41}{10}$

Given the function $f(x) = (x+3)(x^2+6x+8)^4$ and $F(x) = \int f(x) \, dx$ with the condition $F(-3) = 2$ .

To solve this integral, we use the substitution method.

Let $u = x^2 + 6x + 8$ .

The derivative of $u$ with respect to $x$ :

\frac{du}{dx} = 2x + 6

Factor:

\frac{du}{dx} = 2(x + 3)

Multiply both sides by $dx$ :

du = 2(x + 3) \, dx

Divide both sides by $2$ :

\frac{du}{2} = (x + 3) \, dx

Substitute into the integral:

F(x) = \int f(x) \, dx = \int (x+3)(x^2+6x+8)^4 \, dx

With substitution $u = x^2 + 6x + 8$ and $(x+3) \, dx = \frac{du}{2}$ :

F(x) = \int u^4 \cdot \frac{du}{2}

= \frac{1}{2} \int u^4 \, du

= \frac{1}{2} \cdot \frac{u^5}{5} + C

= \frac{1}{10} u^5 + C

Substitute back $u = x^2 + 6x + 8$ :

F(x) = \frac{1}{10}(x^2 + 6x + 8)^5 + C

Use the condition $F(-3) = 2$ :

F(-3) = \frac{1}{10}((-3)^2 + 6(-3) + 8)^5 + C = 2

Calculate the value inside the parentheses:

(-3)^2 + 6(-3) + 8 = 9 - 18 + 8 = -1

Substitute:

\frac{1}{10}(-1)^5 + C = 2

\frac{1}{10}(-1) + C = 2

-\frac{1}{10} + C = 2

C = 2 + \frac{1}{10}

C = \frac{20}{10} + \frac{1}{10}

C = \frac{21}{10}

Substitute the value of $C$ into $F(x)$ :

F(x) = \frac{1}{10}(x^2 + 6x + 8)^5 + \frac{21}{10}

Therefore, the function $F(x)$ is $\frac{1}{10}(x^2 + 6x + 8)^5 + \frac{21}{10}$ .

Command Palette