2. - / 12.5 points find f. f(x) = 6 - 6x - 60x³, f(0) = 9, f(0) = 6 f(x) = blank submit answer 3. - / 12.5 points write as a single integral in the form $\int_{a}^{b} f(x) \, dx$. $\int_{-5}^{2} f(x) \, dx + \int_{2}^{3} f(x) \, dx - \int_{-5}^{-3} f(x) \, dx$ some boxes with $f(x) \, dx$

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### Problem 2: Find $ f(x) $ given $ f''(x) = 6 - 6x - 60x^3 $, $ f(0) = 9 $, $ f'(0) = 6 $

# Explanation:
## Step 1: Integrate $ f''(x) $ to find $ f'(x) $
We know that to find the first derivative $ f'(x) $ from the second derivative $ f''(x) $, we integrate $ f''(x) $ with respect to $ x $. The integral of a sum is the sum of the integrals, and we use the power rule for integration $ \int x^n dx=\frac{x^{n + 1}}{n+1}+C$ (for $ n
eq - 1$).

$$
\begin{align*}
f'(x)&=\int(6 - 6x - 60x^3)dx\
&=\int6dx-\int6xdx-\int60x^3dx\
&=6x-6\times\frac{x^{2}}{2}-60\times\frac{x^{4}}{4}+C_1\
&=6x - 3x^{2}-15x^{4}+C_1
\end{align*}
$$

Now we use the initial condition $ f'(0) = 6 $. Substitute $ x = 0 $ and $ f'(0)=6 $ into the equation for $ f'(x) $:

$$
6=6(0)-3(0)^{2}-15(0)^{4}+C_1
$$

Simplifying the right - hand side, we get $ C_1 = 6 $. So, $ f'(x)=- 15x^{4}-3x^{2}+6x + 6 $

## Step 2: Integrate $ f'(x) $ to find $ f(x) $
Now we integrate $ f'(x) $ with respect to $ x $ to find $ f(x) $. Again, we use the power rule for integration.

$$
\begin{align*}
f(x)&=\int(-15x^{4}-3x^{2}+6x + 6)dx\
&=\int-15x^{4}dx-\int3x^{2}dx+\int6xdx+\int6dx\
&=-15\times\frac{x^{5}}{5}-3\times\frac{x^{3}}{3}+6\times\frac{x^{2}}{2}+6x + C_2\
&=- 3x^{5}-x^{3}+3x^{2}+6x + C_2
\end{align*}
$$

Now we use the initial condition $ f(0)=9 $. Substitute $ x = 0 $ and $ f(0) = 9 $ into the equation for $ f(x) $:

$$
9=-3(0)^{5}-(0)^{3}+3(0)^{2}+6(0)+C_2
$$

Simplifying the right - hand side, we get $ C_2=9 $.

# Answer:
$ f(x)=-3x^{5}-x^{3}+3x^{2}+6x + 9 $

### Problem 3: Combine the integrals $ \int_{-5}^{2}f(x)dx+\int_{2}^{3}f(x)dx-\int_{-5}^{-3}f(x)dx $ into a single integral of the form $ \int_{a}^{b}f(x)dx $

# Explanation:
## Step 1: Use the property of definite integrals $ \int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx $
First, consider $ \int_{-5}^{2}f(x)dx+\int_{2}^{3}f(x)dx $. By the property of definite integrals that the sum of integrals with adjacent limits is the integral over the combined interval, we have:

$ \int_{-5}^{2}f(x)dx+\int_{2}^{3}f(x)dx=\int_{-5}^{3}f(x)dx $

## Step 2: Use the property $ \int_{a}^{b}f(x)dx-\int_{a}^{c}f(x)dx=\int_{c}^{b}f(x)dx $ (which is equivalent to $ \int_{a}^{b}f(x)dx+\int_{c}^{a}f(x)dx=\int_{c}^{b}f(x)dx $)
Now we have $ \int_{-5}^{3}f(x)dx-\int_{-5}^{-3}f(x)dx $. We can rewrite the subtraction of the integral as the addition of the integral with the limits reversed (since $ \int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx $):

$ \int_{-5}^{3}f(x)dx-\int_{-5}^{-3}f(x)dx=\int_{-5}^{3}f(x)dx+\int_{-3}^{-5}f(x)dx $

But using the property $ \int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx $ (or the equivalent for subtraction $ \int_{a}^{b}f(x)dx-\int_{a}^{c}f(x)dx=\int_{c}^{b}f(x)dx $ when $ c < b $ and $ a$ is the lower limit for both), we know that $ \int_{-5}^{3}f(x)dx-\int_{-5}^{-3}f(x)dx=\int_{-3}^{3}f(x)dx $

# Answer:
$ \int_{-3}^{3}f(x)dx $

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Question

Explanation:

Problem 2: Find \( f(x) \) given \( f''(x) = 6 - 6x - 60x^3 \), \( f(0) = 9 \), \( f'(0) = 6 \)

Step 1: Integrate \( f''(x) \) to find \( f'(x) \)

Step 2: Integrate \( f'(x) \) to find \( f(x) \)

Step 1: Use the property of definite integrals \( \int_{a}^{b}f(x)dx+\int_{b}^{c}f(x)dx=\int_{a}^{c}f(x)dx \)

Step 2: Use the property \( \int_{a}^{b}f(x)dx-\int_{a}^{c}f(x)dx=\int_{c}^{b}f(x)dx \) (which is equivalent to \( \int_{a}^{b}f(x)dx+\int_{c}^{a}f(x)dx=\int_{c}^{b}f(x)dx \))

Answer:

Problem 3: Combine the integrals \( \int_{-5}^{2}f(x)dx+\int_{2}^{3}f(x)dx-\int_{-5}^{-3}f(x)dx \) into a single integral of the form \( \int_{a}^{b}f(x)dx \)