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[tex]\[ f(x)=6 x^2+\frac{1}{x} \][/tex], where [tex]\( x \)[/tex] is positive.
[tex]\( F \)[/tex] is the antiderivative of [tex]\( f \)[/tex] and [tex]\( F(1)=9 \)[/tex]. What is [tex]\( F(4) \)[/tex]?
Round your answer to two decimal places.

Sagot :

GinnyAnswer
To solve for [tex]\( F(4) \)[/tex], where [tex]\( F \)[/tex] is the antiderivative of the function [tex]\( f(x) = 6x^2 + \frac{1}{x} \)[/tex] and it is given that [tex]\( F(1) = 9 \)[/tex], we follow these steps:

### Step 1: Find the antiderivative of [tex]\( f(x) \)[/tex]
The given function [tex]\( f(x) \)[/tex] is:
[tex]\[ f(x) = 6x^2 + \frac{1}{x} \][/tex]

To find the antiderivative [tex]\( F(x) \)[/tex], integrate [tex]\( f(x) \)[/tex]:
[tex]\[ \int f(x) \, dx = \int (6x^2 + \frac{1}{x}) \, dx \][/tex]

### Step 2: Integrate term by term
Integrate each term separately:
[tex]\[ \int 6x^2 \, dx \][/tex]
and
[tex]\[ \int \frac{1}{x} \, dx \][/tex]

For the first term:
[tex]\[ \int 6x^2 \, dx = 6 \int x^2 \, dx = 6 \left( \frac{x^3}{3} \right) = 2x^3 \][/tex]

For the second term:
[tex]\[ \int \frac{1}{x} \, dx = \ln|x| \][/tex]

So, the antiderivative [tex]\( F(x) \)[/tex] is:
[tex]\[ F(x) = 2x^3 + \ln|x| + C \][/tex]
where [tex]\( C \)[/tex] is the constant of integration.

### Step 3: Determine the constant [tex]\( C \)[/tex]
We are given that [tex]\( F(1) = 9 \)[/tex]. Plug [tex]\( x = 1 \)[/tex] and [tex]\( F(1) = 9 \)[/tex] into the antiderivative:
[tex]\[ F(1) = 2(1)^3 + \ln|1| + C = 2 \cdot 1 + \ln(1) + C = 2 + 0 + C = 2 + C \][/tex]

Since [tex]\( F(1) = 9 \)[/tex], we have:
[tex]\[ 2 + C = 9 \][/tex]
[tex]\[ C = 9 - 2 \][/tex]
[tex]\[ C = 7 \][/tex]

### Step 4: Write the complete antiderivative
Substitute [tex]\( C = 7 \)[/tex] back into the antiderivative:
[tex]\[ F(x) = 2x^3 + \ln|x| + 7 \][/tex]

### Step 5: Evaluate [tex]\( F(4) \)[/tex]
Now we need to find [tex]\( F(4) \)[/tex]:
[tex]\[ F(4) = 2(4)^3 + \ln|4| + 7 \][/tex]
[tex]\[ F(4) = 2 \cdot 64 + \ln(4) + 7 \][/tex]
[tex]\[ F(4) = 128 + \ln(4) + 7 \][/tex]

Using the value of [tex]\( \ln(4) \approx 1.386294361120 \)[/tex]:
[tex]\[ F(4) = 128 + 1.386294361120 + 7 \][/tex]
[tex]\[ F(4) = 136.386294361120 \][/tex]

Rounded to two decimal places, [tex]\( F(4) \)[/tex] is:
[tex]\[ F(4) \approx 136.39 \][/tex]

Thus, the value of [tex]\( F(4) \)[/tex] is approximately [tex]\( 136.39 \)[/tex].
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