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A pump is delivering water into a tank at a rate of [tex]R(x) = 8x^3 + 2[/tex] liters per minute, where [tex]x[/tex] is time in minutes since the pump is turned on.

1. Find the function representing the total volume of water pumped [tex]V(x)[/tex].
2. Determine the total volume of water pumped between 3 and 4 minutes.

[tex]V(x) = \square[/tex]

The total volume of water pumped between 3 and 4 minutes is [tex]\square[/tex].

Round the answer to 2 decimal places as needed.

Sagot :

GinnyAnswer
To solve the problem, we start with determining the function that represents the total volume of water pumped over time, [tex]\( V(x) \)[/tex].

### Step 1: Define the rate function
The rate at which water is pumped into the tank is given by:
[tex]\[ R(x) = 8x^3 + 2 \][/tex]
This represents the rate in liters per minute.

### Step 2: Find the volume function [tex]\( V(x) \)[/tex]
To find the volume function [tex]\( V(x) \)[/tex], we need to integrate the rate function [tex]\( R(x) \)[/tex] with respect to time [tex]\( x \)[/tex]:
[tex]\[ V(x) = \int R(x) \, dx = \int (8x^3 + 2) \, dx \][/tex]

### Step 3: Integrate [tex]\( R(x) \)[/tex]
Perform the integration:
[tex]\[ V(x) = \int (8x^3 + 2) \, dx \][/tex]
[tex]\[ V(x) = \int 8x^3 \, dx + \int 2 \, dx \][/tex]
[tex]\[ V(x) = 8 \int x^3 \, dx + 2 \int 1 \, dx \][/tex]
[tex]\[ V(x) = 8 \left( \frac{x^4}{4} \right) + 2x \][/tex]
[tex]\[ V(x) = 2x^4 + 2x + C \][/tex]
Here, [tex]\( C \)[/tex] is the constant of integration. Since we are interested in the volume pumped from the start, we can consider [tex]\( C = 0 \)[/tex] if we are dealing with definite integrals based on initial conditions.

So, the volume function is:
[tex]\[ V(x) = 2x^4 + 2x \][/tex]

### Step 4: Determine the total volume of water pumped between 3 and 4 minutes
We need to calculate the definite integral of [tex]\( R(x) \)[/tex] from [tex]\( x = 3 \)[/tex] to [tex]\( x = 4 \)[/tex]. Alternatively, we can use the volume function [tex]\( V(x) \)[/tex] to find this difference:
[tex]\[ \text{Total volume} = V(4) - V(3) \][/tex]

Evaluate [tex]\( V(x) \)[/tex] at [tex]\( x = 4 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ V(4) = 2(4)^4 + 2(4) \][/tex]
[tex]\[ V(4) = 2(256) + 8 \][/tex]
[tex]\[ V(4) = 512 + 8 \][/tex]
[tex]\[ V(4) = 520 \][/tex]

[tex]\[ V(3) = 2(3)^4 + 2(3) \][/tex]
[tex]\[ V(3) = 2(81) + 6 \][/tex]
[tex]\[ V(3) = 162 + 6 \][/tex]
[tex]\[ V(3) = 168 \][/tex]

Now, subtract [tex]\( V(3) \)[/tex] from [tex]\( V(4) \)[/tex]:
[tex]\[ \text{Total volume} = V(4) - V(3) \][/tex]
[tex]\[ \text{Total volume} = 520 - 168 \][/tex]
[tex]\[ \text{Total volume} = 352 \][/tex]

Thus, the total volume of water pumped between 3 and 4 minutes is [tex]\( 352.0 \)[/tex] liters.

### Final Answer
[tex]\[ V(x) = 2x^4 + 2x \][/tex]
The total volume of water pumped between 3 and 4 minutes is [tex]\( 352.0 \)[/tex] liters.
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