Answer :
To find the function representing the total volume of water pumped, [tex]\( V(x) \)[/tex], we need to integrate the rate function, [tex]\( R(x) = 7x^5 + 5 \)[/tex]. The integration of [tex]\( R(x) \)[/tex] with respect to [tex]\( x \)[/tex] gives us the total volume of water pumped up to time [tex]\( x \)[/tex].
To find [tex]\( V(x) \)[/tex]:
1. Integrate [tex]\( R(x) = 7x^5 + 5 \)[/tex]:
[tex]\[ V(x) = \int (7x^5 + 5) \, dx \][/tex]
2. Calculate the integral:
[tex]\[ \int 7x^5 \, dx = \frac{7}{6}x^6 \][/tex]
[tex]\[ \int 5 \, dx = 5x \][/tex]
3. Combine the results:
[tex]\[ V(x) = \frac{7}{6} x^6 + 5x + C \][/tex]
Since the pump starts when time [tex]\( x \)[/tex] starts (typically at [tex]\( x = 0 \)[/tex]), the constant of integration [tex]\( C \)[/tex] is zero in the context of this problem. Thus:
[tex]\[ V(x) = \frac{7}{6} x^6 + 5x \][/tex]
Next, to determine the total volume of water pumped between 3 and 7 minutes, we need to compute [tex]\( V(7) - V(3) \)[/tex].
1. Evaluate [tex]\( V(7) \)[/tex]:
[tex]\[ V(7) = \frac{7}{6} (7^6) + 5(7) \][/tex]
2. Evaluate [tex]\( V(3) \)[/tex]:
[tex]\[ V(3) = \frac{7}{6} (3^6) + 5(3) \][/tex]
3. Calculate the difference [tex]\( V(7) - V(3) \)[/tex]:
[tex]\[ V(7) - V(3) = \left( \frac{7}{6} (7^6) + 5(7) \right) - \left( \frac{7}{6} (3^6) + 5(3) \right) \][/tex]
After evaluating these expressions and computing the values:
[tex]\[ V(7) = 136423.5 \][/tex]
[tex]\[ V(3) = -3.17 \][/tex]
Thus, the total volume of water pumped between 3 and 7 minutes is approximately:
[tex]\[ 136426.67 \ liters \][/tex]
So, the function [tex]\( V(x) \)[/tex] is:
[tex]\[ V(x) = \frac{7}{6} x^6 + 5x \][/tex]
And the total volume of water pumped between 3 and 7 minutes is:
[tex]\[ 136426.67 \ liters \][/tex]
Therefore:
- [tex]\( V(x) = \frac{7}{6} x^6 + 5x \)[/tex]
- The total volume of water pumped between 3 and 7 minutes is [tex]\( 136426.67 \)[/tex] liters
To find [tex]\( V(x) \)[/tex]:
1. Integrate [tex]\( R(x) = 7x^5 + 5 \)[/tex]:
[tex]\[ V(x) = \int (7x^5 + 5) \, dx \][/tex]
2. Calculate the integral:
[tex]\[ \int 7x^5 \, dx = \frac{7}{6}x^6 \][/tex]
[tex]\[ \int 5 \, dx = 5x \][/tex]
3. Combine the results:
[tex]\[ V(x) = \frac{7}{6} x^6 + 5x + C \][/tex]
Since the pump starts when time [tex]\( x \)[/tex] starts (typically at [tex]\( x = 0 \)[/tex]), the constant of integration [tex]\( C \)[/tex] is zero in the context of this problem. Thus:
[tex]\[ V(x) = \frac{7}{6} x^6 + 5x \][/tex]
Next, to determine the total volume of water pumped between 3 and 7 minutes, we need to compute [tex]\( V(7) - V(3) \)[/tex].
1. Evaluate [tex]\( V(7) \)[/tex]:
[tex]\[ V(7) = \frac{7}{6} (7^6) + 5(7) \][/tex]
2. Evaluate [tex]\( V(3) \)[/tex]:
[tex]\[ V(3) = \frac{7}{6} (3^6) + 5(3) \][/tex]
3. Calculate the difference [tex]\( V(7) - V(3) \)[/tex]:
[tex]\[ V(7) - V(3) = \left( \frac{7}{6} (7^6) + 5(7) \right) - \left( \frac{7}{6} (3^6) + 5(3) \right) \][/tex]
After evaluating these expressions and computing the values:
[tex]\[ V(7) = 136423.5 \][/tex]
[tex]\[ V(3) = -3.17 \][/tex]
Thus, the total volume of water pumped between 3 and 7 minutes is approximately:
[tex]\[ 136426.67 \ liters \][/tex]
So, the function [tex]\( V(x) \)[/tex] is:
[tex]\[ V(x) = \frac{7}{6} x^6 + 5x \][/tex]
And the total volume of water pumped between 3 and 7 minutes is:
[tex]\[ 136426.67 \ liters \][/tex]
Therefore:
- [tex]\( V(x) = \frac{7}{6} x^6 + 5x \)[/tex]
- The total volume of water pumped between 3 and 7 minutes is [tex]\( 136426.67 \)[/tex] liters
