Question

The total cost of purchasing and maintaining a piece of equipment for \(x\) years can be modeled by \(C=5000\left(25+3 \int_{0}^{x} t^{1 / 4} d t\right)\) Find the total cost after (a) 1 year, (b) 5 years, and (c) 10 years.

Short answer

(a) The total cost after 1 year is $137,000. (b) The total cost after 5 years is $247,127.08. (c) The total cost after 10 years is $425,264.82

Step-by-step solution

Solving the Integral

First solve the given integral \(\int_{0}^{x} t^{1 / 4} dt\) using the power rule for integration. The power rule states that the integral of \(x^n\) is \(\frac{1}{n+1}x^{n+1}\), thus applying this rule yields \(\int t^{1 / 4} dt = \frac{4}{5}t^{5 / 4} + C\). Evaluating this from 0 to x gives \(\frac{4}{5}x^{5 / 4} - \frac{4}{5}(0)^{5 / 4} = \frac{4}{5}x^{5 / 4}\)

Substituting the Integral into the Cost Function

Now, substitute the solution of the integral back into the cost function \(C = 5000(25 + 3 \int_{0}^{x} t^{1 / 4} dt)\). This gives \(C = 5000(25 + 3(\frac{4}{5}x^{5 / 4})) = 5000(25 + \frac{12}{5}x^{5 / 4}) = 125000 + 12000x^{5 / 4}\)

Calculate the value of C for x = 1 year

Substitute x = 1 into \(C = 125000 + 12000x^{5 / 4}\), which results in \(C = 125000 + 12000(1)^{5 / 4} = 137000\)

Calculate the value of C for x = 5 years

Substitute x = 5 into \(C = 125000 + 12000x^{5 / 4}\), which results in \(C = 125000 + 12000(5)^{5 / 4} = 247127.083\)

Calculate the value of C for x = 10 years

Substitute x = 10 into \(C = 125000 + 12000x^{5 / 4}\), which results in \(C = 125000 + 12000(10)^{5 / 4} = 425264.822\)

Key concepts

Understanding Integration

Integration is an essential concept in calculus, focusing on finding the collective accumulation of quantities. Imagine it as adding up all the little parts to get a total amount. It plays a critical role in areas where calculating total size or value from a varying rate of change is needed.

In the scenario of a cost function, integration helps in determining the accumulated cost over time. Here, we're using definite integration, which means evaluating the integral over a specific range from a lower limit to an upper limit. In this case, the range is from 0 to \(x\), the number of years.

The integral \(\int_{0}^{x} t^{1 / 4} dt\) involves a power of \(t\), and the result, derived using the power rule, represents a cumulative value that contributes to the understanding of costs over time.

Applying the Power Rule

The Power Rule is a fundamental tool in calculus for finding the antiderivative of polynomial functions. It's like a magical formula to solve integrals quickly and efficiently. The general form of the Power Rule for integration is: if you have \(x^n\), its integral is \(\frac{1}{n+1}x^{n+1}\).

In the given exercise, we deal with the integral \(\int t^{1 / 4} dt\). Using the Power Rule, the result becomes \(\frac{4}{5}t^{5/4} + C\), where \(C\) is the constant of integration. However, because we are evaluating a definite integral from 0 to \(x\), the \(C\) is not necessary, and the integration yields \(\frac{4}{5}x^{5/4}\).

This antiderivative function is essential for calculating costs over a continuous time frame, showing how the cost grows with respect to the duration \(x\).

Analyzing Cost Function

Cost Function Analysis is crucial for understanding how costs accumulate over time, which aids businesses in financial forecasting and decision-making. In this exercise, the cost function is represented as \(C=5000(25+3 \int_{0}^{x} t^{1 / 4} d t)\).

The purpose of this function is to reveal how total cost depends on the time span \(x\) in years. After solving the integral and substituting it back into the cost function, the equation simplifies to \(C = 125000 + 12000x^{5 / 4}\).

This configuration allows us to see the fixed initial cost and the variable cost that increases with the time power of \(x^{5/4}\).

• For \(x=1\), this results in a total cost of 137,000.
• For \(x=5\), it's 247,127.083.
• For \(x=10\), it's 425,264.822.

This shows that as the time period lengthens, the cost grows significantly, emphasizing the exponential nature of the cost changes over time.