Integral Approach to Cumulative Growth

What an accumulation problem is

Accumulation problems are solved by calculating the total build-up of a quantity over time or space. This involves summing small incremental changes, like how rain accumulates in a bucket. Using calculus, we add up these tiny amounts continuously, giving us the total quantity accumulated, whether it’s distance, growth, volume, or other measurable quantities.

Setting up and evaluating the integral

To solve an accumulation problem mathematically, you integrate a rate of change function. This involves finding the integral of a function \( f(x) \), which represents the rate at which a quantity is changing. The integral \( \int f(x) \, dx \) calculates the total accumulation of the quantity described by \( f(x) \) over its domain. This process essentially sums up all the infinitesimal changes in the quantity, providing the total amount accumulated.

Let’s find the total accumulation of the function \( f(t) = 3t^3 – 2t \) from \( t = 0 \) to \( t = 4 \):

1. Identify the Function and Interval:

• The function given is \( f(t) = 3t^3 – 2t \).
• The interval for accumulation is from \( t = 0 \) to \( t = 4 \).

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2. Set Up the Integral:

• The integral to calculate the total accumulation is:
  \( \int_{0}^{4} (3t^3 – 2t) \, dt \)

3. Calculate the Integral:

• Find the antiderivative of \( 3t^3 – 2t \).
• The antiderivative of \( 3t^3 \) is \( \frac{3}{4}t^4 \), and the antiderivative of \( -2t \) is \( -t^2 \).
• So, the antiderivative of \( 3t^3 – 2t \) is \( \frac{3}{4}t^4 – t^2 \).

4. Evaluate the Integral:

• Substitute \( t = 4 \) and \( t = 0 \) into \( \frac{3}{4}t^4 – t^2 \) and calculate the difference:
  \( \left( \frac{3}{4}(4)^4 – (4)^2 \right) – \left( \frac{3}{4}(0)^4 – (0)^2 \right) \)
• Simplify and calculate:
  \( \frac{3}{4} \cdot 256 – 16 \)
  \( 192 – 16 = 176 \)
• The result, \( 176 \), represents the total accumulation of the quantity described by \( f(t) = 3t^3 – 2t \) over the interval from \( t = 0 \) to \( t = 4 \).