Problem 18 - Biocalculus Calculus for the Life Sciences

Question

$$\lim _{n \rightarrow \infty} \sum_{i=1}^{n}\left[4-3\left(x_{i}^{*}\right)^{2}+6\left(x_{i}^{*}\right)^{5}\right] \Delta x, \quad[0,2]$$

Step-by-Step Solution

Answer

The value of the limit of the Riemann sum is 64.

Step 1: Interpret the Exercise as a Riemann Sum

The given expression represents a Riemann sum, which involves taking the limit as the number of subintervals approaches infinity. This Riemann sum approximates the integral of a function over an interval. The sum \(\sum_{i=1}^{n}\left[4-3\left(x_{i}^{*}\right)^{2}+6\left(x_{i}^{*}\right)^{5}\right]\) with \(\Delta x\) indicates the function we are integrating is \(f(x) = 4 - 3x^2 + 6x^5\) over the interval \([0, 2]\).

Step 2: Set Up the Integral

To evaluate the limit of the Riemann sum, set it up as a definite integral. Convert the Riemann sum \(\lim_{n \rightarrow \infty} \sum_{i=1}^{n} f(x_i^*) \Delta x\) to the integral form: \[\int_{0}^{2} (4 - 3x^2 + 6x^5)\, dx.\]

Step 3: Integrate the Function

Now, integrate each term of the function individually:- The integral of \(4\) is \(4x\).- The integral of \(-3x^2\) is \(-x^3\) (as \(-3\times\frac{1}{3}x^3\)).- The integral of \(6x^5\) is \(\frac{6}{6}x^6\) or \(x^6\).Thus, the integral is:\[4x - x^3 + x^6.\]

Step 4: Evaluate the Definite Integral

Evaluate the integral \(4x - x^3 + x^6\) from \(0\) to \(2\):1. Substitute \(2\) into the integrated function: \[ 4(2) - (2)^3 + (2)^6 = 8 - 8 + 64 = 64. \]2. Substitute \(0\) into the integrated function: \[ 4(0) - (0)^3 + (0)^6 = 0. \]3. Take the difference: \[ 64 - 0 = 64. \]

Step 5: Conclusion

Hence, the value of the limit of the Riemann sum, which is the integral of the function over the interval \([0, 2]\), is 64.

Key Concepts

Definite Integral

A definite integral is a key concept in calculus. It allows us to find the exact accumulation of quantities, such as areas under curves or total displacement.

When we talk about the definite integral of a function, we are essentially looking to compute the "net area" under the curve represented by the function from one point to another.

• In the given problem, the function is integrated over the interval from 0 to 2.
• This means we are considering the total impact of the function between these two points.

Specifically, the problem involves the transition from a Riemann sum to a definite integral. This tells us that the sum, as the number of subintervals approaches infinity, is equivalent to evaluating the definite integral.
Understanding this provides a deeper grasp of how calculus can be used to measure and calculate continuous changes and accumulations. Furthermore, the area calculated (like in this exercise) can represent various real-world quantities depending on context.

Function Integration

Function integration involves finding an antiderivative or an integral of a given function. During this process, we determine a new function whose derivative matches the original function.

The step-by-step solution breaks down the integration of each term for the function:

• The integral of a constant like 4 results in terms such as 4x, because integrating a constant over a variable adds one to the exponent and divides it by the new exponent.
• For powers of x, such as \(-3x^2\), we add 1 to the exponent and divide by the new power, hence obtaining \(-x^3\).
• Similarly, for \(6x^5\), the process accounts for each term separately, resulting in complete integration as \(x^6\).

Each small integration step is significant so that, regardless of the function's complexity, the integration rules let us handle one term at a time. This approach simplifies evaluating functions, making it more comprehensible.

Limit Evaluation

Limit evaluation is an integral part of understanding the behavior of functions, particularly as they extend towards infinity.

In the context of our exercise, the limit \(\lim _{n \rightarrow \infty}\) signifies evaluating the sum as the number of terms n approaches infinity, essentially transforming it into the integral. This step is vital because a Riemann sum approximates the integral, becoming the precise value once limits are applied.

• By calculating the limit of the sum, we transition from an approximation to an exact area or value.
• This concept of limits bridges the gap between discrete and continuous mathematics, showing how sums of many tiny areas (or values) can translate to a well-defined total.

Understanding limit evaluation in limits helps in grasping the perfect transition from a basic sum to a complete integral, as seen in the original exercise. This transforms large sums into manageable, useful results, like finding the exact answer of 64 for the integral.