Problem 3 - Biocalculus Calculus for the Life Sciences

Question

Evaluate the integral. \(\int_{0}^{2}\left(x^{4}-\frac{3}{4} x^{2}+\frac{2}{3} x-1\right) d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{56}{15} \).

Step 1: Identify the Integral

The integral to evaluate is \( \int_{0}^{2} \left(x^{4} - \frac{3}{4}x^{2} + \frac{2}{3}x - 1\right) \, dx \). We need to find the antiderivative of the function within the integral and then apply the definite integral limits.

Step 2: Find the Antiderivative

To find the antiderivative, integrate each term separately:- The antiderivative of \(x^4\) is \(\frac{x^5}{5}\).- The antiderivative of \(-\frac{3}{4}x^2\) is \(-\frac{3}{4} \cdot \frac{x^3}{3} = -\frac{1}{4}x^3\).- The antiderivative of \(\frac{2}{3}x\) is \(\frac{2}{3} \cdot \frac{x^2}{2} = \frac{1}{3}x^2\).- The antiderivative of \(-1\) is \(-x\).Thus, the antiderivative is:\[ F(x) = \frac{x^5}{5} - \frac{1}{4}x^3 + \frac{1}{3}x^2 - x \].

Step 3: Evaluate the Antiderivative at the Bounds

Now, substitute the bounds 0 and 2 into the antiderivative function \( F(x) \):\[ F(2) = \frac{2^5}{5} - \frac{1}{4}(2^3) + \frac{1}{3}(2^2) - 2 \].Simplify:- \(\frac{2^5}{5} = \frac{32}{5}\)- \(-\frac{1}{4}(2^3) = -\frac{1}{4} \cdot 8 = -2\)- \(\frac{1}{3}(2^2) = \frac{1}{3} \cdot 4 = \frac{4}{3}\)- \(-2 = -2\)So, \[ F(2) = \frac{32}{5} - 2 + \frac{4}{3} - 2 \].

Step 4: Evaluate Simplified Expression at Upper Limit

Evaluate the simplified expression from the previous step:\[ F(2) = \frac{32}{5} - 2 + \frac{4}{3} - 2 \]Convert all terms to have a common denominator (15):- \(\frac{32}{5} = \frac{96}{15}\)- \(-2 = -\frac{30}{15}\)- \(\frac{4}{3} = \frac{20}{15}\)- \(-2 = -\frac{30}{15}\)Summing these fractions gives:\[ \frac{96}{15} - \frac{30}{15} + \frac{20}{15} - \frac{30}{15} = \frac{56}{15} \].

Step 5: Evaluate the Antiderivative at Lower Limit

Now, evaluate \( F(x) \) at the lower limit, 0. Since all terms in \( F(x) \) contain \( x \), \( F(0) = 0 \).

Step 6: Compute the Definite Integral

The definite integral is the difference between \( F(2) \) and \( F(0) \):\[ F(2) - F(0) = \frac{56}{15} - 0 = \frac{56}{15} \].

Step 7: Provide Final Answer

Thus, the integral \( \int_{0}^{2} \left(x^{4} - \frac{3}{4} x^{2} + \frac{2}{3} x - 1\right) \, dx \) evaluates to \( \frac{56}{15} \).

Key Concepts

Antiderivative

An antiderivative, often termed an indefinite integral, is a function whose derivative is equal to the original function we want to integrate. When we find the antiderivative, we're essentially reversing the process of differentiation.

In our exercise, we need to integrate the polynomial expression \[ x^4 - \frac{3}{4}x^2 + \frac{2}{3}x - 1 . \]The antiderivative is a new function such that if we differentiate it, we get back our original function in the integral.

To find the antiderivative, integrate each term separately. For example:

• The function \( x^4 \) becomes \( \frac{x^5}{5} \) because integrating \( x^n \) gives \( \frac{x^{n+1}}{n+1} \).
• The antiderivative of a constant like \(-1\) is straightforward: it integrates to \(-x\).

This reversal from derivative back to original function is crucial for solving integrals, especially in calculus.

Polynomial Integration

Polynomial integration involves finding the antiderivative of a polynomial expression. Polynomials are easy to integrate because you apply the power rule for integration to each term.

Consider the polynomial in our exercise\[ x^4 - \frac{3}{4}x^2 + \frac{2}{3}x - 1 . \]To integrate:

• Apply the power rule: add one to the power of \( x \) and divide by the new power.
• For instance, \( x^4 \) becomes \( \frac{x^5}{5} \) and \( x^2 \) becomes \( \frac{x^3}{3} \).
• Coefficients outside the terms, like \(-\frac{3}{4}\), stay with their respective terms.

After integrating each term, you get the antiderivative function as a sum of all integrated terms. This polynomial integration simplifies our task of evaluating definite integrals.

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus connects differentiation and integration, showing that they are essentially inverse processes. It consists of two profound parts:

• The first part allows us to evaluate a definite integral using the antiderivative.
• The second part tells us that the derivative of an integral function yields the original function within the integral.

For a definite integral, like in our exercise, we use the antiderivative we calculated:Find \( F(x) \), the antiderivative of the function, and evaluate it at the upper and lower limits.

• First, compute \( F(2) \).
• Then, compute \( F(0) \).
• The definite integral is given by \( F(2) - F(0) \).

In our case, this results in evaluating the difference as \[ \int_{0}^{2} \left(x^{4} - \frac{3}{4}x^{2} + \frac{2}{3}x - 1\right) dx = \frac{56}{15} . \]This theorem elegantly demonstrates how differentiation and integration complement each other in calculus.