Problem 4 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-26\) $$ \int_{-2}^{2}\left(x^{3}-2 x+3\right) d x $$

Step-by-Step Solution

Answer

The integral evaluates to 12.

Step 1: Set up the integral equation

We are given the integral \(\int_{-2}^{2}(x^3 - 2x + 3) \, dx\). Before solving, we should recognize that this is a definite integral, and it will give us the net area between the curve \(x^3 - 2x + 3\) and the x-axis from \(x = -2\) to \(x = 2\).

Step 2: Find the antiderivative

To evaluate this integral, we first need to find the antiderivative of the integrand, \(x^3 - 2x + 3\). The antiderivative is obtained by integrating term by term:- The antiderivative of \(x^3\) is \(\frac{x^4}{4}\).- The antiderivative of \(-2x\) is \(-x^2\).- The antiderivative of \(3\) is \(3x\).Putting these together, the antiderivative of \(x^3 - 2x + 3\) is \(\frac{x^4}{4} - x^2 + 3x + C\), where \(C\) is the constant of integration.

Step 3: Evaluate the antiderivative at the bounds

Now, evaluate the antiderivative at the upper and lower bounds of the integral: \(x = 2\) and \(x = -2\).Compute at \(x = 2\):\[ \left(\frac{2^4}{4} - 2^2 + 3 \times 2\right) = (4 - 4 + 6) = 6 \]Compute at \(x = -2\):\[ \left(\frac{(-2)^4}{4} - (-2)^2 + 3 \times (-2)\right) = (4 - 4 - 6) = -6 \]

Step 4: Subtract to find the definite integral

To find the definite integral, subtract the value of the antiderivative at the lower bound from the value at the upper bound:\[ 6 - (-6) = 6 + 6 = 12 \]This result represents the net area between the polynomial curve and the x-axis from \(x = -2\) to \(x = 2\).

Key Concepts

Antiderivative

An antiderivative of a function is essentially the reverse of finding the derivative. It's what you might think of as "undoing" the derivative process. In our given problem, the task was to find the antiderivative of the polynomial function \(x^3 - 2x + 3\). This process requires integrating term by term.

• For \(x^3\), the antiderivative is \(\frac{x^4}{4}\).
• For \(-2x\), the antiderivative is \(-x^2\).
• For the constant \(3\), it's simply \(3x\).

So, after finding each term's antiderivative, we combine them to get \(\frac{x^4}{4} - x^2 + 3x + C\), with \(C\) as the constant of integration, representing any constant function, as this constant's derivative is zero.

Net Area

The net area between a curve and the x-axis is a central concept in definite integrals. Unlike ordinary areas, the net area takes into account the parts of the curve that lie below the x-axis, treating these as negative areas. In our problem, the integral of \(x^3 - 2x + 3\) over the interval from \(-2\) to \(2\) represents this net area.
Instead of just computing "additive" areas, you track how much is "above" versus "below" the x-axis. This might involve:

• Summing up positive areas above the x-axis.
• Subtracting negative areas from below it.

Ultimately, the process provides the "net" effect over the specific bounds, which was calculated as \(12\) in this example, meaning there was more positive area than negative.

Polynomial Integration

Polynomial integration is a method applied specifically to polynomials, breaking them down into individual terms and integrating each term separately. It's straightforward due to its simplicity and reliability. In our exercise with the polynomial \(x^3 - 2x + 3\), the polynomial's nature made integration a matter of integrating each power of \(x\). This method simplifies because:

• For any term like \(x^n\), the antiderivative is \(\frac{x^{n+1}}{n+1}\).
• Constant terms translate into linear terms in integration.

Breaking it down made seeing the distinct contributions of different powers of \(x\) easy. This is why polynomial integration is foundational in calculus.

Bound Evaluation

Bound evaluation is the final stage in computing a definite integral and takes the problem-solving process from an abstract algebraic form to concrete numbers. After finding the antiderivative, you need to calculate it at specific points: the bounds of integration.For our exercise, we first evaluate the antiderivative at the upper bound \(x = 2\): \[ \left(\frac{2^4}{4} - 2^2 + 3 \times 2\right) = 6 \]Then at the lower bound \(x = -2\):\[ \left(\frac{(-2)^4}{4} - (-2)^2 + 3 \times (-2)\right) = -6 \]Finally, subtracting these values, \(6 - (-6)\) gives us the total result of \(12\). This indicates the "net area" from \(-2\) to \(2\). Bound evaluation helps in translating the abstract concept into numerical results, giving the final answer its concrete form.