Problem 2 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{17}{3}\).

Step 1: Break Down the Integral

The integral to be evaluated is given by \[\int_{-1}^{1} \left(x^{2} - 2x + 3\right)\, dx.\]This is a definite integral over the interval \([-1, 1]\) with the integrand being a polynomial function.

Step 2: Integrate the Polynomial

The integral of a polynomial \(ax^n\) is generally \(\frac{a}{n+1}x^{n+1}\), plus the constant of integration which we ignore for definite integrals.- For \(x^2\), we have \(\int x^2 \, dx = \frac{1}{3}x^3\).- For \(-2x\), we have \(\int -2x \, dx = -x^2\).- For the constant term \(3\), \(\int 3 \, dx = 3x\).Therefore, the integral of the entire function is \[F(x) = \frac{1}{3}x^3 - x^2 + 3x.\]

Step 3: Evaluate the Definite Integral

Using the Fundamental Theorem of Calculus, evaluate the antiderivative \(F(x)\) at the bounds of the integral:\[\int_{-1}^{1} \left(x^{2} - 2x + 3\right) \, dx = F(1) - F(-1).\]Substitute \(x = 1\) and \(x = -1\) into the integrated function:- \(F(1) = \frac{1}{3}(1)^3 - (1)^2 + 3(1) = \frac{1}{3} - 1 + 3 = \frac{7}{3}\).- \(F(-1) = \frac{1}{3}(-1)^3 - (-1)^2 + 3(-1) = -\frac{1}{3} - 1 - 3 = -\frac{10}{3}\).Thus, \[F(1) - F(-1) = \frac{7}{3} - (-\frac{10}{3}) = \frac{7}{3} + \frac{10}{3} = \frac{17}{3}.\]

Key Concepts

Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus is a critical concept that links the operations of differentiation and integration. This theorem serves as a bridge between the definite integral and the antiderivative, allowing us to evaluate the area under a curve over a certain interval.

In simple terms, the theorem states if you have a continuous function \( f \) on an interval \([a, b]\), and \( F \) is an antiderivative of \( f \), then:

• \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \)

Here, \( F(b) - F(a) \) represents the net area between the curve of the function and the \( x \)-axis from \( a \) to \( b \).

This theorem is powerful because it provides a straightforward way to compute definite integrals. Instead of adding up infinitely many rectangles under a curve, you can simply evaluate the antiderivative at the upper and lower bounds of the interval.

Polynomial Integration

Polynomial integration focuses on integrating polynomial expressions, which are sums of terms like \( ax^n \).

The process of integration is essentially reversing differentiation, where each term of the polynomial is integrated according to a specific rule:

• For \( x^n \), the integral is \( \frac{1}{n+1}x^{n+1} \).
• Constants like \( 3 \) are simply multiplied by \( x \) in their integral form (e.g., \( \int 3 \, dx = 3x \)).

These rules are applied term by term to find the antiderivative of the entire polynomial. For instance, in the problem \( x^2 - 2x + 3 \), each part—\( x^2 \), \(-2x \), and \( 3 \)—is integrated separately:

- \( \int x^2 \, dx = \frac{1}{3}x^3 \)
- \( \int -2x \, dx = -x^2 \)
- \( \int 3 \, dx = 3x \)

The complete antiderivative for a polynomial expression is formed by summing these individual integrals.

Evaluating Integrals

Evaluating integrals is the process of finding the numerical value of a definite integral over a given interval. This provides the total accumulated value, such as area or displacement, which is quite meaningful in practical applications.

In our case, after computing the antiderivative \( F(x) \), we evaluate it at two points: the upper and lower limits of the integral. These limits are typically denoted as \( a \) and \( b \). For our example:

• Calculate \( F(1) = \frac{1}{3}(1)^3 - 1^2 + 3(1) = \frac{7}{3} \)
• Calculate \( F(-1) = \frac{1}{3}(-1)^3 - (-1)^2 + 3(-1) = -\frac{10}{3} \)

The difference \( F(1) - F(-1) \) yields the exact value of the integral:

• \( \frac{7}{3} - (-\frac{10}{3}) = \frac{7}{3} + \frac{10}{3} = \frac{17}{3} \)

Understanding these steps simplifies the task of evaluating integrals, providing both the method and result in a clean process.