Problem 41 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

The integral evaluates to \(\frac{8}{3}\).

Step 1: Integrate Term-by-Term

We need to find the integral of the polynomial \(1+x+x^2+x^3\) over the interval \([-1, 1]\). Integrate each term separately:\[\int (1) \, dx = x + C_1\]\[\int (x) \, dx = \frac{x^2}{2} + C_2\]\[\int (x^2) \, dx = \frac{x^3}{3} + C_3\]\[\int (x^3) \, dx = \frac{x^4}{4} + C_4\]

Step 2: Combine Integral Expressions

Combine the individual integrals from Step 1 into a single expression:\[\int (1+x+x^2+x^3) \, dx = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + C\]

Step 3: Evaluate Definite Integral Limits

Now, evaluate the definite integral from -1 to 1 by substituting the limits:\[F(1) = 1 + \frac{1^2}{2} + \frac{1^3}{3} + \frac{1^4}{4} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\]\[F(-1) = -1 + \frac{(-1)^2}{2} + \frac{(-1)^3}{3} + \frac{(-1)^4}{4} = -1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4}\]

Step 4: Calculate the Final Result

Subtract \(F(-1)\) from \(F(1)\):\[\begin{align*}F(1) - F(-1) &= \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) - \left(-1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4}\right) \&= \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) + 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} \&= 2 + \frac{2}{3}\end{align*}\]Convert \(\frac{2}{3}\) to decimals or keep as a fraction. Thus, the final result is \(\frac{8}{3}\).

Key Concepts

Polynomial Integration

Polynomial integration is the process of finding the antiderivative of a polynomial function. A polynomial is an expression made up of variables and coefficients, where variables do not appear in the denominator. The degree of the polynomial indicates the highest power of the variable. Common examples are linear, quadratic, and cubic polynomials.

When integrating a polynomial, we apply the power rule, which states that for any real number order exponent, the integral of a term of the form \(x^n\) is \(\frac{x^{n+1}}{n+1} + C\), where \(C\) is the constant of integration.

In our example, the polynomial \(1 + x + x^2 + x^3\) is integrated term-by-term:

• The integral of \(1\) is \(x\).
• The integral of \(x\) is \(\frac{x^2}{2}\).
• The integral of \(x^2\) is \(\frac{x^3}{3}\).
• The integral of \(x^3\) is \(\frac{x^4}{4}\).

After combining each integral, we have the expression \(x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + C\). Remember, the constant \(C\) is eliminated when evaluating definite integrals, making this only relevant for indefinite integrals.

Integral Calculus

Integral calculus is one of the two main branches of calculus, with the other being differential calculus. While differential calculus focuses on rates of change and slopes of curves, integral calculus is concerned with accumulation and the area under curves.

In integral calculus, there are two types of integrals: definite and indefinite. An indefinite integral, often represented as \(\int f(x) \, dx\), results in a function that describes a family of curves, each differing by a constant value \(C\). This is useful for finding general solutions to problems.

On the other hand, a definite integral, represented as \(\int_a^b f(x) \, dx\), provides a number that represents the total accumulation or area between the x-axis, the curve, and the vertical lines at \(x = a\) and \(x = b\). It uses the limits of integration to evaluate the difference between the antiderivative values at these endpoints. Sum (accumulation) of data is an essential part of definite integration.

In the given exercise, we are dealing with a definite integral from \(-1\) to \(1\). This involves substituting these limits into the antiderivative and subtracting the resulting values.

Evaluation of Limits

Evaluation of limits in the context of definite integrals involves calculating the antiderivative function at specified boundary points, known as limits of integration, and determining their difference.

For a definite integral \(\int_a^b f(x) \, dx\), the process generally involves:

• Finding the antiderivative \(F(x)\) of \(f(x)\).
• Substituting the upper limit \(b\) into \(F(x)\).
• Substituting the lower limit \(a\) into \(F(x)\).
• Subtracting the result \(F(a)\) from \(F(b)\).

For our polynomial exercise, we calculated \(F(1)\) as \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\) and \(F(-1)\) as \(-1 + \frac{1}{2} - \frac{1}{3} + \frac{1}{4}\). The definite result came by computing \(F(1) - F(-1)\), which provides the integral’s net area over the given interval \([-1, 1]\). This final subtraction yields the value of \(\frac{8}{3}\), indicating the net accumulation under the polynomial curve.