Problem 447 - Calculus Volume 1

Question

Evaluate the following integrals. $$\int_{-1}^{1}\left(x^{3}-2 x^{2}+4 x\right) d x$$

Step-by-Step Solution

Answer

The integral evaluates to \(-\frac{4}{3}\).

Step 1: Integrate the Function

The function given is \( f(x) = x^3 - 2x^2 + 4x \). In this step, integrate each term separately:- The integral of \( x^3 \) is \( \frac{x^4}{4} \).- The integral of \( -2x^2 \) is \( \frac{-2x^3}{3} \).- The integral of \( 4x \) is \( 2x^2 \).So, the indefinite integral \( F(x) \) is:\[ F(x) = \frac{x^4}{4} - \frac{2x^3}{3} + 2x^2 \]

Step 2: Apply the Definite Integral Limits

Now evaluate the definite integral using the limits from \( -1 \) to \( 1 \).- Calculate \( F(1) \): \[ F(1) = \frac{1^4}{4} - \frac{2(1)^3}{3} + 2(1)^2 = \frac{1}{4} - \frac{2}{3} + 2 \] Simplify: \[ F(1) = \frac{1}{4} - \frac{8}{12} + 2 = \frac{1}{4} - \frac{2}{3} + 2 \] Use a common denominator (12): \[ F(1) = \frac{3}{12} - \frac{8}{12} + \frac{24}{12} = \frac{19}{12} \]

Step 3: Evaluate at the Lower Limit

Evaluate the function at the lower limit (-1):- Calculate \( F(-1) \): \[ F(-1) = \frac{(-1)^4}{4} - \frac{2(-1)^3}{3} + 2(-1)^2 = \frac{1}{4} + \frac{2}{3} + 2 \] Simplify: \[ F(-1) = \frac{1}{4} + \frac{2}{3} + 2 \] Use a common denominator (12): \[ F(-1) = \frac{3}{12} + \frac{8}{12} + \frac{24}{12} = \frac{35}{12} \]

Step 4: Subtract the Evaluations

The value of the definite integral is the difference between \( F(1) \) and \( F(-1) \):- Calculate: \[ F(1) - F(-1) = \frac{19}{12} - \frac{35}{12} = -\frac{16}{12} \] Simplify: \[ -\frac{16}{12} = -\frac{4}{3} \]

Key Concepts

Indefinite Integral

Before delving into the calculation of a definite integral, it's important to understand the concept of an indefinite integral. Indefinite integrals, also known as antiderivatives, are functions that reverse the process of differentiation.
Given a function, the goal of taking an indefinite integral is to find a new function whose derivative is the original function. This is denoted as \( \int f(x) \, dx \), and it includes a constant of integration, typically represented as \( C \), because differentiation of constants yields zero.
For the polynomial \( f(x) = x^3 - 2x^2 + 4x \), its indefinite integral is calculated by integrating each term separately:

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

Thus, the indefinite integral is: \[F(x) = \frac{x^4}{4} - \frac{2x^3}{3} + 2x^2 + C \]In the context of definite integration, the constant \( C \) cancels out, as it would eventually be subtracted.

Polynomial Integration

Polynomial integration is the process of integrating functions that are composed of terms involving powers of the variable, like \( ax^n \). This process is typically straightforward, as each term is integrated separately following a general rule, which is:
When integrating \( x^n \), if \( n eq -1 \), the formula used is:\[\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\]In our example, the function \( x^3 - 2x^2 + 4x \) was dealt with term by term:

• For \( x^3 \), we increased the exponent by 1 to obtain \( \frac{x^4}{4} \).
• For \( -2x^2 \), we obtained \( \frac{-2x^3}{3} \).
• For \( 4x \), we got \( 2x^2 \).

Polyintegration is useful because many practical problems can be simplified into these polynomial forms, making them much easier to solve analytically.
Remember, capturing the constant of integration \( C \) is key for indefinite integrals, but it gets eliminated when dealing with definite integrals.

Limits of Integration

When calculating definite integrals, you must evaluate the antiderivative at specified limits. This process uses the fundamental theorem of calculus, which states that if \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral from \( a \) to \( b \) of \( f(x) \) is:\[\int_{a}^{b} f(x) \, dx = F(b) - F(a)\]In the given problem, our limits are \( -1 \) and \( 1 \). After finding the indefinite integral, which is \( F(x) \), you place the upper limit 1 and the lower limit -1 into the function:

• Calculate \( F(1) \).
• Calculate \( F(-1) \).

Finally, you find the definite integral by subtracting these two results:\[F(1) - F(-1)\]This gives you the accumulated value of the function from \( -1 \) to \( 1 \). In this exercise, this subtraction resulted in the value \( -\frac{4}{3} \), demonstrating how areas calculated in definite integration can be positive or negative based on the function's behavior across the interval. The limits define the bounds of integration and ensure the function's complete behavior is considered within these bounds.