Problem 19 - Calculus: Early Transcendentals

Question

Evaluate the integral. \( \displaystyle \int^3_1 (x^2 + 2x - 4) \,dx \)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{26}{3} \).

Step 1: Identify the Integral

We need to evaluate the definite integral of the function \( f(x) = x^2 + 2x - 4 \) from \( x = 1 \) to \( x = 3 \). This can be written as: \[ \int_{1}^{3} (x^2 + 2x - 4) \, dx \]

Step 2: Antiderivative Calculation

Find the antiderivative of the integrand \( f(x) = x^2 + 2x - 4 \). The antiderivative is calculated as follows:\( \int (x^2) \, dx = \frac{x^3}{3} \)\( \int (2x) \, dx = x^2 \)\( \int (-4) \, dx = -4x \)Thus, the antiderivative \( F(x) \) is: \[ F(x) = \frac{x^3}{3} + x^2 - 4x \]

Step 3: Apply the Fundamental Theorem of Calculus

According to the Fundamental Theorem of Calculus, \( \int_{a}^{b} f(x) \,dx = F(b) - F(a) \) where \( F(x) \) is the antiderivative of \( f(x) \). Here, evaluate \( F(x) \) at \( x = 3 \) and \( x = 1 \) and subtract:\[ F(3) = \frac{3^3}{3} + 3^2 - 4(3) = 9 + 9 - 12 = 6 \]\[ F(1) = \frac{1^3}{3} + 1^2 - 4(1) = \frac{1}{3} + 1 - 4 = \frac{-8}{3} \]

Step 4: Calculate the Definite Integral

Subtract the values obtained: \( F(3) - F(1) \).\[ F(3) - F(1) = 6 - \left( \frac{-8}{3} \right) = 6 + \frac{8}{3} \]Convert 6 to fractions: \( 6 = \frac{18}{3} \)\[ \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \]Thus, the value of the integral is \( \frac{26}{3} \).

Key Concepts

Antiderivative

The process of finding the antiderivative is like turning the clock backwards in calculus. When you have a function that resulted from an operation like differentiation, you seek what original function provided this result. This operation is known as finding the antiderivative or the indefinite integral.

For instance, when we look at the function \( f(x) = x^2 + 2x - 4 \), we find the antiderivative by evaluating each term.

• The antiderivative of \( x^2 \) is \( \frac{x^3}{3} \).
• The antiderivative of \( 2x \) is \( x^2 \).
• Lastly, for the constant -4, the antiderivative is \( -4x \).

By adding these results together, we get the antiderivative \( F(x) = \frac{x^3}{3} + x^2 - 4x \).

Understanding this process helps us build the foundation for solving definite integrals, as the antiderivative plays a crucial role in the next step, which is employing the fundamental theorem of calculus.

Fundamental Theorem of Calculus

The fundamental theorem of calculus bridges the world of integration and differentiation. It gives us a powerful tool to evaluate definite integrals. This theorem tells us that if \( F(x) \) is the antiderivative of \( f(x) \), then the definite integral of \( f(x) \) from \( a \) to \( b \) equals \( F(b) - F(a) \).

In our exercise, this means using the antiderivative \( F(x) = \frac{x^3}{3} + x^2 - 4x \) to find:

• \( F(3) = 9 + 9 - 12 = 6 \)
• \( F(1) = \frac{1}{3} + 1 - 4 = \frac{-8}{3} \)

Then we simply subtract: \( F(3) - F(1) = 6 - \frac{-8}{3} = 6 + \frac{8}{3} \).

This theorem elegantly unifies differentiation and integration, providing an efficient way to find the area under a curve between two points, without requiring the arduous process of adding infinitesimal slices together.

Integral Evaluation

Evaluating an integral, like the one given in the exercise, brings together all we've learned about the antiderivative and the fundamental theorem of calculus. The purpose is to find the overall value from a starting point \( a \) to an ending point \( b \) for a given function.

For this integral, the steps are straightforward once we have found the antiderivative and understood the theorem.

• First, compute \( F(b) \) and \( F(a) \) using the antiderivative \( F(x) \).
• In our example: \( F(3) = 6 \) and \( F(1) = \frac{-8}{3} \).
• Then, subtract these results: \( F(3) - F(1) = 6 + \frac{8}{3} \).
• Convert and combine into a single fraction: \( \frac{18}{3} + \frac{8}{3} = \frac{26}{3} \).

Thus, the definite integral evaluates to \( \frac{26}{3} \).

This process confirms the area under the curve of \( f(x) = x^2 + 2x - 4 \) from \( x = 1 \) to \( x = 3 \) and demonstrates how calculus tools provide insightful solutions.