Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

use the Second Fundamental Theorem of Calculus to evaluate each definite integral. $$ \int_{1}^{2}\left(4 x^{3}+7\right) d x $$

Step-by-Step Solution

Answer

The definite integral evaluates to 22.

Step 1: Identify the Problem

We need to evaluate the definite integral \( \int_{1}^{2}(4x^{3} + 7) \, dx \) using the Second Fundamental Theorem of Calculus, which states that if \( F \) is an antiderivative of \( f \) on an interval \( [a, b] \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \).

Step 2: Find the Antiderivative

Start by finding the antiderivative of the integrand \( 4x^{3} + 7 \). The antiderivative of \( 4x^{3} \) is \( \frac{4}{4}x^{4} = x^{4} \), and the antiderivative of \( 7 \) is \( 7x \). Therefore, the antiderivative \( F(x) \) is \( x^{4} + 7x \).

Step 3: Apply the Limits of Integration

Using the Second Fundamental Theorem of Calculus, evaluate \( F(2) \) and \( F(1) \). Compute \( F(2) = 2^{4} + 7 \times 2 = 16 + 14 = 30 \). Similarly, compute \( F(1) = 1^{4} + 7 \times 1 = 1 + 7 = 8 \).

Step 4: Calculate the Definitive Answer

Find the value of the definite integral by subtracting \( F(1) \) from \( F(2) \). The calculation is \( 30 - 8 = 22 \). Hence, the definite integral evaluates to 22.

Key Concepts

Definite Integral

A definite integral represents the exact area under a curve, between two points along the x-axis. In simpler terms, it's about finding out how much 'space' is enclosed by a curve and the x-axis from one point to another.

Here's how it works:

• We take a function, like the one in our example, which is \(4x^3 + 7\).
• We then look at the range over which we want to find the area, which in this case is from \(x = 1\) to \(x = 2\).
• By calculating the definite integral, we find out exactly how much area exists under the curve between these two x-values.

The beauty of definite integrals comes into play with the help of the Second Fundamental Theorem of Calculus. This theorem links the concept of the antiderivative with the area under the curve, allowing us to get the result in an effective way.

Antiderivative

An antiderivative is a crucial part of solving integrals. It’s essentially the reverse of differentiation. While differentiation gives us the rate of change, finding an antiderivative allows us to recover the original function from that rate.

To find an antiderivative, you need a function that, when differentiated, gives you the integrand (the function you're integrating). In our example:

• The original function is \(4x^3 + 7\).
• The antiderivative for \(4x^3\) is \(x^4\) since the derivative of \(x^4\) is \(4x^3\).
• Similarly, the antiderivative of the constant \(7\) is \(7x\), because the derivative of \(7x\) is \(7\).

Thus, the antiderivative we use here is \(F(x) = x^4 + 7x\). This function helps us evaluate the definite integral effectively, bringing us to the final answer through the Fundamental Theorem of Calculus.

Integral Calculus

Integral calculus forms one half of calculus. It revolves around two main ideas: finding areas (definite integrals) and recovering original functions from their rates of change (antiderivatives). This branch is vital for a deeper understanding of mathematical and natural phenomena.

Integral calculus is powerful with applications in:

• Physics: Calculating the consequences of variable forces, or understanding the dynamics of systems.
• Engineering: Optimizing processes and understanding fluid dynamics or electrical currents.
• Economics: Estimating economic activities such as consumption and growth over time.

In our specific problem, integral calculus allowed us to take a snapshot of the entire picture. While the integration process can seem complex, the Fundamental Theorems of Calculus simplify finding both definite integrals and antiderivatives, making diverse applications possible.