Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-14, use the Second Fundamental Theorem of Calculus to evaluate each definite integral. \(\int_{0}^{2} x^{3} d x\)

Step-by-Step Solution

Answer

The integral \( \int_{0}^{2} x^{3} \, dx \) evaluates to 4.

Step 1: Understand the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \( F \) is an antiderivative of \( f \) on an interval \([a, b]\), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). This means we need to find an antiderivative of the function to evaluate the definite integral.

Step 2: Find the Antiderivative of the Function

The function given is \( x^{3} \). To find an antiderivative, we use the power rule of integration: \( \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \), where \( C \) is a constant. For \( x^{3} \), apply the rule: \( \int x^{3} \, dx = \frac{x^{4}}{4} + C \). So, the antiderivative is \( F(x) = \frac{x^{4}}{4} \).

Step 3: Evaluate the Antiderivative at the Bounds

Using the antiderivative found, evaluate it at the upper and lower bounds of the integral: \( F(2) = \frac{2^{4}}{4} = \frac{16}{4} = 4 \) and \( F(0) = \frac{0^{4}}{4} = 0 \).

Step 4: Apply the Second Fundamental Theorem of Calculus

Subtract the value of the antiderivative at the lower bound from the value at the upper bound: \( F(2) - F(0) = 4 - 0 = 4 \). Therefore, the value of the definite integral \( \int_{0}^{2} x^{3} \, dx \) is 4.

Key Concepts

Definite Integral

When you think of a definite integral, think of it as a way to calculate the total area under a curve between two points on the x-axis. In other words, it's about measuring accumulated change. A definite integral is represented by the integral sign with limits of integration. For instance, in the integral \( \int_{0}^{2} x^{3} \, dx \), the limits are 0 and 2.

The process involves several steps:

• Identify the function you want to integrate, which is called the integrand, like \( x^{3} \) in this case.
• Find the antiderivative of this function. This is crucial because the definite integral uses this to compute the area.
• Evaluate the antiderivative at the upper limit and at the lower limit, then take the difference.

The result gives the net area under the curve from the lower limit to the upper limit.

Antiderivative

The antiderivative is the reverse process of differentiation. If differentiation helps you find the slope of a function, finding the antiderivative essentially retraces those steps to determine the original function. For instance, if you differentiate \( F(x) = \frac{x^{4}}{4} \), you end up with \( f(x) = x^{3} \).

Knowing the antiderivative is crucial in solving definite integrals. Here's why:

• You need the antiderivative to apply the Second Fundamental Theorem of Calculus effectively.
• For polynomials like \( x^{3} \), we typically use the power rule of integration.
• Once you have the antiderivative, you can evaluate it at specific bounds to find the area under the curve.

The calculated difference between these evaluations gives you the definite integral's value.

Power Rule of Integration

The power rule of integration is one of the most used rules when finding the antiderivatives of functions. It states that to integrate a function of the form \( x^{n} \), you need to add 1 to the exponent and then divide by the new exponent, which converts \( \int x^{n} \, dx \) to \( \frac{x^{n+1}}{n+1} + C \). This rule simplifies the task of finding antiderivatives by providing a straightforward formula for polynomials.

Using this rule:

• For \( x^{3} \), increase the exponent to 4, making it \( x^{4} \).
• Divide by the new exponent, resulting in \( \frac{x^{4}}{4} \).
• Always remember to add a constant \( C \) for indefinite integrals, though you typically don't need it for definite integrals.

This rule is especially helpful because it turns what might look like a daunting integral into a simple calculation. For our example, the power rule lets us swiftly find that the antiderivative of \( x^{3} \) is \( \frac{x^{4}}{4} \).