Problem 4 - Calculus

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 value of the definite integral is 22.

Step 1: Identify the Function to Integrate

The function inside the integral is \( f(x) = 4x^3 + 7 \). We'll use this function to find the antiderivative.

Step 2: Find the Antiderivative

To apply the Second Fundamental Theorem of Calculus, we need to find the antiderivative of \( f(x) \). The power rule gives us \( \int 4x^3 \, dx = x^4 \) and \( \int 7 \, dx = 7x \). Thus, the antiderivative is \( F(x) = x^4 + 7x + C \), where \( C \) is the constant of integration. However, for definite integrals, we don't need to consider \( C \).

Step 3: Apply the Second Fundamental Theorem of Calculus

The Second Fundamental Theorem of Calculus states that if \( F(x) \) is the antiderivative of \( f(x) \), then \( \int_{a}^{b} f(x) \, dx = F(b) - F(a) \). In this case, \( F(x) = x^4 + 7x \).

Step 4: Evaluate the Antiderivative at the Upper Limit

Substitute \( x = 2 \) into \( F(x) \):\[F(2) = (2)^4 + 7(2) = 16 + 14 = 30.\]

Step 5: Evaluate the Antiderivative at the Lower Limit

Substitute \( x = 1 \) into \( F(x) \):\[F(1) = (1)^4 + 7(1) = 1 + 7 = 8.\]

Step 6: Subtract to Find the Definite Integral

Subtract the result from Step 5 from the result from Step 4 to find the value of the definite integral:\[F(2) - F(1) = 30 - 8 = 22.\]

Key Concepts

Definite Integral

A definite integral is a way to calculate the area under a curve from one point to another on the x-axis. In this exercise, we're looking at the definite integral \( \int_{1}^{2} (4x^3 + 7) \, dx \). This tells us that we're interested in finding the area under the curve of the function \( 4x^3 + 7 \) between \( x = 1 \) and \( x = 2 \).Unlike indefinite integrals, which give a family of functions with a constant \( C \), definite integrals provide a specific numerical value. This is because the limits of integration—lower limit \( a \) and upper limit \( b \)—are specified, making it possible to use the Second Fundamental Theorem of Calculus to directly find the difference \( F(b) - F(a) \). This approach helps in evaluating the precise area associated with the function over the specified interval. Definite integrals are essential in real-world applications, such as calculating distances, areas, and in physics for finding quantities like work and energy.

Antiderivative

An antiderivative is a function whose derivative gives back to the original function. In simpler terms, it's the reverse process of differentiation. In this exercise, we need to find the antiderivative of \( f(x) = 4x^3 + 7 \).To find the antiderivative of polynomial functions like \( 4x^3 \) and constants like \( 7 \), we add one to the power of \( x \) and then divide by the new power for each term:

• For \( 4x^3 \), the antiderivative is \( \frac{4}{4}x^4 = x^4 \).
• For \( 7 \), the antiderivative is simply \( 7x \) because the integral of a constant \( c \) is \( cx \).

Thus, the antiderivative for the given function is \( F(x) = x^4 + 7x \). Notice we omit the constant of integration \( C \) because it's irrelevant for definite integrals, as it cancels out when evaluating \( F(b) - F(a) \).

Power Rule

The power rule is a fundamental technique used to find the antiderivative of terms where the variable is raised to a power. It simplifies the process of integration, especially for polynomial expressions. The rule states that for any function \( x^n \), where \( n eq -1 \), the antiderivative is given by:\[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \]In the context of our exercise, the power rule helps us solve the integral of \( 4x^3 \). If we apply the power rule:

• Add 1 to the exponent (3 + 1) to get 4.
• Divide by the new exponent: \( \frac{4}{4}x^4 = x^4 \).

For constant terms like \( 7 \), we remember that the integral simply becomes \( 7x \) since there's no variable to raise to any particular power. The power rule is a handy shortcut instead of manually reversing differentiation, making the integration process much more efficient for students.