Problem 15 - Applied Calculus

Question

Using the Fundamental Theorem, evaluate the definite integrals in Problems \(1-20\) exactly. $$\int_{2}^{5}\left(x^{3}-\pi x^{2}\right) d x$$

Step-by-Step Solution

Answer

The definite integral evaluates to \( \frac{609}{4} - \frac{117\pi}{3} \).

Step 1: Identify the Function to Integrate

The function we need to integrate is given by \( f(x) = x^3 - \pi x^2 \). Our goal is to find the integral over the interval from \( x = 2 \) to \( x = 5 \).

Step 2: Find the Indefinite Integral

To evaluate the definite integral, we must first compute the indefinite integral \( \int \left(x^3 - \pi x^2\right) \, dx \). The indefinite integral of \( x^3 \) is \( \frac{x^4}{4} \) and the indefinite integral of \( \pi x^2 \) is \( \frac{\pi x^3}{3} \). Therefore, the indefinite integral is \( F(x) = \frac{x^4}{4} - \frac{\pi x^3}{3} + C \), where \( C \) is the constant of integration.

Step 3: Apply the Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus states that to evaluate the definite integral \( \int_{a}^{b} f(x) \, dx \), where \( F(x) \) is an antiderivative of \( f(x) \), the result is \( F(b) - F(a) \). In this case, since \( F(x) = \frac{x^4}{4} - \frac{\pi x^3}{3} \), we need to compute \( F(5) - F(2) \).

Step 4: Calculate \( F(5) \) and \( F(2) \)

Substitute \( x = 5 \) into \( F(x) \): \[ F(5) = \frac{5^4}{4} - \frac{\pi \cdot 5^3}{3} = \frac{625}{4} - \frac{125\pi}{3} \].Now substitute \( x = 2 \) into \( F(x) \):\[ F(2) = \frac{2^4}{4} - \frac{\pi \cdot 2^3}{3} = \frac{16}{4} - \frac{8\pi}{3} = 4 - \frac{8\pi}{3} \].

Step 5: Find the Definite Integral Value

Compute the definite integral by calculating \( F(5) - F(2) \):\[\int_{2}^{5} (x^3 - \pi x^2) \, dx = \left( \frac{625}{4} - \frac{125\pi}{3} \right) - \left( 4 - \frac{8\pi}{3} \right) \].Simplifying this expression yields:\[\left( \frac{625}{4} - 4 \right) - \left( \frac{125\pi}{3} - \frac{8\pi}{3} \right) = \frac{609}{4} - \frac{117\pi}{3} \].

Key Concepts

Definite Integral

A definite integral is a number that represents the area under a curve between two points on the x-axis. It's like taking a slice between these points and measuring the area in between. The definite integral is one of the key concepts in calculus and is calculated using the Fundamental Theorem of Calculus.

This calculation involves evaluating an antiderivative, which we'll discuss in another section, at the upper and lower limits of integration. Here, our exercise focuses on the function \( f(x) = x^3 - \pi x^2 \) and it asks us to find the area under the curve from \( x = 2 \) to \( x = 5 \).

When computing a definite integral, you apply the upper and lower limits to the antiderivative function, then subtract the lower limit result from the upper limit result. In our example, the definite integral of \( x^3 - \pi x^2 \) from 2 to 5 required calculating \( F(5) \) and \( F(2) \), which we then subtracted to find the result.

The fundamental part of solving a definite integral is evaluating this subtraction because it gives us the exact numerical answer. This is different from an indefinite integral, where the result includes an arbitrary constant rather than a specific number.

Indefinite Integral

An indefinite integral is essentially finding the antiderivative of a given function. This gives you a new function that represents a family of all possible areas under the original curve, but without specific numerical bounds. It's represented with a \( + C \) at the end of the function, where \( C \) is any constant.

For our function \( f(x) = x^3 - \pi x^2 \), the indefinite integral would be calculated as follows:

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

Therefore, the indefinite integral is \( F(x) = \frac{x^4}{4} - \frac{\pi x^3}{3} + C \).

This expression shows us a general shape of the antiderivative. It doesn't focus on the values at specific points but instead gives us a broad solution that can be adjusted (by changing \( C \)) to meet specific conditions in boundary value problems or initial conditions.

Antiderivative

The antiderivative is a function whose derivative is the original function for which it was derived. Put simply, it's the reverse process of differentiation. Antiderivatives are crucial when dealing with integrals, as they help find solutions to integrals, especially when performing definite integration.

In the problem at hand, finding the antiderivative of \( f(x) = x^3 - \pi x^2 \) led us to the function \( F(x) = \frac{x^4}{4} - \frac{\pi x^3}{3} + C \). This formula flips the focus of calculus from slopes (derivatives) back to areas (integrals).

When working on definite integrals, having the antiderivative is mandatory because it allows us to use the Fundamental Theorem of Calculus. This theorem states that if \( F(x) \) is an antiderivative of \( f(x) \), then the definite integral is evaluated as \( F(b) - F(a) \), where \( a \) and \( b \) are the boundaries or limits of integration.

While "antiderivative" sounds technical, it's just about undoing the process of differentiation. By doing so, integrals tell us the accumulated quantity, such as area or distance, rather than rate of change.