Problem 18 - Precalculus Mathematics for Calculus

Question

Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=4 x^{3}, \quad 2 \leq x \leq 5$$

Step-by-Step Solution

Answer

The area under the curve from \( x = 2 \) to \( x = 5 \) is 609.

Step 1: Understand the Problem

We need to find the area under the curve of the function \( f(x) = 4x^3 \) from \( x = 2 \) to \( x = 5 \). This is a problem of finding the definite integral of the function over the given interval.

Step 2: Set Up the Integral

To find the area under the curve, we need to compute the definite integral of \( f(x) = 4x^3 \) from 2 to 5. Set up the integral as follows: \[ \int_{2}^{5} 4x^3 \, dx \].

Step 3: Find the Antiderivative

Find the antiderivative, or indefinite integral, of the function \( 4x^3 \). Using the power rule for integration, the antiderivative of \( 4x^3 \) is \( x^4 \). Thus, the integral of \( 4x^3 \) is \( x^4 + C \), where \( C \) is the constant of integration. Since we are finding a definite integral, we do not need to include the constant \( C \).

Step 4: Evaluate the Definite Integral

Plug the limits of integration into the antiderivative:\[ \left[ x^4 \right]_{2}^{5} = 5^4 - 2^4 \].Calculate each power: - \( 5^4 = 625 \) - \( 2^4 = 16 \)Subtract these values: \( 625 - 16 = 609 \).

Step 5: Interpret the Result

The value \( 609 \) represents the area under the curve of \( f(x) = 4x^3 \) from \( x = 2 \) to \( x = 5 \). This is the solution to the problem.

Key Concepts

Antiderivative

In calculus, the antiderivative is a critical concept. It is essentially the reverse of a derivative. Think of it as a way to "undo" differentiation. When you find an antiderivative, you're seeking a function that, when differentiated, returns the original function. In our exercise, we explored the function \(4x^3\).

• The antiderivative of \(4x^3\) is \(x^4 + C\), where \(C\) is a constant called the constant of integration.
• The constant \(C\) results from the fact that differentiation erases constants, so every function \(F(x) + C\) will yield the same derivative \(f(x)\).
• In definite integrals, like our problem finding the area under the curve from \(x = 2\) to \(x = 5\), we don't need the constant \(C\) because the limits serve as boundaries, making \(C\) irrelevant.

Antiderivatives are foundational to definite integrals. Rearranges them provides a way to calculate exact areas under curves, unlike approximations from graphs.

Power Rule for Integration

The power rule for integration is one of the simplest and most powerful tools in calculus. It lets you find antiderivatives for functions in the form of a power of \(x\). The power rule states:

• If \(n eq -1\), the integral \(\int x^n \, dx = \frac{x^{n+1}}{n+1} + C\).
• For our function \(4x^3\), apply the power rule: \(\int 4x^3 \, dx = 4 \cdot \frac{x^{3+1}}{3+1} = x^4 + C\).

This straightforward technique makes it easy to find antiderivatives for any power of \(x\), transforming tasks that could be labor-intensive into simple calculations.
The power rule simplifies integration, including finding definite integrals, by enabling quick evaluation of even complex expressions. In our specific exercise, it led us directly from the function \(4x^3\) to its antiderivative \(x^4\), without complication.

Area Under a Curve

Calculating the area under a curve between two points is a vital application of integration in calculus. The concept might sound abstract, but it's about summing up tiny rectangles under the curve from one point to the next. In our case, to find the area under the curve of \(f(x) = 4x^3\) from \(x = 2\) to \(x = 5\), we used a definite integral.

• We set up the integral as \(\int_{2}^{5} 4x^3 \, dx\), and by evaluating it, found that the area was \(609\).
• The definite integral precisely calculates this area by considering infinitely small parts, giving an exact value.

Integration, through definite integrals, helps us understand spaces under curves in a precise numeric way. It has practical applications, such as physics where it might relate to distance, or economics in terms of accumulated costs or profits.
In essence, the definite integral doesn't just give a number; it describes a space, a quantity that is often very meaningful in real-world problems.