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 square units.

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 area can be calculated using the definite integral of the function over the given interval.

Step 2: Set Up the Integral

The area under the curve of the function \( f(x) = 4x^3 \) from \( x = 2 \) to \( x = 5 \) is given by the definite integral \( \int_{2}^{5} 4x^3 \, dx \).

Step 3: Integrate the Function

The integral of \( f(x) = 4x^3 \) is \( \int 4x^3 \, dx = x^4 + C \), where \( C \) is the constant of integration. Since we are evaluating a definite integral, we can ignore \( C \).

Step 4: Substitute the Limits

Evaluate the integral at the upper limit and the lower limit: \[ \left[ x^4 \right]_2^5 = (5^4) - (2^4). \]

Step 5: Calculate the Area

Compute the numerical values: \( 5^4 = 625 \) and \( 2^4 = 16 \). Hence, the area under the curve is \( 625 - 16 = 609 \).

Key Concepts

Area Under the Curve

The area under the curve is a fundamental concept in calculus, crucial for understanding how to calculate the space enclosed between a curve and the x-axis over a particular interval. Essentially, it's all about measuring the total "shadow" a curve might cast between two points on the x-axis. This gives us a way to effectively see how much "space" a function occupies.
To find the area under a curve when given a specific function, we use the definite integral. The integral provides the accumulated value of a function over a given range, representing the total area. This accumulated value can be thought of as slicing the area into tiny pieces and summing them up. By doing so, we obtain a precise measurement of that area.
For the function \(f(x) = 4x^3\) from \(x = 2\) to \(x = 5\), the definite integral \(\int_{2}^{5} 4x^3 \, dx\) allows us to calculate the precise area between the curve and the x-axis over the interval \(2 \, \text{to} \, 5\). Always remember, the definite integral gives you an exact "net area," considering any portions below the x-axis as negative contributions.

Polynomial Function Integration

Integration of polynomial functions like \(f(x) = 4x^3\) involves finding an antiderivative or a "reverse derivative." This means we determine a new function whose derivative would give us the original polynomial function.
When integrating \(4x^3\), we apply simple power rules. Each time we integrate a term of \(x^n\), we increase the power by one and divide by the new power. So:

• Integrate \(4x^{3} \, dx = \frac{4}{4}{x^{4}} = x^{4}\).

The constant multiple 4 in front of \(x^{3}\) allows us to treat the integration process straightforwardly by focusing only on the exponent rules, confirming how the coefficients multiply back into place after integrating.
This integration concept is central in dealing with polynomials, as each term follows this power rule progression, making polynomial integration one of the more approachable topics in calculus.

Evaluation of Limits

Once we have integrated a function, the next step in solving a definite integral involves evaluating the function with its limits. This means we look at the integrated function, substituting in the upper and lower boundary points (limit points) specified in the integral.
For \( \left[ x^4 \right]_2^5\), we have:

• First, substitute the upper limit (5) into the integrated function, obtaining \(5^4 = 625\).
• Next, substitute the lower limit (2) into the function, obtaining \(2^4 = 16\).
• Finally, compute the difference: \(625 - 16\), which gives \(609\).

This subtraction method importantly calculates the net area accumulated under the curve between these two bounds.
Understanding how to evaluate these limits allows us to transition from the abstract concept of an integral into the tangible result of an area value, essential in applying calculus principles to real-world problems.