Problem 17 - Precalculus: Mathematical for Calculus

Question

Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=x+6 x^{2}, \quad 1 \leq x \leq 4$$

Step-by-Step Solution

Answer

The area is 133.5.

Step 1: Understanding the Problem

To find the area under the curve of a function over a given interval, we need to perform integration of the function over that interval. The function given is \(f(x) = x + 6x^2\) and the interval is \([1, 4]\).

Step 2: Setting up the Integral

The integral to find the area under \(f(x) = x + 6x^2\) from \(x = 1\) to \(x = 4\) is: \[ \int_{1}^{4} (x + 6x^2) \, dx \]

Step 3: Integrating the Function

To integrate \(x + 6x^2\), we integrate each term separately:1. The integral of \(x\) is \(\frac{1}{2}x^2\).2. The integral of \(6x^2\) is \(6 \cdot \frac{1}{3}x^3 = 2x^3\).Thus, the antiderivative of \(f(x)\) is:\[ \frac{1}{2}x^2 + 2x^3 \]

Step 4: Evaluating the Integral at the Bounds

We evaluate the antiderivative \(\frac{1}{2}x^2 + 2x^3\) at \(x = 4\) and \(x = 1\), and then subtract:- Evaluate at \(x = 4\): \[ \frac{1}{2}(4)^2 + 2(4)^3 = 8 + 128 = 136 \]- Evaluate at \(x = 1\): \[ \frac{1}{2}(1)^2 + 2(1)^3 = \frac{1}{2} + 2 = 2.5 \]Subtracting these results gives:\[ 136 - 2.5 = 133.5 \]

Step 5: Conclusion

The area of the region under the graph of \(f(x) = x + 6x^2\) from \(x = 1\) to \(x = 4\) is \(133.5\).

Key Concepts

Definite Integral

The concept of a definite integral combines ideas of limits, sums, and functions to calculate the exact accumulation of quantities, such as areas under curves. In this context, the definite integral is used to find the area under the graph of a function over a specific interval. In simpler terms, it tells us how much space lies beneath a curve, between two boundaries on the x-axis.

• The integration process considers small slices or partitions of the area, gradually getting smaller to approach the true value.
• Symbolically, the definite integral is expressed as \( \int_{a}^{b} f(x)\, dx \), where \(a\) and \(b\) are the boundaries.
• To solve it, we first need to find the antiderivative of the function.

The limits, or bounds of integration, are crucial as they define the segment of the curve we are focusing on. Evaluating the definite integral of a function over a given interval yields the accumulated area, which in this exercise results in 133.5 units.

Area Under a Curve

The main goal of working with definite integrals is often to find the area under a curve, a common task in calculus. This area represents a summation of all the infinite tiny rectangles that can fit beneath a curve, within the specified limits. Imagine slicing a region under the curve into little sections, adding up the areas of all these sections exactly.

• To calculate it, the definite integral determines total area by accounting for every point on the curve within the bounds.
• It is a geometric representation, transforming complex curves into manageable calculations.
• For our example, the area between \(x = 1\) and \(x = 4\) under \(f(x) = x + 6x^2\) was found to be 133.5 square units.

This approach is not only about adding slices but refining them so the combined sum accurately reflects the curve's behavior from one limit to another.

Antiderivative

An antiderivative, also called an indefinite integral, is a function whose derivative gives back the original function. In the context of definite integrals, finding an antiderivative is an essential step to deriving the area under a curve.

• The antiderivative provides a way to reverse differentiation. We need this to see behind the rate of change back to the original amount.
• For example, integrating \(x + 6x^2\) gives us \(\frac{1}{2}x^2 + 2x^3\).
• Once the antiderivative is known, we use the evaluation step, applying the Fundamental Theorem of Calculus to substitute the upper and lower limits.

In simple terms, the antiderivative enables us to find the exact "sum of parts" over a specific section of the curve, making it foundational for evaluating definite integrals and understanding accumulation.