Problem 17 - Precalculus Mathematics for Calculus

Question

Find the area of the region that lies under the graph of \(f\) over the given interval. $$f(x)=x^{3}+2, \quad 0 \leq x \leq 5$$

Step-by-Step Solution

Answer

The area under the curve is 166.25.

Step 1: Define the Problem

We need to find the area under the curve of the function \( f(x) = x^3 + 2 \) over the interval \( [0, 5] \). This requires us to evaluate the definite integral of \( f(x) \) from 0 to 5.

Step 2: Set Up the Integral

The area under the curve can be found by calculating the definite integral: \[ \int_{0}^{5} (x^3 + 2) \, dx \]

Step 3: Integrate the Function

Integrate \( f(x) = x^3 + 2 \): - The integral of \( x^3 \) is \( \frac{x^4}{4} \). - The integral of \( 2 \) is \( 2x \). Thus, the indefinite integral becomes: \[ \frac{x^4}{4} + 2x \]

Step 4: Evaluate the Definite Integral

Substitute the bounds of the interval into the integrated function and evaluate: \[ \left[ \frac{x^4}{4} + 2x \right]_{0}^{5} = \left( \frac{5^4}{4} + 2 \times 5 \right) - \left( \frac{0^4}{4} + 2 \times 0 \right) \] This simplifies to: \[ \frac{625}{4} + 10 - 0 = \frac{625}{4} + 10 \]

Step 5: Simplify the Result

Convert \( \frac{625}{4} \) to a decimal or mixed number: \[ \frac{625}{4} = 156.25 \] Add 10 to 156.25 to get: \[ 156.25 + 10 = 166.25 \] Thus, the area under the curve is 166.25.

Key Concepts

Area Under a Curve

Finding the area under a curve is an essential concept in calculus, especially when analyzing functions and their graphical representations. Imagine the curve as part of a graph of a function, in this case, the curve of the function \( f(x) = x^3 + 2 \) over the interval \([0, 5]\).

Why is this area important? Here are a few reasons:

• It represents the accumulation of quantities, such as distance, volume, or time.
• It can also indicate total growth or decay in certain applications like economics or physics.

To determine this area, the definite integral is used, symbolized by the integral sign \( \int \), with specific bounds (in this case, 0 and 5). This integral calculation produces a numerical value that represents the total area beneath the curve between these two limits.

Calculus

Calculus encompasses a wide range of techniques and purposes, solving complex problems by dealing with changing quantities. It is divided into two primary branches:

• Differential Calculus: Concerned with rates of change, such as slopes of curves.
• Integral Calculus: Deals with accumulation of quantities, such as areas under curves.

Calculus provides the backbone for tackling problems involving dynamic systems and continuous change. The process of finding the area under the curve of \( f(x) = x^3 + 2 \) employs integral calculus. This showcases how integrals help us measure overall quantities from rates of changes, unlike differential calculus that handles instant rates at points on a curve.

Integration Techniques

The solution of an integral often involves several integration techniques, which are methods for solving integrals analytically. These techniques simplify the functions and make integration feasible.

• The integral of \( x^3 \) results in \( \frac{x^4}{4} \), applying the power rule where the exponent is increased by 1, and then divided by the new exponent.
• The constant \( 2 \) is integrated to \( 2x \), using a basic rule of integration which states constants become their product with the variable.

These techniques culminate in determining the integral: \( \int (x^3 + 2) \, dx = \frac{x^4}{4} + 2x \).

Subsequent evaluation of this at specified bounds (0 and 5) offers a concrete area value under the curve. Mastery of various integration techniques allows tackling diverse problems and obtaining precise solutions in real-world contexts.