Problem 6 - Calculus and its applications

Question

Find the area under the graph of \(f\) over [-6,4] . $$ f(x)=\left\\{\begin{array}{ll} -x-1, & \text { for } \quad x<-1 \\ -x^{2}+4 x+5, & \text { for } \quad x \geq-1 \end{array}\right. $$

Step-by-Step Solution

Answer

The area under the graph of \( f(x) \) from \( -6 \) to \( 4 \) is approximately 63.67.

Step 1: Understanding the Problem

The given function is piecewise. We need to calculate the area under the curve of \( f(x) \) from \( x = -6 \) to \( x = 4 \). This involves integrating \( f(x) \) over its two different definitions.

Step 2: Determine Integration Intervals

Since the function has two different expressions divided at \( x = -1 \), our integration needs to be split. We'll integrate \( f(x) = -x - 1 \) from \( -6 \) to \( -1 \) and \( f(x) = -x^2 + 4x + 5 \) from \( -1 \) to \( 4 \).

Step 3: Integrate \( f(x) = -x - 1 \) over [-6, -1]

The integral of \( -x - 1 \) is \( -\frac{x^2}{2} - x \). Evaluating from \( x = -6 \) to \( x = -1 \): \[ \int_{-6}^{-1} (-x-1) \, dx = \left[ -\frac{x^2}{2} - x \right]_{-6}^{-1} \]After computation, the result is 12.5.

Step 4: Integrate \( f(x) = -x^2 + 4x + 5 \) over [-1, 4]

The integral of \( -x^2 + 4x + 5 \) is \( -\frac{x^3}{3} + 2x^2 + 5x \). Evaluating from \( x = -1 \) to \( x = 4 \): \[ \int_{-1}^{4} (-x^2 + 4x + 5) \, dx = \left[ -\frac{x^3}{3} + 2x^2 + 5x \right]_{-1}^{4} \]After computation, the result is approximately 51.17.

Step 5: Sum the Areas from Each Interval

Add the results from each integral to find the total area. The area from Step 3 is 12.5 and from Step 4 is 51.17. Adding these gives \( 12.5 + 51.17 = 63.67 \).

Step 6: Provide the Final Answer

The total area under the graph of \( f(x) \) from \( x = -6 \) to \( x = 4 \) is approximately 63.67.

Key Concepts

Piecewise Functions

A piecewise function is a function that utilizes different rules or expressions in distinct parts of its domain. This allows the function's behavior to change depending upon which part of the domain you are examining. For instance, functions can exhibit one linear behavior in one section of the domain and another quadratic behavior in another. In the current problem, the function \( f(x) \) is defined by two different expressions:

• The expression \(-x - 1\) applies for \(x < -1\).
• The expression \(-x^2 + 4x + 5\) is valid for \(x \geq -1\).

Understanding where these pieces of the function apply is crucial, especially for problems involving integration, where the area under these curves needs to be found. Each expression often results in different shapes which contribute to the total area calculation.

Definite Integrals

Definite integrals are a fundamental concept in calculus used to calculate the area under a function's curve over a specific interval. When dealing with piecewise functions, it’s essential to integrate each segment separately over its respective interval and then add the results. The definite integral of a function \( f(x) \) from \( a \) to \( b \), denoted as \( \int_{a}^{b} f(x) \, dx \), gives us the exact area under the function's graph between the vertical lines \( x = a \) and \( x = b \) along the x-axis.In our exercise, we performed two separate integrations:

• The first integral of \(-x - 1\) between \(-6\) and \(-1\), resulting in an area of 12.5.
• The second integral of \(-x^2 + 4x + 5\) between \(-1\) and \(4\), yielding an area of approximately 51.17.

By summing these two results, we achieve the total area under the curve for the entire range from \(-6\) to \(4\).

Area Under a Curve

The area under a curve refers to the region bounded by the curve itself and the x-axis between two specified points. Calculating this area is a central task in calculus which often involves definite integrals. The result can have practical applications, such as determining real-world quantities like distance, time, or resources over a given period or range.In our problem, the goal was to find the area under the graph of the function \( f(x) \) over the interval \([-6, 4]\). By understanding how the curve behaves (using its piecewise definitions) and applying the principles of calculus, particularly definite integrals, we could compute this total area as 63.67. Each section of the function's graph contributes differently, impacted by whether portions of it lie above or below the x-axis. Integrals allow us to capture these subtleties and compute meaningful solutions from complex mathematical expressions.