Problem 66 - Calculus Single Variable

Question

A sum of integrals of the form \(\int_{a}^{b} f(x) d x\) is given. Express the sum as a single integral of form \(\int_{c}^{d} g(y) d y\). $$ \int_{-2}^{0} \sqrt{x+2} d x+\int_{0}^{2}(\sqrt{x+2}-x) d x $$

Step-by-Step Solution

Answer

Combine integrals as \(\int_{-2}^{2} (\sqrt{x+2} - x_+) \, dx\) with \(x_+ = x\) when \(x > 0\) and \(0\) otherwise.

Step 1: Analyze the Function Limits

First, identify the function limits that cover both integrals. The first integral \[ \int_{-2}^{0} \sqrt{x+2} \, dx \] starts from \(-2\) to \(0\), and the second integral \[ \int_{0}^{2} (\sqrt{x+2} - x) \, dx \] covers from \(0\) to \(2\). Together, these integrals span from \(-2\) to \(2\).

Step 2: Rewrite the Sum of Integrals as One

Combine the two integrals to express them as a single integral over the combined interval. Since the first integral runs from \(-2\) to \(0\), and the second from \(0\) to \(2\), rewrite as:\[ \int_{-2}^{2} g(x) \, dx \] where \[ g(x) = \begin{cases} \sqrt{x+2} & \text{if } x \in [-2, 0] \ \sqrt{x+2} - x & \text{if } x \in (0, 2] \end{cases} \]

Step 3: Simplify and Express as a Single Integral

Note that \( \sqrt{x+2} \) appears in both integral expressions, while \(-x\) only appears in the second. Therefore, we can express the combined integral as:\[ \int_{-2}^{2} (\sqrt{x+2} - x_+) \, dx \] where \(x_+\) is defined as \(x\) for \(x > 0\) and \(0\) otherwise. This means splitting behaviors are incorporated in one expression.

Key Concepts

Definite Integral

When we talk about a definite integral, we're looking at a tool in calculus that helps us calculate the total accumulation of a quantity over a certain interval. This means you'll be finding the exact area under the curve of a function between two points. Consider it as stacking up tiny, invisible slices from one point to another on a graph, and efficiently summing up all these slices.

The definite integral is represented by:

• The integration symbol: \( \int \)
• The limits of integration: the lower limit \( a \) and the upper limit \( b \)
• The integrand: the function \( f(x) \) that represents what you are integrating

Together, you represent this as \( \int_{a}^{b} f(x) \, dx \). The "dx" indicates you are integrating with respect to the variable \( x \).

In our exercise, each integral part \( \int_{-2}^{0} \sqrt{x+2} \, dx \) and \( \int_{0}^{2} (\sqrt{x+2} - x) \, dx \) sums the area under their respective function curves from \(-2\) to \(0\) and from \(0\) to \(2\), respectively. Hence, the outcome is a cumulative measure of the enclosed region within these bounds.

Piecewise Functions

Piecewise functions can be a bit like a piece of art made up of parts joined together. They're functions defined by multiple sub-functions, each applying to a certain interval of the main function's domain. This means that the function has different rules depending on the value of \( x \).

Think of a piecewise function as a sequence of separate "pieces" of functions put together, and that's exactly what's happening in the exercise.

• For \( x \) between \(-2\) and \(0\), you use \( f(x) = \sqrt{x+2} \).
• For \( x \) between \(0\) and \(2\), you use \( f(x) = \sqrt{x+2} - x \).

This structure makes it possible to handle more complex functions that might not be expressible with just one formula. It provides the flexibility to apply different rules in different intervals, ensuring a complete integration when combining multiple functions together.

In the given problem, the piecewise nature allows the integration to reflect accurately the behavior of the total function \( g(x) \) across the whole interval.

Integral Limits

Integral limits in calculus specify the start and end points over which you're integrating your function. These limits are crucial because they tell you exactly over which part of your function you're summing up the area. Think of it as setting boundaries for the area you're interested in measuring.

In the context of the problem:

• The integral \( \int_{-2}^{0} \sqrt{x+2} \, dx \) signifies you're considering the area from \(-2\) to \(0\).
• The integral \( \int_{0}^{2} (\sqrt{x+2} - x) \, dx \) extends that measurement from \(0\) to \(2\).

By combining these, we set our limits from \(-2\) to \(2\), bridging together the domain and allowing the integration to encompass the entire interval and the complete behavior of the function:\[\int_{-2}^{2} g(x) \, dx\]Understanding the importance of these limits helps in correctly applying them when you move from splitting pieces to their combined form. Once these limits are well-placed, they guide your integration along the path you need, ensuring precision and proper evaluation of the defined regions.