Problem 17 - Calculus Single Variable

Question

Sketch the integrand of the given definite integral over the interval of integration. Evaluate the integral by calculating the area it represents. $$ \int_{1}^{3} \sqrt{2} d x $$

Step-by-Step Solution

Answer

The integral evaluates to \( 2\sqrt{2} \).

Step 1: Understand the Integrand

The integrand given is a constant function, \( \sqrt{2} \), which means it yields a horizontal line on the graph.

Step 2: Sketch the Function

On a coordinate plane, plot the horizontal line \( y = \sqrt{2} \). This line is constant over the interval \( x = 1 \) to \( x = 3 \).

Step 3: Define the Interval

The interval of integration is from \( x = 1 \) to \( x = 3 \). On our graph, this corresponds to limiting the horizontal line from these x-values.

Step 4: Visualize the Area

The area under the curve \( y = \sqrt{2} \) from \( x = 1 \) to \( x = 3 \) is a rectangle. The height of the rectangle is \( \sqrt{2} \) and the width is \( 3 - 1 = 2 \).

Step 5: Calculate the Area

The area of a rectangle is given by the product of its height and width. Thus, the area of our rectangle is \( \sqrt{2} \times 2 = 2\sqrt{2} \).

Step 6: Evaluate the Integral

By calculating the area, we have found the value of the integral: \( \int_{1}^{3} \sqrt{2} \, dx = 2\sqrt{2} \).

Key Concepts

Constant Function

A constant function is one of the simplest types of functions you can encounter in mathematics. It is called a "constant" because it gives the same value for any input in its domain. In simpler terms, the output does not change regardless of the input you choose. Consider a constant function like \( f(x) = \sqrt{2} \). This indicates that every value of \( x \) will have an output of \( \sqrt{2} \). This results in a horizontal line on the graph.

• Constant values mean horizontal lines in graphs.
• Every point on the graph lies at the same height.
• The horizontal nature means it's level, creating an easy-to-visualize area under the curve.

When dealing with definite integrals, recognizing a constant function helps simplify the problem considerably, since calculating the integral essentially involves finding the area of a rectangle.

Area Under Curve

The concept of the "area under the curve" is fundamental in calculus, especially when working with definite integrals. The term refers to the total region that lies between a curve (which can be linear, nonlinear, or even just a flat line) and the x-axis, within a specific interval. For our specific problem involving a constant function, the curve is a horizontal line, making this area easy to determine.

• For a horizontal line, the area forms a rectangle.
• The height of the rectangle is the constant value of the function, \( \sqrt{2} \).
• The width is determined by the length of the interval of integration, which is the difference between the upper and lower limits.

So, in our example, sketch a rectangle from \( x = 1 \) to \( x = 3 \) under the line \( y = \sqrt{2} \). Its area (and thus the value of the definite integral) is the height multiplied by the width, yielding \( 2\sqrt{2} \) as the total area.

Interval of Integration

The interval of integration is key to solving definite integrals, as it specifies the bounds over which we need to calculate the area. It's represented with limits in integral notation, showing where the area begins and ends on the x-axis. For our example:

• The interval of integration is from 1 to 3.
• These endpoints create boundaries for the region whose area is calculated.
• This determines how much of the function's horizontal line is included in the area calculation.

Imagine a ruler along the x-axis stretching from \( x=1 \) to \( x=3 \). The segment of the horizontal line between these points is exactly what's captured by the bounds. Calculating integrals over specific intervals often means dissecting functions into manageable pieces and calculating the area of each bounded piece, just like measuring an area of land within fences.