Question

In Exercises \(7-12,\) evaluate the integral. $$\int_{\sqrt{2}}^{\sqrt{18}} \sqrt{2} d r$$

Short answer

The value of the integral is \( 6\sqrt{2} - 2 \).

Step-by-step solution

Identify the function and bounds

The function is \( \sqrt{2} \) and is constant, and the limits of integration are \( \sqrt{2} \) and \( \sqrt{18} \).

Integrate the function

Recall that the integral of a constant function \( c \) over an interval \([a, b]\) is given by \( c* (b - a) \). So, the indefinite integral of \( \sqrt{2} \) in the interval from \( \sqrt{2} \) to \( \sqrt{18} \) is \( \sqrt{2}*(\sqrt{18} - \sqrt{2}) \).

Evaluate the integral

The integral can be calculated as \( \sqrt{2}*(\sqrt{18} - \sqrt{2}) = \sqrt{2} *\sqrt{18} - 2\), which gives us \( 6\sqrt{2} - 2 \).

Key concepts

Integral Calculus

Integral calculus is one of the two principal branches of calculus, with the other being differential calculus. It is concerned with the accumulation of quantities, such as areas under a curve, volumes of solids, and other quantities that add up infinitesimally small pieces to determine a whole. In essence, integration is the inverse process of differentiation.

When we look at the integral calculus problem presented in the exercise, our goal is to find the total accumulation of the constant function as its value increases from one point to another, specifically from \( \sqrt{2} \) to \( \sqrt{18} \). This concept can be applied to various fields, including physics, engineering, economics, and beyond, making it an indispensable tool for solving real-world problems.

Definite Integral

A definite integral has clear start and end values, which mean it is evaluated over a specific interval. This kind of integral provides a numerical value, which represents the exact area under the curve of a function over a given interval on a graph. In the context of the original problem, the definite integral of \( \sqrt{2} \) from \( \sqrt{2} \) to \( \sqrt{18} \) tells us the accumulated value of our constant function within these bounds.

When solving for a definite integral, we employ the fundamental theorem of calculus, which states that if a function is continuous over an interval and has an antiderivative, its definite integral over that interval can be found by evaluating the antiderivative at the endpoints and subtracting.

Integration Techniques

Integration techniques encompass various methods for finding the antiderivatives of functions. Learning different techniques is crucial because not all integrals can be solved with a single method. Some common techniques include substitution, integration by parts, partial fractions, and trigonometric integration. For the constant function in our exercise, the integration process is straightforward since the antiderivative of a constant function \( c \) is simply \( c\cdot x \) plus a constant of integration.

However, for more complex functions, identifying the correct technique to apply is key to simplifying and correctly solving the integral. Always start by analyzing the integrand and then selecting the integration method that matches the form and complexity of the function.

Constant Function Integration

When we integrate a constant function, the process is greatly simplified because the derivative of a constant is zero, and thus the antiderivative is simply the constant multiplied by the variable of integration. This is why the solution to our problem is a straightforward application of the rule that \( \int_a^b c \, dx = c\cdot (b - a) \), where \( c \) is the constant and \( a \) and \( b \) are the limits of integration.

In the given exercise, \( \sqrt{2} \) is the constant function and is integrated over the interval from \( \sqrt{2} \) to \( \sqrt{18} \). The calculation using this rule yields the final answer quite easily, demonstrating that while some integrals can be complex and challenging, constant function integration often offers a welcome respite with its simplicity.