Question

In Exercises 31-36, evaluate the integral using the following values. $$\int_{2}^{4} x^{3} d x=60, \quad \int_{2}^{4} x d x=6, \quad \int_{2}^{4} d x=2$$ $$ \int_{2}^{4}(x-8) d x $$

Short answer

The value of \(\int_{2}^{4}(x-8) dx = -10\)

Step-by-step solution

Break Down the Integral

Let's break down the integral expression into two integrals, splitting it into terms that we already have the values for: \(\int_{2}^{4}(x-8) dx = \int_{2}^{4}x dx - \int_{2}^{4} 8 dx\).

Substitute the Provided Values

Now, we substitute the values given. We know from the problem that \(\int_{2}^{4}x dx = 6\) and \(\int_{2}^{4} dx = 2\). So our equation becomes : \(\int_{2}^{4}(x-8) dx = 6 - 8*\int_{2}^{4} dx\)

Simplify for the Answer

Now, plug the known values and simplify: \(\int_{2}^{4}(x-8) dx = 6 - 8*2 = 6 - 16 = -10\)

Key concepts

Integral Calculus

Integral calculus is a branch of mathematics focused on finding the total size or value, such as area or volume, by stitching together infinitely many infinitesimally small pieces. When we evaluate an integral, we are essentially adding up these infinitesimal parts to find a whole.

An integral represented by \[\begin{equation} \[\int\limits_a^b f(x) \, dx\]\end{equation}\]indicates that we are summing the values of a function f(x) from the starting point a to the endpoint b along the x-axis. This process is akin to summing up the area under the curve of f(x) between a and b. The function \[\begin{equation} f(x) \, dx\end{equation}\]is known as the 'integrand', a and b are the 'limits of integration', and \[\begin{equation} dx\end{equation}\]indicates a minute width in the x-direction over which the function is integrated.

Integrating involves a variety of techniques such as substitution, integration by parts, and using the properties of integrals. Understanding these methods allows us to tackle more complex integrals in a structured and systematic manner.

Definite Integral

The definite integral is an integral that has specific start and end values, giving us a concrete number as a result. It represents the exact area under the curve between two points along the x-axis. In contrast, an indefinite integral, which lacks limits, results in a family of functions plus a constant of integration.

For the example provided, the definite integral{\[\begin{equation} \int_{2}^{4} x^{3} \, dx = 60\end{equation}\]} represents the accumulated area under the curve of the function \[\begin{equation} x^{3}\end{equation}\] between x = 2 and x = 4. A definite integral's value is independent of the variable of integration, which is why its result is a number, not a function.

When solving, it's often beneficial to look for symmetry, apply the properties of integrals, or break down a complex integral into simpler parts whose values are known or easier to calculate, as seen in the provided exercise solution.

Properties of Integrals

Integrals have specific properties that make the process of evaluating them more straightforward. Let's consider these important properties illustrated by the exercise:

The property of linearity allows us to break down integrals over sum or difference of functions, such as{\[ \int_{a}^{b} (f(x) \pm g(x)) \, dx = \int_{a}^{b} f(x) \, dx \pm \int_{a}^{b} g(x) \, dx \]}This means we can split one complex integral into simpler parts, often those we already know how to integrate or have previously evaluated values for.

Another property, the constant factor rule, lets us pull constants out of the integral, simplifying evaluation, as seen by{\[ \int_{a}^{b} c \cdot f(x) \, dx = c \cdot \int_{a}^{b} f(x) \, dx \]}where c is a constant.

Applying these properties allows us to see the integral\[\begin{equation} \int_{2}^{4}(x-8) dx\end{equation}\]as the difference of the integrals\[ \int_{2}^{4}x dx - \int_{2}^{4} 8 dx \]and calculate each part separately. This is particularly useful when the values of these separate integrals are already provided, as in the textbook exercise solution where integration properties are effectively used to find the result efficiently.