Question

Find the area of the following regions. The region bounded by the graph of \(f(x)=\frac{x}{\sqrt{x^{2}-9}}\) and the \(x\) -axis between \(x=4\) and \(x=5\)

Short answer

Question: Find the area of the region bounded by the graph of \(f(x)=\frac{x}{\sqrt{x^{2}-9}}\) and the x-axis between \(x=4\) and \(x=5\). Answer: The area of the region is \(4 - \sqrt{7}\) square units.

Step-by-step solution

Integrate the function with respect to x

The integral of \(f(x)\) with respect to x between limits 4 and 5 is: $$ \int_{4}^{5} \frac{x}{\sqrt{x^{2}-9}} dx $$

Perform substitution

Let \(u = x^{2} - 9\). Then, \(du = 2x dx\) and \(x dx = \frac{1}{2} du\). The limits of integration also change. When \(x = 4\), \(u = 4^2 - 9 = 7\). When \(x = 5\), \(u = 5^2 - 9 = 16\). The integral now looks like: $$ \int_{7}^{16} \frac{1}{2} \frac{1}{\sqrt{u}} du $$

Integrate with respect to u

Now, we integrate \(\frac{1}{2\sqrt{u}}\) with respect to u between limits 7 and 16: $$ \frac{1}{2} \int_{7}^{16} u^{-\frac{1}{2}} du $$ Using the power rule for integration, we get: $$ \left[ \frac{1}{2} \cdot 2 \sqrt{u} \right]_{7}^{16} $$

Evaluate the integral using the limits

Now, we plug in the limits of integration: $$ \left[ \sqrt{u} \right]_7^{16} = \sqrt{16} - \sqrt{7} = 4 - \sqrt{7} $$

Write the final answer

The area of the region bounded by the graph of \(f(x)=\frac{x}{\sqrt{x^{2}-9}}\) and the x-axis between \(x=4\) and \(x=5\) is: $$ 4 - \sqrt{7} \text{ square units} $$

Key concepts

Area under a curve

When we talk about finding the area under a curve, we're essentially trying to calculate how much "space" lies beneath a function graph over a specific interval on the x-axis. For functions that sit above the x-axis, the calculated area is considered positive. Conversely, if the function dips below the x-axis, the area is negative.
To determine this area, we utilize definite integration. A definite integral gives us the exact area under a curve from one point to another. It's important to visualize: if you imagine plotting the curve of a function, the area under this curve between two points on the x-axis can be thought of like cutting out a specific slice beneath the graph.
For instance, in the exercise, finding the area under the curve of the function \(f(x)=\frac{x}{\sqrt{x^{2}-9}}\) from \(x=4\) to \(x=5\) involves calculating the integration of this function over that specific interval.

Substitution method

In calculus, especially when dealing with definite integrals, the substitution method is a powerful technique to simplify complex integrals. It's somewhat akin to a change of variables, which makes the integral easier to solve.
The core idea is to replace a complex expression within the integral with a simpler variable. This allows us to rewrite the integral into a form that's easier to integrate.
During substitution, remember to also adjust the limits of integration. As shown in the solution, using the substitution \(u = x^2 - 9\) helped transform the integral into a simpler form. Here’s how it works step by step:

• First, substitute \(u = x^2 - 9\), which simplifies the expression under the square root.
• Next, compute \(du = 2x dx\), rearranging gives \(x dx = \frac{1}{2} du\).
• Adjust the limits: when \(x=4\), \(u=7\), and when \(x=5\), \(u=16\).

This new integral is easier to solve, leading naturally to the next step: using the power rule for integration.

Power rule for integration

The power rule for integration is a fundamental technique that's quite handy in integration, particularly with polynomial terms. The rule provides a straightforward way to integrate expressions that are simple powers of a variable.
The power rule states: For any function \(x^n\) where \(n eq -1\), the integral is:\[\int x^n dx = \frac{x^{n+1}}{n+1} + C\]
Here, \(C\) is the constant of integration, but in definite integrals, it cancels out as we are evaluating over specific limits.
In our particular example, we utilized this rule to integrate \(u^{-rac{1}{2}}\). Applying the power rule, we found that:\[\frac{1}{2} \int u^{-rac{1}{2}} du = \frac{1}{2} \cdot 2 \sqrt{u}\]
This simplifies to \(\sqrt{u}\), which we then evaluate from \(u=7\) to \(u=16\). The power rule is invaluable for solving such integrals quickly and efficiently.