Question

Using the integral of sec \(^{3} u\) By reduction formula 4 in Section 8.3 $$ \int \sec ^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C $$ Graph the following functions and find the area under the curve on the given interval. $$f(x)=\left(x^{2}-25\right)^{1 / 2},[5,10]$$

Short answer

Answer: The area under the curve on the given interval is $$2(4+\ln(2+\sqrt{3}))$$.

Step-by-step solution

Graph the Function and Identify the Function on the Interval Tmin and Tmax

The given function is $$f(x) = \sqrt{x^2 - 25}$$. First, we will graph this function and observe it in the interval [5, 10]. We find that it looks like the right half of a horizontal hyperbola, with the function being defined and continuous in the given interval.

Find the Integral of the Function

Since the given integral formula is for sec³u, we will perform a change of variables to transform our function into an expression involving sec³u. Let's take $$x = 5 \sec{u}.$$ Then $$dx = 5 \sec{u} \tan{u} du.$$ Now, when $$x = 5$$, $$u = \sec^{-1}{1} = 0$$ and when $$x = 10$$, $$u = \sec^{-1}{2}.$$ We now substitute these values into the integral to find the area. Our new integral is: $$\int_{0}^{\sec^{-1}(2)} \sqrt{ (5\sec{u})^2 - 25} \cdot 5\sec{u}\tan{u} du$$ Simplifying, we get: $$\int_{0}^{\sec^{-1}(2)} 5\sec{u}\tan{u}(\sec{u}-1)(\sec{u}+1) du$$

Integrate the Function and Find the Area

Now, we can use the provided formula for the integral of sec³u, which is: $$\int \sec^{3} u d u=\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|)+C$$ We can now integrate our expression (keeping in mind that there is an extra factor of 5 in our function): $$5\int_{0}^{\sec^{-1}(2)} \sec^{3}{u} du = 5\left[\frac{1}{2}(\sec u \tan u+\ln |\sec u+\tan u|) \right]_{0}^{\sec^{-1}(2)}$$ Now we evaluate the expression at the limits of integration: $$5\left[\frac{1}{2}(\sec (\sec^{-1}(2)) \tan (\sec^{-1}(2))+\ln |\sec (\sec^{-1}(2))+\tan (\sec^{-1}(2))|) - \frac{1}{2}(\sec (\sec^{-1}(1)) \tan (\sec^{-1}(1))+\ln |\sec (\sec^{-1}(1))+\tan (\sec^{-1}(1))|)\right]$$ Simplifying, we get: $$5\left[\frac{1}{2}(2 \cdot 2+\ln |2+\sqrt{3}|) - \frac{1}{2}(1 \cdot 0+\ln |1+0|)\right]$$ $$= 5\left[\frac{1}{2}(4 + \ln (2+\sqrt{3}))\right]$$ The area under the curve on the given interval is $$2(4+\ln(2+\sqrt{3}))$$.

Key concepts

Integration Techniques

Gaining proficiency in integration techniques is essential for solving a wide range of problems in calculus. Integration is the process of finding the integral of a function, which represents the accumulated area under the curve of the function over a certain interval. There are various strategies one can employ, including substitution, integration by parts, partial fractions, and trigonometric integrals.

In the context of the exercise, we see a technique involving a change of variables to express the original function in terms that match a known integral form. Our function \( f(x) = \sqrt{x^2 - 25} \) becomes a trigonometric expression by letting \( x = 5 \sec{u} \), hence transforming the integrand into a form that involves \( \sec^3{u} \), for which a direct integration formula is provided. Mastery of these techniques allows for the evaluation of complex integrals without the need for direct antiderivative formulas.

Area Under a Curve

Determining the area under a curve between two points is a fundamental application of integral calculus. The integral of a function \(f(x)\) from \(a\) to \(b\) geometrically represents the area under the curve of \(f(x)\) over the interval \[a, b\]. A positive function creates an area above the x-axis.

In the exercise, we are interested in the area under the curve \(f(x) = \sqrt{x^2 - 25}\) between \(x = 5\) and \(x = 10\). By performing the integration correctly with respect to a variable \(u\), after altering the original integral into a suitable form, we can find the exact area encapsulated between the curve, the x-axis, and the vertical lines \(x = 5\) and \(x = 10\). This approach not only gives a numerical answer but also illuminates the concept of integral as accumulation.

Hyperbolic Functions

While hyperbolic functions are not explicitly mentioned in the exercise, they bear a resemblance to the function we are evaluating. Hyperbolas find their significance in both geometry and calculus. Hyperbolic functions, such as \(\sinh x\) and \(\cosh x\), are analogs of trigonometric functions but for a hyperbola, instead of a circle.

Our function, \(f(x) = \sqrt{x^2 - 25}\), resembles the right half of a hyperbola. Though not precisely a hyperbolic function, its graph and integrand post variable change \(x = 5 \sec{u}\) are reminiscent of the properties that define the shape and behavior of hyperbolic functions. Understanding the characteristics of hyperbolic functions and their inverses can be quite helpful, for instance, in simplifying the integrand or when calculating the inverse hyperbolic functions encountered in integration.